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| author | hardythe1 | 2015-05-05 14:21:39 +0530 |
|---|---|---|
| committer | hardythe1 | 2015-05-05 14:21:39 +0530 |
| commit | fba055ce5aa0955e22bac2413c33493b10ae6532 (patch) | |
| tree | be70ef4fccd07c9c88de778014219201b4ea971f /Applied_Physics | |
| parent | 67068710030ddd6b6c809518c34af2e04e0bf7ca (diff) | |
| download | Python-Textbook-Companions-fba055ce5aa0955e22bac2413c33493b10ae6532.tar.gz Python-Textbook-Companions-fba055ce5aa0955e22bac2413c33493b10ae6532.tar.bz2 Python-Textbook-Companions-fba055ce5aa0955e22bac2413c33493b10ae6532.zip | |
add books
Diffstat (limited to 'Applied_Physics')
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diff --git a/Applied_Physics/Chapter1.ipynb b/Applied_Physics/Chapter1.ipynb new file mode 100755 index 00000000..e7e0fc16 --- /dev/null +++ b/Applied_Physics/Chapter1.ipynb @@ -0,0 +1,248 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:7b7dfd33b7ab2e69421d0f26bfbfaa20f35bc1084164b445bbbc400215113c08"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 1:Bonding in Solids"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.3 , Page no:15"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "r=2; #in angstrom(distance)\n",
+ "e=1.6E-19; # in C (charge of electron)\n",
+ "E_o= 8.85E-12;# absolute premittivity\n",
+ "\n",
+ "#calculate\n",
+ "r=2*1*10**-10; # since r is in angstrom\n",
+ "V=-e**2/(4*3.14*E_o*r); # calculate potential\n",
+ "V1=V/e; # changing to eV\n",
+ "\n",
+ "#result\n",
+ "print \"\\nThe potential energy is V = \",V,\"J\";\n",
+ "print \"In electron-Volt V = \",round(V1,3),\"eV\"; \n",
+ "print \"Note: the answer in the book is wrong due to calculation mistake\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The potential energy is V = -1.15153477995e-18 J\n",
+ "In electron-Volt V = -7.197 eV\n",
+ "Note: the answer in the book is wrong due to calculation mistake\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.4 , Page no:15"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "#given\n",
+ "r0=0.236; #in nanometer(interionic distance)\n",
+ "e=1.6E-19; # in C (charge of electron)\n",
+ "E_o= 8.85E-12;# absolute premittivity\n",
+ "N=8; # Born constant\n",
+ "IE=5.14;# in eV (ionisation energy of sodium)\n",
+ "EA=3.65;# in eV (electron affinity of Chlorine)\n",
+ "pi=3.14; # value of pi used in the solution\n",
+ "\n",
+ "#calculate\n",
+ "r0=r0*1E-9; # since r is in nanometer\n",
+ "PE=(e**2/(4*pi*E_o*r0))*(1-1/N); # calculate potential energy\n",
+ "PE=PE/e; #changing unit from J to eV\n",
+ "NE=IE-EA;# calculation of Net energy\n",
+ "BE=PE-NE;# calculation of Bond Energy\n",
+ "\n",
+ "#result\n",
+ "print\"The potential energy is PE= \",round(PE,2),\"eV\";\n",
+ "print\"The net energy is NE= \",round(NE,2),\"eV\";\n",
+ "print\"The bond energy is BE= \",round(BE,2),\"eV\";\n",
+ "# Note: (1)-In order to make the answer prcatically feasible and avoid the unusual answer, I have used r_0=0.236 nm instead of 236 nm. because using this value will give very much irrelevant answer.\n",
+ "# Note: (2) There is slight variation in the answer due to round off."
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The potential energy is PE= 5.34 eV\n",
+ "The net energy is NE= 1.49 eV\n",
+ "The bond energy is BE= 3.85 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.5 , Page no:16"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "r_0=.41; #in mm(lattice constant)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "E_o= 8.85E-12; #absolute premittivity\n",
+ "n=0.5; #repulsive exponent value\n",
+ "alpha=1.76; #Madelung constant\n",
+ "pi=3.14; # value of pi used in the solution\n",
+ "\n",
+ "#calculate\n",
+ "r=.41*1E-3; #since r is in mm\n",
+ "Beta=72*pi*E_o*r**4/(alpha*e**2*(n-1)); #calculation compressibility\n",
+ "\n",
+ "#result\n",
+ "print\"The compressibility is\tBeta=\",round(Beta);"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The compressibility is\tBeta= -2.50967916144e+15\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.6 , Page no:16"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "r_0=3.56; #in Angstrom\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "IE=3.89; #in eV (ionisation energy of Cs)\n",
+ "EA=-3.61; #in eV (electron affinity of Cl)\n",
+ "n=10.5; #Born constant\n",
+ "E_o= 8.85E-12; #absolute premittivity\n",
+ "alpha=1.763; #Madelung constant\n",
+ "pi=3.14; #value of pi used in the solution\n",
+ "\n",
+ "#calculate\n",
+ "r_0=r_0*1E-10; #since r is in nanometer\n",
+ "U=-alpha*(e**2/(4*pi*E_o*r_0))*(1-1/n); #calculate potential energy\n",
+ "U=U/e; #changing unit from J to eV\n",
+ "ACE=U+EA+IE; #calculation of atomic cohesive energy\n",
+ "\n",
+ "#result\n",
+ "print\"The ionic cohesive energy is \",round(U),\"eV\";\n",
+ "print\"The atomic cohesive energy is\",round(ACE),\"eV\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The ionic cohesive energy is -6.0 eV\n",
+ "The atomic cohesive energy is -6.0 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.7 , Page no:17"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "r_0=2.81; #in Angstrom\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "n=9; #Born constant\n",
+ "E_o= 8.85E-12; #absolute premittivity\n",
+ "alpha=1.748; #Madelung constant\n",
+ "pi=3.14; #value of pi used in the solution\n",
+ "\n",
+ "#calculate\n",
+ "r_0=r_0*1E-10; #since r is in nanometer\n",
+ "V=-alpha*(e**2/(4*pi*E_o*r_0))*(1-1/n); #calculate potential energy\n",
+ "V=V/e; #changing unit from J to eV\n",
+ "V_1=V/2; #Since only half of the energy contribute per ion to the cohecive energy therfore\n",
+ "\n",
+ "#result\n",
+ "print\"The potential energy is V=\",round(V,2),\"eV\";\n",
+ "print\"The energy contributing per ions to the cohesive energy is \",round(V_1,2),\"eV\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The potential energy is V= -7.96 eV\n",
+ "The energy contributing per ions to the cohesive energy is -3.98 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Applied_Physics/Chapter10.ipynb b/Applied_Physics/Chapter10.ipynb new file mode 100755 index 00000000..7b7470a8 --- /dev/null +++ b/Applied_Physics/Chapter10.ipynb @@ -0,0 +1,316 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:03ee5374207ea96e9a4f31e16f771d187a11f07daf287d94fc54e352b5db8bff"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 10:Superconductivity"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.1 , Page no:313"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "Tc=7.2; #in K (critical temperature)\n",
+ "T=5; #in K (given temperature)\n",
+ "H0=6.5E3; #in A/m (critical magnetic field at 0K)\n",
+ "\n",
+ "#calculate\n",
+ "Hc=H0*(1-(T/Tc)**2); #calculation of magnitude of critical magnetic field\n",
+ "\n",
+ "#result\n",
+ "print\"The magnitude of critical magnetic field is Hc=\",round(Hc,2),\"A/m\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The magnitude of critical magnetic field is Hc= 3365.35 A/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.2 , Page no:313"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "r=0.02; #in m (radius of ring)\n",
+ "Hc=2E3; #in A/m (critical magnetic field at 5K)\n",
+ "pi=3.14; #value of pi used in the solutiion\n",
+ "\n",
+ "#calculate\n",
+ "Ic=2*pi*r*Hc; #calculation of critical current value\n",
+ "\n",
+ "#result\n",
+ "print\"The critical current value is Ic=\",Ic,\"A\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The critical current value is Ic= 251.2 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.3 , Page no:313"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "M1=199.5; #in amu (isotropic mass at 5K)\n",
+ "T1=5; #in K (first critical temperature)\n",
+ "T2=5.1; #in K (second critical temperature)\n",
+ "#calculate\n",
+ "#since Tc=C*(1/sqrt(M)\n",
+ "#therefore T1*sqrt(M1)=T2*sqrt(M2)\n",
+ "#therefore we have M2=(T1/T2)^2*M1\n",
+ "M2=(T1/T2)**2*M1; #calculation of isotropic mass at 5.1K\n",
+ "\n",
+ "#result\n",
+ "print\"The isotropic mass at 5.1K is M2=\",round(M2,3),\"a.m.u.\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The isotropic mass at 5.1K is M2= 191.753 a.m.u.\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.4 , Page no:314"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "T=6; #in K (given temperature)\n",
+ "Hc=5E3; #in A/m (critical magnetic field at 5K)\n",
+ "H0=2E4; #in A/m (critical magnetic field at 0K)\n",
+ "\n",
+ "#calculate\n",
+ "#since Hc=H0*(1-(T/Tc)^2)\n",
+ "#therefor we have Tc=T/sqrt(1-(Hc/H)^2)\n",
+ "Tc=T/math.sqrt(1-(Hc/H0)); #calculation of transition temperature\n",
+ "\n",
+ "#result\n",
+ "print\"The transition temperature is Tc=\",round(Tc,3),\"K\";\n",
+ "print \"NOTE: The answer in the textbook is wrong\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The transition temperature is Tc= 6.928 K\n",
+ "NOTE: The answer in the textbook is wrong\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.5 , Page no:314"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "T=5; #in K (given temperature)\n",
+ "d=3; #in mm (diameter of the wire)\n",
+ "Tc=8; #in K (critical temperature for Pb)\n",
+ "H0=5E4; #in A/m (critical magnetic field at 0K)\n",
+ "pi=3.14; #value of pi used in the solution\n",
+ "\n",
+ "#calculate\n",
+ "Hc=H0*(1-(T/Tc)**2); #calculation of critical magnetic field at 5K\n",
+ "r=(d*1E-3)/2; #calculation of radius in m\n",
+ "Ic=2*pi*r*Hc; #calculation of critical current at 5K\n",
+ "\n",
+ "#result\n",
+ "print\"The critical magnetic field at 5K is Hc=\",Hc,\"A/m\";\n",
+ "print\"The critical current at 5K is Ic=\",Ic,\"A\";\n",
+ "print \" (roundoff error)\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The critical magnetic field at 5K is Hc= 30468.75 A/m\n",
+ "The critical current at 5K is Ic= 287.015625 A\n",
+ " (roundoff error)\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.6 , Page no:314"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "V=8.50; #in micro V (voltage across Josephson junction )\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "h=6.626E-34; #in J/s (Planck\u2019s constant)\n",
+ "\n",
+ "#calculate\n",
+ "V=V*1E-6; #changing unit from V to microVolt\n",
+ "v1=2*e*V/h; #calculation of frequency of EM waves\n",
+ "\n",
+ "#result\n",
+ "print\"The frequency of EM waves is v=\",v1,\"Hz\";\n",
+ "print \"NOTE: The answer in the textbook is wrong\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The frequency of EM waves is v= 4105040748.57 Hz\n",
+ "NOTE: The answer in the textbook is wrong\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.7 , Page no:315"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "M1=200.59; #in amu (average atomic mass at 4.153K)\n",
+ "Tc1=4.153; #in K (first critical temperature)\n",
+ "M2=204; #in amu (average atomic mass of isotopes)\n",
+ "\n",
+ "#calculate\n",
+ "#since Tc=C*(1/sqrt(M)\n",
+ "#therefore T1*sqrt(M1)=T2*sqrt(M2)\n",
+ "#therefore we have Tc2=Tc1*sqrt(M1/M2)\n",
+ "Tc2=Tc1*math.sqrt(M1/M2); #calculation of transition temperature of the isotopes\n",
+ "\n",
+ "#result\n",
+ "print\"The transition temperature of the isotopes is Tc2=\",round(Tc2,3),\"K\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The transition temperature of the isotopes is Tc2= 4.118 K\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Applied_Physics/Chapter12.ipynb b/Applied_Physics/Chapter12.ipynb new file mode 100755 index 00000000..95067254 --- /dev/null +++ b/Applied_Physics/Chapter12.ipynb @@ -0,0 +1,365 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:13927a6aa5f4b8929243c9ca2c470f4da4ee5b45564e616337e68523dd18d625"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 12:Fibre Optics"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.1 , Page no:360"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "u1=1.563; #refractive index of core\n",
+ "u2=1.498; #refractive index of cladding\n",
+ "\n",
+ "#calculate\n",
+ "d=(u1-u2)/u1; #calculation of fractional index change\n",
+ "\n",
+ "#result\n",
+ "print\"The fractional index change for a given optical fibre is =\",round(d,4);"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The fractional index change for a given optical fibre is = 0.0416\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.2 , Page no:360"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "u1=1.55; #refractive index of core\n",
+ "u2=1.50; #refractive index of cladding\n",
+ "\n",
+ "#calculate\n",
+ "d=(u1-u2)/u1; #calculation of fractional index change\n",
+ "NA=u1*math.sqrt(2*d); #calculation of numerical aperture\n",
+ "theta=math.asin(NA); #calculation of acceptance angle\n",
+ "theta1=theta*180/3.14;\n",
+ "\n",
+ "#result\n",
+ "print\"The numerical aperture of the fibre is NA=\",round(NA,3);\n",
+ "print\"The acceptance angle of the optical fibre is =\",round(theta1,2),\"degree\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The numerical aperture of the fibre is NA= 0.394\n",
+ "The acceptance angle of the optical fibre is = 23.2 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.3 , Page no:360"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "u1=1.563; #refractive index of core\n",
+ "u2=1.498; #refractive index of cladding\n",
+ "\n",
+ "#calculate\n",
+ "NA=math.sqrt(u1**2-u2**2); #calculation of numerical aperture\n",
+ "theta=math.asin(NA); #calculation of acceptance angle\n",
+ "theta1=theta*180/3.14;\n",
+ "#result\n",
+ "print\"The numerical aperture of the fibre is NA=\",round(NA,4);\n",
+ "print\"The acceptance angle of the optical fibre is =\",round(theta1,2),\"degree\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The numerical aperture of the fibre is NA= 0.4461\n",
+ "The acceptance angle of the optical fibre is = 26.5 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.4 , Page no:360"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "NA=0.39; #numerical aperture of the optical fibre\n",
+ "d=0.05; #difference in the refractive index of the material of the core and cladding\n",
+ "\n",
+ "#calculate\n",
+ "#since NA=u1*sqrt(2*d)\n",
+ "#we have u1=NA/sqrt(2*d)\n",
+ "u1= NA/math.sqrt(2*d); #calculation of refractive index of material of the core\n",
+ "\n",
+ "#result\n",
+ "print\"The refractive index of material of the core is u1=\",round(u1,3);"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The refractive index of material of the core is u1= 1.233\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.5 , Page no:361"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "u1=1.50; #refractive index of core\n",
+ "u2=1.45; #refractive index of cladding\n",
+ "\n",
+ "#calculate\n",
+ "d=(u1-u2)/u1; #calculation of fractional index change\n",
+ "NA=u1*math.sqrt(2*d); #calculation of numerical aperture\n",
+ "theta_0=math.asin(NA); #calculation of acceptance angle\n",
+ "theta_01=theta_0*180/3.14;\n",
+ "theta_c=math.asin(u2/u1); #calculation of critical angle\n",
+ "theta_c1=theta_c*180/3.14;\n",
+ "\n",
+ "#result\n",
+ "print\"The numerical aperture of the fibre is NA=\",round(NA,3);\n",
+ "print\"The acceptance angle of the optical fibre is =\",round(theta_01,2),\"degree\";\n",
+ "print\"The critical angle of the optical fibre is =\",round(theta_c1,2),\"degree\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The numerical aperture of the fibre is NA= 0.387\n",
+ "The acceptance angle of the optical fibre is = 22.8 degree\n",
+ "The critical angle of the optical fibre is = 75.2 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.6 , Page no:361"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "NA=0.33; #numerical aperture\n",
+ "d=0.02; #difference in the refractive index of the core and cladding of the material\n",
+ "\n",
+ "#calculate\n",
+ "#since NA=u1*sqrt(2*d)\n",
+ "#therefore we have\n",
+ "u1=NA/math.sqrt(2*d); #calculation of refractive index of the core\n",
+ "#since d=(u1-u2)/u2\n",
+ "#therefore we have\n",
+ "u2=(1-d)*u1; #calculation of refractive index of the cladding\n",
+ "\n",
+ "#result\n",
+ "print\"The refractive index of the core is u1=\",round(u1,2);\n",
+ "print\"The refractive index of the cladding is u2=\",round(u2,3);\n",
+ "print \"NOTE: The answer in the textbook is wrong\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The refractive index of the core is u1= 1.65\n",
+ "The refractive index of the cladding is u2= 1.617\n",
+ "NOTE: The answer in the textbook is wrong\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.7 , Page no:361"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "u1=3.5; #refractive index of core\n",
+ "u2=3.45; #refractive index of cladding\n",
+ "u0=1; #refractive index of the air\n",
+ "\n",
+ "#calculate\n",
+ "NA=math.sqrt(u1**2-u2**2); #calculation of numerical aperture\n",
+ "NA1=NA/u0;\n",
+ "alpha=math.asin(NA); #calculation of acceptance angle\n",
+ "alpha1=alpha*180/3.14;\n",
+ "#result\n",
+ "print\"The numerical aperture of the fibre is NA=\",round(NA1,2);\n",
+ "print\"The acceptance angle of the optical fibre is =\",round(alpha1,2),\"degree\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The numerical aperture of the fibre is NA= 0.59\n",
+ "The acceptance angle of the optical fibre is = 36.14 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.8 , Page no:361"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "u1=1.48; #refractive index of core\n",
+ "u2=1.45; #refractive index of cladding\n",
+ "\n",
+ "#calculate\n",
+ "NA=math.sqrt(u1**2-u2**2); #calculation of numerical aperture\n",
+ "theta=math.asin(NA); #calculation of acceptance angle\n",
+ "theta1=theta*180/3.14;\n",
+ "#result\n",
+ "print\"The numerical aperture of the fibre is NA=\",round(NA,3);\n",
+ "print\"The acceptance angle of the optical fibre is =\",round(theta1,2),\"degree\";\n",
+ "print \" (roundoff error)\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The numerical aperture of the fibre is NA= 0.296\n",
+ "The acceptance angle of the optical fibre is = 17.26 degree\n",
+ " (roundoff error)\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Applied_Physics/Chapter2.ipynb b/Applied_Physics/Chapter2.ipynb new file mode 100755 index 00000000..2d0b7a14 --- /dev/null +++ b/Applied_Physics/Chapter2.ipynb @@ -0,0 +1,291 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:f14517ab3de72640bce8c38f7d46947caa54ae7067bb37477a81849b7b917982"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 2:Crystal Structure"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.1 , Page no:40"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "N=6.02*10**26; #in /Kg-molecule (Avogadro's number)\n",
+ "n=4; #number of molecules per unit cell ofr NaCl\n",
+ "M=58.5; #in Kg/Kg-molecule (molecular weight of NaCl)\n",
+ "p=2189; #in Kg/m^3 (density)\n",
+ "\n",
+ "#calculate\n",
+ "a=pow(((n*M)/(N*p)),0.3333333);\n",
+ "a1=a*10**10; #changing unit to Angstrom\n",
+ "\n",
+ "#result\n",
+ "print\"The lattice constant is a=\",a,\"m\";\n",
+ "print\"\\t\\t\\ta=\",a1,\"Angstrom\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The lattice constant is a= 5.62072226997e-10 m\n",
+ "\t\t\ta= 5.62072226997 Angstrom\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2 , Page no:41"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "N=6.02*10**23; #in /gram-atom (Avogadro's number)\n",
+ "n=4; #number of atom per unit cell for fcc structure\n",
+ "M=63.5; #in gram/gram-atom (atomic weight of Cu)\n",
+ "p=8.96; #in g/cm^3 (density)\n",
+ "\n",
+ "#calculate\n",
+ "a=pow((n*M/(N*p)),0.3333333);\n",
+ "a=a*1E8; #changing unit from cm to Angstrom\n",
+ "d=a/math.sqrt(2); #distance infcc lattice\n",
+ "\n",
+ "#result\n",
+ "print\"lattice constant is a=\",a,\"E-08 cm\";\n",
+ "print\"\\t\\t a=\",round(a,4),\"Angstrom\";\n",
+ "print\"distance between two nearest Cu atoms is d=\",round(d,2),\"Angstrom\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "lattice constant is a= 3.61113576149 E-08 cm\n",
+ "\t\t a= 3.6111 Angstrom\n",
+ "distance between two nearest Cu atoms is d= 2.55 Angstrom\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.3 , Page no:41"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "N=6.02E26; #in /Kg-atom (Avogadro's number)\n",
+ "n=2; #number of molecules per unit cell for bcc lattice\n",
+ "M=55.85; #in Kg/Kg-atom (atomic weight of Iron)\n",
+ "p=7860; #in Kg/m^3 (density)\n",
+ "\n",
+ "#calculate\n",
+ "a=pow((n*M/(N*p)),0.33333);\n",
+ "a1=a*1E10; #changing unit to Angstrom\n",
+ "\n",
+ "#result\n",
+ "print\"The lattice constant is a=\",a,\"m\";\n",
+ "print\"\\t\\t\\ta=\",round(a1,3),\"Angstrom\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The lattice constant is a= 2.86928355016e-10 m\n",
+ "\t\t\ta= 2.869 Angstrom\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.4 , Page no:42"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "N=6.02*10**26; #in /Kg-atom (Avogadro's number)\n",
+ "n=2; #number of molecules per unit cell for bcc lattice\n",
+ "M=6.94; #in Kg/Kg-atom (atomic weight of Iron)\n",
+ "p=530; #in Kg/m^3 (density)\n",
+ "\n",
+ "#calculate\n",
+ "a=pow((n*M/(N*p)),0.33333);\n",
+ "a1=a*1E10; #changing unit to Angstrom\n",
+ "\n",
+ "#result\n",
+ "print\"The lattice constant is a=\",a,\"m\";\n",
+ "print\"\\t\\t\\ta=\",round(a1,3),\"Angstrom\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The lattice constant is a= 3.51776567326e-10 m\n",
+ "\t\t\ta= 3.518 Angstrom\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.5 , Page no:42"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "N=6.02*10**23; #in /gram-molecule (Avogadro's number)\n",
+ "M=58.5; #in gram/gram-molecule (atomic weight of NaCl)\n",
+ "p=2.17; #in g/cm^3 (density)\n",
+ "\n",
+ "#calculate\n",
+ "#since V=M/p\n",
+ "#(1/d)^-3=2N/V=2Np/M\n",
+ "#therefore d= (M/2Np)^-3\n",
+ "d=pow((M/(2*N*p)),0.33333333);\n",
+ "d1=d*1*10**8; #changing unit from cm to Angstrom\n",
+ "\n",
+ "#result\n",
+ "print\"The distance between two adjacent atoms of NaCl is d=\",d,\"m\";\n",
+ "print\"\\t\\t\\t\\t\\t\\t d=\",round(d1,3),\"Angstrom\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The distance between two adjacent atoms of NaCl is d= 2.81853408124e-08 m\n",
+ "\t\t\t\t\t\t d= 2.819 Angstrom\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.6 , Page no:43"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "r_Na=0.98; #in Angstrom (radius of sodium ion)\n",
+ "r_Cl=1.81; #in Angstrom (radius of chloride ion)\n",
+ "M_Na=22.99; #in amu (atomic mass of sodium)\n",
+ "M_Cl=35.45; #in amu (atomic mass of chlorine)\n",
+ "\n",
+ "#calculate\n",
+ "a=2*(r_Na+r_Cl); #lattice parameter\n",
+ "#PF=volume of ions present in the unit cell/volume of unit cell\n",
+ "PF=((4*(4/3)*3.14)*r_Na**3+(4*(4/3)*3.14)*r_Cl**3)/a**3;\n",
+ "#Density=mass of unit cell/volume of unit cell\n",
+ "p=4*(M_Na+M_Cl)*1.66E-27/(a*1E-10)**3;\n",
+ "p1=p*1E-3; #changing unit to gm/cm^-3\n",
+ "\n",
+ "#result\n",
+ "print\"Lattice constant is a=\",round(a,3),\"Angstrom\";\n",
+ "print\"Packing fraction is =\",round(PF,3); \n",
+ "print\"Density is p=\",round(p),\"Kg/m^3\";\n",
+ "print\"Density is p=\",round(p1,2),\"g/cm^3\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Lattice constant is a= 5.58 Angstrom\n",
+ "Packing fraction is = 0.662\n",
+ "Density is p= 2233.0 Kg/m^3\n",
+ "Density is p= 2.23 g/cm^3\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Applied_Physics/Chapter3.ipynb b/Applied_Physics/Chapter3.ipynb new file mode 100755 index 00000000..763a53c2 --- /dev/null +++ b/Applied_Physics/Chapter3.ipynb @@ -0,0 +1,144 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:1ec2372be7a9974010a31557c9658c85707025e7a87f82e6ef26be63dbcfb0e7"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 3:Planes in Crystals"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.11 , Page no:61"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "h=3;\n",
+ "k=2;\n",
+ "l=1; #miller indices\n",
+ "a=4.2E-8; #in cm (lattice constant)\n",
+ "\n",
+ "#calculate\n",
+ "d=a/math.sqrt(h**2+k**2+l**2); #calculation for interplanar spacing\n",
+ "d=d*1E8; #changing unit from cm to Angstrom\n",
+ "\n",
+ "#result\n",
+ "print\"The interplanar spacing is d=\",round(d,2),\"cm\";\n",
+ "print\"\\t\\t\\t d=\",round(d,2),\"Angstrom\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The interplanar spacing is d= 1.12 cm\n",
+ "\t\t\t d= 1.12 Angstrom\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.12 , Page no:61"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "h=1;k=1;l=1; #miller indices\n",
+ "a=2.5;b=2.5;c=1.8; #in Angstrom (lattice constants for tetragonal lattice )\n",
+ "\n",
+ "#calculate\n",
+ "d=1/math.sqrt((h/a)**2+(k/b)**2+(l/c)**2); #calculation for interplanar spacing\n",
+ "\n",
+ "#result\n",
+ "print\"The lattice spacing is d=\",round(d,2),\"Angstrom\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The lattice spacing is d= 1.26 Angstrom\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.15 , Page no:63"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "h=1;k=0;l=0; #miller indices\n",
+ "a=2.5; #in Angstrom (lattice constant)\n",
+ "\n",
+ "#calculate\n",
+ "a=a*1E-10; #hence a is in Angstrom\n",
+ "d=a/math.sqrt(h**2+k**2+l**2); #calculation for interplanar spacing\n",
+ "p=d/a**3;\n",
+ "\n",
+ "#result\n",
+ "print\"The density is p=\",round(p),\"lattice points/m^2\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The density is p= 1.6e+19 lattice points/m^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Applied_Physics/Chapter4.ipynb b/Applied_Physics/Chapter4.ipynb new file mode 100755 index 00000000..6c9c83eb --- /dev/null +++ b/Applied_Physics/Chapter4.ipynb @@ -0,0 +1,494 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:177e44357417b2cfb0b476ada5b36a934225ade51d50763b09586ba8679f08fc"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 4:Crystal Diffraction"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.1 , Page no:75"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "lambda1=2.6; #in Angstrom (wavelength)\n",
+ "theta=20; #in Degree (angle)\n",
+ "n=2;\n",
+ "\n",
+ "#calculate\n",
+ "lambda1=lambda1*1E-10; #since lambda is in Angstrom\n",
+ "#Since 2dsin(theta)=n(lambda)\n",
+ "#therefore d=n(lambda)/2sin(theta)\n",
+ "d=n*lambda1/(2*math.sin(math.radians(theta)));\n",
+ "\n",
+ "#result\n",
+ "print\"The spacing constant is d=\",d,\"m\";\n",
+ "print\"\\t\\t\\td=\",d*10**10,\"Angstrom\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The spacing constant is d= 7.60189144042e-10 m\n",
+ "\t\t\td= 7.60189144042 Angstrom\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.2 , Page no:75"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "h=1;k=1;l=0; #miller indices\n",
+ "a=0.26; #in nanometer (lattice constant)\n",
+ "lambda1=0.065; #in nanometer (wavelength)\n",
+ "n=2; #order\n",
+ "\n",
+ "#calculate\n",
+ "d=a/math.sqrt(h**2+k**2+l**2); #calculation of interlattice spacing\n",
+ "#Since 2dsin(theta)=n(lambda)\n",
+ "#therefore we have\n",
+ "theta=math.asin(n*lambda1/(2*d));\n",
+ "theta1=theta*180/3.14;\n",
+ "\n",
+ "#result\n",
+ "print\"The glancing angle is =\",round(theta1,2),\"degree\";\n",
+ "print \"Note: there is slight variation in the answer due to round off error\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The glancing angle is = 20.72 degree\n",
+ "Note: there is slight variation in the answer due to round off error\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.3 , Page no:75"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "d=3.04E-10; #in mm (spacing constant)\n",
+ "lambda1=0.79; #in Angstrom (wavelength)\n",
+ "n=3; #order\n",
+ "\n",
+ "#calculate\n",
+ "#Since 2dsin(theta)=n(lambda)\n",
+ "#therefore we have\n",
+ "lambda2=lambda1*1E-10; #since lambda is in angstrom\n",
+ "theta=math.asin(n*lambda2/(2*d));\n",
+ "theta1=theta*180/3.14;\n",
+ "\n",
+ "#result\n",
+ "print\"The glancing angle is\",round(theta1,3),\"degree\";\n",
+ "print \"Note: In question the value of d=3.04E-9 cm but in solution is using d=3.04E-10 m. So I have used d=3.04E-10 cm as used in the solution\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The glancing angle is 22.954 degree\n",
+ "Note: In question the value of d=3.04E-9 cm but in solution is using d=3.04E-10 m. So I have used d=3.04E-10 cm as used in the solution\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4 , Page no:76"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "d=0.282; #in nanometer (spacing constant)\n",
+ "n=1; #order\n",
+ "theta=8.35; #in degree (glancing angle)\n",
+ "\n",
+ "#calculate\n",
+ "d=d*1E-9; #since d is in nanometer\n",
+ "#Since 2dsin(theta)=n(lambda)\n",
+ "#therefore we have\n",
+ "lambda1=2*d*math.sin(math.radians(theta))/n; \n",
+ "lambda_1=lambda1*1E10; #changing unit from m to Angstrom\n",
+ "theta_1=90; #in degree (for maximum order theta=90)\n",
+ "n_max=2*d*math.sin(math.radians(theta_1))/lambda1; #calculation of maximum order. \n",
+ "\n",
+ "#result\n",
+ "print\"The wavelength =\",lambda1,\"m\";\n",
+ "print\"\\t\\t=\" ,round(lambda_1,3),\"Angstrom\";\n",
+ "print\"The maximum order possible is n=\",round(n_max);\n",
+ "print \"Note: In question value of theta=8 degree and 35 minutes but solution uses theta=8.35 degree.So I am using theta=8.35 degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The wavelength = 8.19038939236e-11 m\n",
+ "\t\t= 0.819 Angstrom\n",
+ "The maximum order possible is n= 7.0\n",
+ "Note: In question value of theta=8 degree and 35 minutes but solution uses theta=8.35 degree.So I am using theta=8.35 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.5 , Page no:76"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "theta=6; #in degree (glancing angle)\n",
+ "p=2170; #in Kg/m^3 (density)\n",
+ "M=58.46; #Molecular weight of NaCl\n",
+ "N=6.02E26; #in Kg-molecule (Avogadro's number)\n",
+ "n=1; #order\n",
+ "XU=1E-12; #since 1X.U.= 1E-12m\n",
+ "\n",
+ "#calculate\n",
+ "d=(M/(2*N*p))**(1/3); #calclation of lattice constant\n",
+ "#Since 2dsin(theta)=n(lambda)\n",
+ "#therefore we have\n",
+ "lambda1=2*d*math.sin(math.radians(theta))/n; #calculation of wavelength\n",
+ "lambda2=lambda1/XU;\n",
+ "\n",
+ "#result\n",
+ "print\"The spacing constant is d=\",d,\"m\";\n",
+ "print\"The wavelength is =\",lambda1,\"m\";\n",
+ "print\"\\t\\t =\",lambda2,\"X.U\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The spacing constant is d= 2.81789104396e-10 m\n",
+ "The wavelength is = 5.89099640962e-11 m\n",
+ "\t\t = 58.9099640962 X.U\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.6 , Page no:77"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "h=1;k=1;l=1; #miller indices\n",
+ "a=5.63; #in Angstrom (lattice constant)\n",
+ "theta=27.5; #in degree (Glancing angle)\n",
+ "n=1; #order\n",
+ "H=6.625E-34; #in J-s (Plank's constant)\n",
+ "c=3E8; #in m/s (velocity of light)\n",
+ "e=1.6E-19; #charge of electron\n",
+ "\n",
+ "#calculate\n",
+ "d=a/math.sqrt(h**2+k**2+l**2); #calculation for interplanar spacing\n",
+ "#Since 2dsin(theta)=n(lambda)\n",
+ "#therefore we have\n",
+ "lambda1=2*d*math.sin(math.radians(theta))/n; #calculation for wavelength\n",
+ "E=H*c/(lambda1*1E-10); #calculation of Energy\n",
+ "E1=E/e; #changing unit from J to eV\n",
+ "\n",
+ "#result\n",
+ "print\"The lattice spacing is d=\",round(d,2),\"Angstrom\";\n",
+ "print\"The wavelength is =\",round(lambda1),\"Angstrom\";\n",
+ "print\"The energy of X-rays is E=\",E,\"J\";\n",
+ "print\"\\t\\t\\tE=\",E1,\"eV\";\n",
+ "print \"Note: c=3E8 m/s but in solution c=3E10 m/s \""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The lattice spacing is d= 3.25 Angstrom\n",
+ "The wavelength is = 3.0 Angstrom\n",
+ "The energy of X-rays is E= 6.62100284311e-16 J\n",
+ "\t\t\tE= 4138.12677695 eV\n",
+ "Note: c=3E8 m/s but in solution c=3E10 m/s \n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.7 , Page no:77"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "V=344; #in V (accelerating voltage)\n",
+ "theta=60; #in degree (glancing angle)\n",
+ "m=9.1E-31; #in Kg (mass of electron)\n",
+ "h=6.625e-34; #in J-s (Plank's constant)\n",
+ "n=1; #order\n",
+ "e=1.6E-19; #charge on electron\n",
+ "\n",
+ "#calculate\n",
+ "#Since K=m*v^2/2=e*V\n",
+ "#therefore v=sqrt(2*e*V/m)\n",
+ "#since lambda=h/(m*v)\n",
+ "#therefore we have lambda=h/sqrt(2*m*e*V)\n",
+ "lambda1=h/math.sqrt(2*m*e*V); #calculation of lambda\n",
+ "lambda2=lambda1*1E10; #changing unit from m to Angstrom\n",
+ "#Since 2dsin(theta)=n(lambda)\n",
+ "#therefore we have\n",
+ "d=n*lambda2/(2*math.sin(math.radians(theta)));\n",
+ "\n",
+ "#result\n",
+ "print\"The wavelength is =\",lambda1,\"m\";\n",
+ "print\"\\t\\t =\",round(lambda2,2),\"Angstrom\";\n",
+ "print\"The interplanar spacing is d=\",round(d,2),\"Angstrom\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The wavelength is = 6.61928340764e-11 m\n",
+ "\t\t = 0.66 Angstrom\n",
+ "The interplanar spacing is d= 0.38 Angstrom\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.8 , Page no:78"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "K=0.02; #in eV (kinetic energy)\n",
+ "d=2.0; #in Angstrom (Bragg's spacing)\n",
+ "m=1.00898; #in amu (mass of neutron)\n",
+ "amu=1.66E-27; #in Kg (1amu=1.66E-27 Kg)\n",
+ "h=6.625e-34; #in J-s (Plank's constant)\n",
+ "n=1; #order\n",
+ "e=1.6E-19; #charge on electron\n",
+ "\n",
+ "#calculate\n",
+ "#Since K=m*v^2/2\n",
+ "#therefore v=sqrt(2*K/m)\n",
+ "#since lambda=h/(m*v)\n",
+ "#therefore we have lambda=h/sqrt(2*m*K)\n",
+ "m=m*amu; #changing unit from amu to Kg\n",
+ "K=K*e; #changing unit to J from eV\n",
+ "lambda1=h/math.sqrt(2*m*K); #calculation of lambda\n",
+ "lambda2=lambda1*1E10; #changing unit from m to Angstrom\n",
+ "theta=math.asin(n*lambda2/(2*d)); #calculation of angle of first order diffraction maximum\n",
+ "theta1=theta*180/3.14;\n",
+ "#result\n",
+ "print\"The wavelength is =\",lambda1,\"m\";\n",
+ "print\"\\t \\t =\",round(lambda2),\"Angstrom\";\n",
+ "print\"The angle of first order diffraction maximum is \",round(theta1),\"Degree\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The wavelength is = 2.02348771817e-10 m\n",
+ "\t \t = 2.0 Angstrom\n",
+ "The angle of first order diffraction maximum is 30.0 Degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.9 , Page no:79"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "lambda1=0.586; #in Angstrom (wavelength of X-rays)\n",
+ "n1=1; n2=2; n3=3; #orders of diffraction\n",
+ "theta1=5+(58/60); #in degree (Glancing angle for first order of diffraction)\n",
+ "theta2=12+(01/60); #in degree (Glancing angle for second order of diffraction)\n",
+ "theta3=18+(12/60); #in degree (Glancing angle for third order of diffraction)\n",
+ "\n",
+ "#calculate\n",
+ "K1=math.sin(math.radians(theta1));\n",
+ "K2=math.sin(math.radians(theta2));\n",
+ "K3=math.sin(math.radians(theta3));\n",
+ "#Since 2dsin(theta)=n(lambda)\n",
+ "#therefore we have\n",
+ "d1=n1*lambda1/(2*K1);\n",
+ "d2=n2*lambda1/(2*K2);\n",
+ "d3=n3*lambda1/(2*K3);\n",
+ "d1=d1*1E-10; #changing unit from Angstrom to m\n",
+ "d2=d2*1E-10; #changing unit from Angstrom to m\n",
+ "d3=d3*1E-10; #changing unit from Angstrom to m\n",
+ "d=(d1+d2+d3)/3;\n",
+ "\n",
+ "#result\n",
+ "print\"The value of sine of different angle of diffraction is \\nK1=\",K1;\n",
+ "print\"K2=\",K2;\n",
+ "print\"K3=\",K3;\n",
+ "#Taking the ratios of K1:K2:K3\n",
+ "#We get K1:K2:K3=1:2:3\n",
+ "#Therefore we have\n",
+ "print\"Or we have K1:K2:K3=1:2:3\";\n",
+ "print\"Hence these angles of incidence are for Ist, 2nd and 3rd order reflections respectively\";\n",
+ "print\"The spacing constants are \\nd1=\",d1;\n",
+ "print\"d2=\",d2;\n",
+ "print\"d3=\",d3;\n",
+ "print\"The mean value of crystal spacing is d=\",d,\"m\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of sine of different angle of diffraction is \n",
+ "K1= 0.103949856225\n",
+ "K2= 0.208196213621\n",
+ "K3= 0.312334918512\n",
+ "Or we have K1:K2:K3=1:2:3\n",
+ "Hence these angles of incidence are for Ist, 2nd and 3rd order reflections respectively\n",
+ "The spacing constants are \n",
+ "d1= 2.81866671719e-10\n",
+ "d2= 2.81465253286e-10\n",
+ "d3= 2.81428667722e-10\n",
+ "The mean value of crystal spacing is d= 2.81586864243e-10 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Applied_Physics/Chapter5.ipynb b/Applied_Physics/Chapter5.ipynb new file mode 100755 index 00000000..cf16b683 --- /dev/null +++ b/Applied_Physics/Chapter5.ipynb @@ -0,0 +1,1258 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:17ded29a54411f524d1901cc530f956cb4706c517e993f5acba4014789eca0bd"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 5:Principles of Quantum Mechanics"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.1 , Page no:102"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "lambda1=1; #in Angstrom (wavelength)\n",
+ "m=1.67E-27; #in Kg (mass of neutron)\n",
+ "h=6.625E-34; #in J-s (Planck's constant)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "lambda2=lambda1*1E-10; #since lambda is in Angstrom\n",
+ "#Since lambda=h/(m*v)\n",
+ "#Therefore we have\n",
+ "v=h/(m*lambda2); #calculation of velocity\n",
+ "K=m*v**2/2; #calculation of kinetic energy\n",
+ "K1=K/e; #changing unit fro J to eV\n",
+ "\n",
+ "#result\n",
+ "print\"The velocity is v=\",v,\"m/s\";\n",
+ "print\"The kinetic energy is K=\",K,\"J\";\n",
+ "print\"\\t\\t =\",K1,\"eV\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The velocity is v= 3967.06586826 m/s\n",
+ "The kinetic energy is K= 1.31409056886e-20 J\n",
+ "\t\t = 0.0821306605539 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.2 , Page no:103"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "K=50; #in eV (Kinetic energy)\n",
+ "m0=9.1E-31; #in Kg (mass of electron)\n",
+ "h=6.625E-34; #in J-s (Planck's constant)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "K=K*e; #changing unit from eV to J\n",
+ "#Since K=m*v^2/2\n",
+ "#Therefore v=sqrt(2*K/m)\n",
+ "#Since lambda=h/(m*v)\n",
+ "#Therefore we have\n",
+ "lambda1=h/math.sqrt(2*m0*K); #calculation of wavelength\n",
+ "lambda2=lambda1*1E10; #changing unit from m to Angstrom\n",
+ "\n",
+ "#result\n",
+ "print\"The wavelength is =\",lambda1,\"m\";\n",
+ "print\"\\t\\t =\",round(lambda2,3),\"Angstrom\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The wavelength is = 1.73622176082e-10 m\n",
+ "\t\t = 1.736 Angstrom\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.3 , Page no:104"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "E=2000; #in eV (Kinetic energy)\n",
+ "m=9.1E-31; #in Kg (mass of electron)\n",
+ "h=6.625E-34; #in J-s (Planck's constant)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "E=E*e; #changing unit from eV to J\n",
+ "#Since E=m*v^2/2\n",
+ "#Therefore v=sqrt(2*E/m)\n",
+ "#Since lambda=h/(m*v)\n",
+ "#Therefore we have\n",
+ "lambda1=h/math.sqrt(2*m*E); #calculation of wavelength\n",
+ "lambda2=lambda1*1E9; #changing unit from m to nanometer\n",
+ "\n",
+ "#result\n",
+ "print\"The wavelength is =\",lambda1,\"m\";\n",
+ "print\"\\t\\t =\",round(lambda2,4),\"nm\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The wavelength is = 2.74520764367e-11 m\n",
+ "\t\t = 0.0275 nm\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.4 , Page no:104"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "m_e=9.1E-31; #in Kg (mass of electron)\n",
+ "m_n=1.676E-27; #in Kg (mass of neutron)\n",
+ "h=6.625E-34; #in J-s (Planck's constant)\n",
+ "c=3E8; #in m/s (velocity of light)\n",
+ "\n",
+ "#calculate\n",
+ "E_e=m_e*c**2; #rest mass energy of electron\n",
+ "E_n=2*E_e; #given (kinetic energy of neutron)\n",
+ "#Since K=m*v^2/2\n",
+ "#Therefore v=sqrt(2*K/m)\n",
+ "#Since lambda=h/(m*v)\n",
+ "#Therefore we have\n",
+ "lambda1=h/math.sqrt(2*m_n*E_n); #calculation of wavelength\n",
+ "lambda2=lambda1*1E10; #changing unit from m to Angstrom\n",
+ "\n",
+ "#result\n",
+ "print\"The wavelength is =\",lambda1,\"m\";\n",
+ "print\"\\t\\t =\",lambda2,\"Angstrom\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The wavelength is = 2.82733278331e-14 m\n",
+ "\t\t = 0.000282733278331 Angstrom\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5 , Page no:104"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "V=1600; #in V (Potential)\n",
+ "\n",
+ "#calculate\n",
+ "lambda1=12.27/math.sqrt(V); #calculation of wavelength in Angstrom\n",
+ "\n",
+ "#result\n",
+ "print\"The wavelength is =\",round(lambda1,3),\"Angstrom\";\n",
+ "print \"Note : The answer in the book is wrong \""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The wavelength is = 0.307 Angstrom\n",
+ "Note : The answer in the book is wrong \n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.6 , Page no:105"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "E=1000; #in eV (Kinetic energy of photon)\n",
+ "K=1000; #in eV (Kinetic energy of electron)\n",
+ "m0=9.1E-31; #in Kg (mass of electron)\n",
+ "h=6.6E-34; #in J-s (Planck's constant)\n",
+ "c=3E8; #in m/s (velocity of light)\n",
+ "e=1.6E-19; #in C (charge on electron)\n",
+ "\n",
+ "#calculate\n",
+ "E=E*e; #changing unit from eV to J\n",
+ "lambda_p=h*c/E; #For photon E=hc/lambda\n",
+ "lambda_p1=lambda_p*1E10; #changing unit from m to Angstrom\n",
+ "#Since K=m*v^2/2\n",
+ "#Therefore v=sqrt(2*K/m)\n",
+ "#Since lambda=h/(m*v)\n",
+ "#Therefore we have\n",
+ "K=K*e; #changing unit from eV to J\n",
+ "lambda_e=h/math.sqrt(2*m0*K); #calculation of wavelength\n",
+ "lambda_e1=lambda_e*1E10; #changing unit from m to Angstrom\n",
+ "\n",
+ "#result\n",
+ "print\"For photon,the wavelength is =\",lambda_p,\"m\";\n",
+ "print\"\\t\\t\\t =\",round(lambda_p1,1),\"Angstrom\";\n",
+ "print\"For electron,the wavelength is =\",lambda_e,\"m\";\n",
+ "print\"\\t\\t\\t =\",round(lambda_e1,2),\"Angstrom\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "For photon,the wavelength is = 1.2375e-09 m\n",
+ "\t\t\t = 12.4 Angstrom\n",
+ "For electron,the wavelength is = 3.86765965524e-11 m\n",
+ "\t\t\t = 0.39 Angstrom\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.7 , Page no:105"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "lambda1=1.66*10**-10; #in m (wavelength)\n",
+ "m=9.1*10**-31; #in Kg (mass of electron)\n",
+ "h=6.626*10**-34; #in J-s (Planck's constant)\n",
+ "e=1.6E-19; #in C (charge on electron)\n",
+ "\n",
+ "#calculate\n",
+ "#Since lambda=h/(m*v)\n",
+ "#Therefore we have\n",
+ "v=h/(m*lambda1); #calculation of velocity\n",
+ "K=m*v**2/2; #calculation of kinetic energy\n",
+ "K1=K/e; #changing unit from J to eV\n",
+ "\n",
+ "#result\n",
+ "print\"The velocity of electron is v=\",v,\"m/s\";\n",
+ "print\"The kinetic energy is K=\",K,\"J\";\n",
+ "print\"\\t\\t =\",round(K1,3),\"eV\";\n",
+ "print \"Note: The answer in the book for kinetic energy is wrong \""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The velocity of electron is v= 4386336.55501 m/s\n",
+ "The kinetic energy is K= 8.75417651009e-18 J\n",
+ "\t\t = 54.714 eV\n",
+ "Note: The answer in the book for kinetic energy is wrong \n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.8 , Page no:106"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "T=400; #in K (temperature)\n",
+ "m=6.7E-27; #in Kg (mass of He-atom)\n",
+ "h=6.6E-34; #in J-s (Planck's constant)\n",
+ "k=1.376E-23; #in J/degree (Boltzmann constant)\n",
+ "\n",
+ "#calculate\n",
+ "#Since lambda=h/(m*v)\n",
+ "#E=mv^2/2;\n",
+ "#Therefore lambda=h/sqrt(2*m*E)\n",
+ "#E=kT\n",
+ "#Therefore lambda=h/sqrt(2*m*k*T)\n",
+ "lambda1=h/math.sqrt(2*m*k*T)\n",
+ "lambda2=lambda1*1E10; #changing unit from m to Angstrom\n",
+ "\n",
+ "#result\n",
+ "print\"The de-Broglie wavelength of He-atom is =\",lambda1,\"m\";\n",
+ "print\"\\t\\t\\t\\t =\",round(lambda2,4),\"Angstrom\"; "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The de-Broglie wavelength of He-atom is = 7.6851495611e-11 m\n",
+ "\t\t\t\t = 0.7685 Angstrom\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.9 , Page no:106"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "m_e=9.1E-31; #in Kg (mass of electron)\n",
+ "m_p=1.6E-27; #in Kg (mass of proton)\n",
+ "h=6.626E-34; #in J-s (Planck's constant)\n",
+ "c=3E8; #in m/s (velocity of light)\n",
+ "\n",
+ "#calculate\n",
+ "E=m_e*c**2; #in J (rest energy of electron)\n",
+ "#Since lambda=h/(m*v)\n",
+ "#E=mv^2/2;\n",
+ "#herefore lambda=h/sqrt(2*m*E)\n",
+ "#Also E=m_e*c^2;\n",
+ "#therefore lambda=h/sqrt(2*m_p*m_e*c^2)\n",
+ "lambda1=h/math.sqrt(2*m_p*m_e*c**2); #calculation of wavelength\n",
+ "lambda2=lambda1*1*10**10; #changing unit from m to Angstrom\n",
+ "\n",
+ "#result\n",
+ "print\"The de-Broglie wavelength of proton is =\",lambda1,\"m\";\n",
+ "print\"\\t\\t\\t\\t =\",lambda2,\"Angstrom\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The de-Broglie wavelength of proton is = 4.0929316435e-14 m\n",
+ "\t\t\t\t = 0.00040929316435 Angstrom\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.10 , Page no:106"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "V=10000; #in V (Potential)\n",
+ "\n",
+ "#calculate\n",
+ "lambda1=12.27/math.sqrt(V); #calculation of wavelength in Angstrom\n",
+ "\n",
+ "#result\n",
+ "print\"The wavelength is =\",round(lambda1,3),\"Angstrom\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The wavelength is = 0.123 Angstrom\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.11 , Page no:107"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "V=100; #in V (Potential)\n",
+ "n=1; #order of diffraction\n",
+ "d=2.15; #in Angstrom (lattice spacing)\n",
+ "\n",
+ "#calculate\n",
+ "lambda1=12.27/math.sqrt(V); #calculation of wavelength in Angstrom\n",
+ "#Since 2*d*sind(theta)=n*lambda\n",
+ "#therefore we have\n",
+ "theta=math.asin(n*lambda1/(2*d)); #calculation of glancing angle\n",
+ "theta1=theta*180/3.14;\n",
+ "\n",
+ "#result\n",
+ "print\"The wavelength is =\",round(lambda1,3),\"Angstrom\";\n",
+ "print\"The glancing angle is =\",round(theta1,1),\"degree\";\n",
+ "print \"Note: In question V=100 eV but the solution is using V=100V \"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The wavelength is = 1.227 Angstrom\n",
+ "The glancing angle is = 16.6 degree\n",
+ "Note: In question V=100 eV but the solution is using V=100V \n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.12 , Page no:107"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "V=344; #in V (Potential)\n",
+ "n=1; #order of diffraction\n",
+ "theta=60; #in degree (glancing angle)\n",
+ "\n",
+ "#calculate\n",
+ "lambda1=12.27/math.sqrt(V); #calculation of wavelength in Angstrom\n",
+ "#Since 2*d*sind(theta)=n*lambda\n",
+ "#therefore we have\n",
+ "d=n*lambda1/(2*math.sin(math.radians(theta))); #calculation of spacing constant\n",
+ "\n",
+ "#result\n",
+ "print\"The wavelength is =\",round(lambda1,3),\"Angstrom\";\n",
+ "print\"The spacing of the crystal is =\",round(d,4),\"Angstrom\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The wavelength is = 0.662 Angstrom\n",
+ "The spacing of the crystal is = 0.3819 Angstrom\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.13 , Page no:107"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "r=0.53*10**-10; #in m (radius of first Bohr orbit)\n",
+ "h=6.6*10**-34; #in J-s (Planck's constant)\n",
+ "m=9.1*10**-31; #in Kg (mass of electron)\n",
+ "n=1; #First Bohr orbit\n",
+ "pi=3.14; #value of pi used in the solution\n",
+ "\n",
+ "#calculate\n",
+ "#Since 2*pi*r=n*lambda and lambda=h/(m*v)\n",
+ "#Threfore we have v=h*n/(2*pi*r*m)\n",
+ "v=h*n/(2*pi*r*m); #calculation of velocity\n",
+ "\n",
+ "#result\n",
+ "print\"The velocity of electron is =\",v,\"m/s\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The velocity of electron is = 2179049.16859 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.14 , Page no:108"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "dx=0.2; #in Angstrom (uncertainty in the position)\n",
+ "h=6.6E-34; #in J-s (Planck's constant)\n",
+ "m0=9.1E-31; #in Kg (mass of electron)\n",
+ "pi=3.14; #value of pi used in the solution\n",
+ "\n",
+ "#calculate\n",
+ "dx=dx*1E-10; #since dx is in Angstrom\n",
+ "#Since dx*dp=h/4*pi (uncertainty relation)\n",
+ "dp=h/(4*pi*dx); #calculation of uncertainty in the momentum\n",
+ "#since dp=m*dv\n",
+ "dv=dp/m0; #calculation of uncertainty in the velocity\n",
+ "\n",
+ "#result\n",
+ "print\"The uncertainty in the momentum is dp=\",dp,\"Kg-m/\";\n",
+ "print\"The uncertainty in the velocity is dv=\",dv,\"m/s\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The uncertainty in the momentum is dp= 2.62738853503e-24 Kg-m/\n",
+ "The uncertainty in the velocity is dv= 2887240.14839 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.15 , Page no:108"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "m_e=9.1E-31; #in Kg (mass of electron)\n",
+ "m_p=1.67E-27; #in Kg (mass of proton)\n",
+ "dx_p=1; #in nanometer (uncertainty in position of electron)\n",
+ "dx_n=1; #in nanometer (uncertainty in position of proton)\n",
+ "\n",
+ "#calculate\n",
+ "#since dp=h/(4*pi*dx)\n",
+ "#since h/(4*pi) is constant and dx is same for electron and proton \n",
+ "#therefor both electron and proton have same uncertainty in the momentum\n",
+ "#since dv=dp/m and dp is same for both\n",
+ "#therefore dv_e/dv_p=m_p/m_e\n",
+ "#therefore\n",
+ "K=m_p/m_e; #ratio of uncertainty in the velocity of electron and proton\n",
+ "\n",
+ "#result\n",
+ "print\"The ratio of uncertainty in the velocity of electron to that of proton is =\",round(K);"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The ratio of uncertainty in the velocity of electron to that of proton is = 1835.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.16 , Page no:108"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "dx=5*10**-15; #in m (radius of nucleus or uncertainty in the position)\n",
+ "h=6.6*10**-34; #in J-s (Planck's constant)\n",
+ "m=1.67E-27; #in Kg (mass of proton)\n",
+ "pi=3.14; #value of pi used in the solution\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "#Since dx*dp=h/4*pi (uncertainty relation)\n",
+ "dp=h/(4*pi*dx); #calculation of uncertainty in the momentum\n",
+ "p=dp; #minimum value of momentum to calculate mimimum kinetic energy\n",
+ "K=(p**2)/(2*m); #calculation of minimum kinetic energy of proton\n",
+ "K1=K/e; #changing unit from J to eV\n",
+ "K2=K1/1E6; #changing unit from eV to MeV\n",
+ "\n",
+ "#result\n",
+ "print\"The minimum uncertainty in the momentum of proton is dp =\",dp,\"Kg-m/s\";\n",
+ "print\"The minimum kinetic energy of proton is K=\",K,\"J\";\n",
+ "print\"\\t\\t\\t\\t \\t =\",K1,\"eV\";\n",
+ "print\"\\t\\t\\t\\t\\t =\",K2,\"MeV\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The minimum uncertainty in the momentum of proton is dp = 1.05095541401e-20 Kg-m/s\n",
+ "The minimum kinetic energy of proton is K= 3.30690803067e-14 J\n",
+ "\t\t\t\t \t = 206681.751917 eV\n",
+ "\t\t\t\t\t = 0.206681751917 MeV\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.17 , Page no:109"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "K=1; #in KeV (kinetic energy of electron)\n",
+ "dx=1; #in Angstrom (uncertainty in the position)\n",
+ "h=6.63E-34; #in J-s (Planck's constant)\n",
+ "m=9.1E-31; #in Kg (mass of electron)\n",
+ "pi=3.14; #value of pi used in the solution\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "dx=dx*1E-10; #since dx is in Angstrom\n",
+ "#Since dx*dp=h/4*pi (uncertainty relation)\n",
+ "dp=h/(4*pi*dx); #calculation of uncertainty in the momentum\n",
+ "K=K*1E3*1.6E-19; #changing unit from KeV to J\n",
+ "p=math.sqrt(2*m*K); #calculation of momentum \n",
+ "poc=(dp/p)*100; #calculation of percentage of uncertainty\n",
+ "\n",
+ "#result\n",
+ "print\"The uncertainty in the momentum of electron is dp=\",dp,\"Kg-m/s\";\n",
+ "print\"The momentum of electron is p=\",p,\"Kg-m/s\";\n",
+ "print\"The percentage of uncertainty in the momentum is =\",round(poc,1);"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The uncertainty in the momentum of electron is dp= 5.27866242038e-25 Kg-m/s\n",
+ "The momentum of electron is p= 1.70645832062e-23 Kg-m/s\n",
+ "The percentage of uncertainty in the momentum is = 3.1\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.18 , Page no:109"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "v=6.6E4; #m/s (speed of electron)\n",
+ "poc=0.01; #percentage of uncertainty\n",
+ "h=6.63E-34; #in J-s (Planck's constant)\n",
+ "m=9E-31; #in Kg (mass of electron)\n",
+ "pi=3.14; #value of pi used in the solution\n",
+ "\n",
+ "#calculate\n",
+ "p=m*v; #calculation of momentum\n",
+ "dp=(poc/100)*p; #calculation of uncertainty in the momentum\n",
+ "#Since dx*dp=h/4*pi (uncertainty relation)\n",
+ "dx=h/(4*pi*dp); #calculation of uncertainty in the position\n",
+ "\n",
+ "#result\n",
+ "print\"The momentum of electron is p=\",p,\"Kg-m/s\";\n",
+ "print\"The uncertainty in the momentum of electron is dp=\",dp,\"Kg-m/s\";\n",
+ "print\"The uncertainty in the position of electron is dx=\",dx,\"Kg-m/s\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The momentum of electron is p= 5.94e-26 Kg-m/s\n",
+ "The uncertainty in the momentum of electron is dp= 5.94e-30 Kg-m/s\n",
+ "The uncertainty in the position of electron is dx= 8.88663707135e-06 Kg-m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.19 , Page no:111"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "lambda1=1; #in Angstrom (wavelength)\n",
+ "pi=3.14; #value of pi used in the solution\n",
+ "dlambda=1E-6; #uncertainty in wavelength\n",
+ "\n",
+ "#calculate\n",
+ "lambda2=lambda1*1E-10; #sinc lambda is in Angstrom\n",
+ "#By uncertainty principle, dx*dp>=h/(4*pi) --(1)\n",
+ "#since p=h/lambda -----(2)\n",
+ "#Or p*lambda=h \n",
+ "#diffrentiting this equation\n",
+ "#p*dlambda+lambda*dp=0\n",
+ "#dp=-p*dlambda/lambda ----(3)\n",
+ "#from (2) and (3) dp=-h*dlambda/lambda^2 ---(4)\n",
+ "#from (1) and(4) dx*dlambda>=lambda^2/4*pi\n",
+ "#Or dx=lambda^2/(4*pi*dlambda)\n",
+ "dx=lambda2**2/(4*pi*dlambda); #calculation of uncertainty in the position\n",
+ "\n",
+ "#result\n",
+ "print\"The uncertainty in the position of X-ray photon is dx=\",dx,\"m\";\n",
+ "print \"Note: wavelength accuracy is given as 1 in 1E8 but in book solution has used 1 in 1E6 \"\n",
+ "print \" ANSWEER IS WRONG IN THE TEXTBOOK\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The uncertainty in the position of X-ray photon is dx= 7.96178343949e-16 m\n",
+ "Note: wavelength accuracy is given as 1 in 1E8 but in book solution has used 1 in 1E6 \n",
+ " ANSWEER IS WRONG IN THE TEXTBOOK\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.20 , Page no:111"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "dt=1*10**-8; #in sec (average life time)\n",
+ "pi=3.14; #value of pi used in the solution\n",
+ "\n",
+ "#calculate\n",
+ "#Since dE*dt>=h/(4*pi) (uncertainty relation for energy)\n",
+ "#and E=h*v v is the frequency\n",
+ "#therefore we have dv>=1/(4*pi*dt)\n",
+ "dv=1/(4*pi*dt); #calculation of minimum uncertainty in the frequency\n",
+ "\n",
+ "#result\n",
+ "print\"The minimum uncertainty in the frequency of the photon is dv=\",dv,\"sec^-1\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The minimum uncertainty in the frequency of the photon is dv= 7961783.43949 sec^-1\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.21 , Page no:111"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "dt=1E-12; #in sec (average life time)\n",
+ "h=6.63E-34; #in J-s (Planck'c constant)\n",
+ "pi=3.14; #value of pi used in the solution\n",
+ "e=1.6*1E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "#Since dE*dt>=h/(4*pi) (uncertainty relation for energy)\n",
+ "dE=h/(4*pi*dt); #calculation of minimum uncertainty in the energy\n",
+ "dE1=dE/e; #changing unit from J to eV\n",
+ "\n",
+ "#result\n",
+ "print\"The uncertainty in the energy of the photon is dE=\",dE,\"J\";\n",
+ "print\"\\t\\t\\t\\t\\t =\",dE1,\"eV\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The uncertainty in the energy of the photon is dE= 5.27866242038e-23 J\n",
+ "\t\t\t\t\t = 0.000329916401274 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.22 , Page no:111"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "dT=2.5E-14; #in sec (average life time)\n",
+ "h=6.63E-34; #in J-s (Planck'c constant)\n",
+ "pi=3.14; #value of pi used in the solution\n",
+ "e=1.6*1E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "#Since dE*dt>=h/(4*pi) (uncertainty relation for energy)\n",
+ "dE=h/(4*pi*dT); #calculation of minimum uncertainty in the energy\n",
+ "dE1=dE/e; #changing unit from J to eV\n",
+ "\n",
+ "#result\n",
+ "print\"The uncertainty in the energy of the photon is dE=\",dE,\"J\";\n",
+ "print\"\\t\\t\\t\\t\\t =\",dE1,\"eV\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The uncertainty in the energy of the photon is dE= 2.11146496815e-21 J\n",
+ "\t\t\t\t\t = 0.013196656051 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.23 , Page no:112"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "a=2; #in Angstrom (length of the box)\n",
+ "m=9.1E-31; #in Kg (mass of electron)\n",
+ "h=6.626E-34; #in J-s (Planck'c constant)\n",
+ "n2=2; n4=4; #two quantum states\n",
+ "e=1.6*1E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "a=a*1E-10; #since a is in Angstrom \n",
+ "#Since E_n=n^2*h^2/(8*m*a^2) (Energy corresponding to nth quantum state)\n",
+ "E2=n2**2*h**2/(8*m*a**2); #calculation of energy corresponding to the 2nd quantum state\n",
+ "E21=E2/e; #changing unit from J to eV\n",
+ "E4=n4**2*h**2/(8*m*a**2); #calculation of energy corresponding to the 4nd quantum state\n",
+ "E41=E4/e; #changing unit from J to eV\n",
+ "\n",
+ "#result\n",
+ "print\"The energy corresponding to the 2nd quantum state is E2=\",E2,\"J\";\n",
+ "print\"\\t\\t\\t\\t\\t\\t =\",E21,\"eV\";\n",
+ "print\"The energy corresponding to the 4nd quantum state is E4=\",E4,\"J\";\n",
+ "print\"\\t\\t\\t\\t\\t\\t =\",E41,\"eV\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The energy corresponding to the 2nd quantum state is E2= 6.0307521978e-18 J\n",
+ "\t\t\t\t\t\t = 37.6922012363 eV\n",
+ "The energy corresponding to the 4nd quantum state is E4= 2.41230087912e-17 J\n",
+ "\t\t\t\t\t\t = 150.768804945 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.24 , Page no:113"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "a=1E-10; #in m (width of the well)\n",
+ "m=9.1E-31; #in Kg (mass of electron)\n",
+ "h=6.626E-34; #in J-s (Planck'c constant)\n",
+ "n1=1; n2=2; n3=3; #ground and first two excited states\n",
+ "e=1.6*1E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "#Since E_n=n^2*h^2/(8*m*a^2) (Energy corresponding to nth quantum state)\n",
+ "E1=n1**2*h**2/(8*m*a**2); #calculation of energy corresponding to the Ground state\n",
+ "E11=E1/e; #changing unit from J to eV\n",
+ "E2=n2**2*h**2/(8*m*a**2); #calculation of energy corresponding to the 1st excited state\n",
+ "E21=E2/e; #changing unit from J to eV\n",
+ "E3=n3**2*h**2/(8*m*a**2); #calculation of energy corresponding to the 2nd excited state\n",
+ "E31=E3/e; #changing unit from J to eV\n",
+ "\n",
+ "#result\n",
+ "print\"The energy corresponding to the ground state is E1=\",E1,\"J\";\n",
+ "print\"\\t\\t\\t\\t\\t =\",E11,\"eV\";\n",
+ "print\"The energy corresponding to the 1st excited state is E2=\",E2,\"\";\n",
+ "print\"\\t\\t\\t\\t\\t =\",E21,\"eV\";\n",
+ "print\"The energy corresponding to the 2nd excited state is E3=\",E3,\"J\";\n",
+ "print\"\\t\\t\\t\\t\\t =\",E31,\"eV\";\n",
+ "print \" There is slight variation in the answer due to round off error\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The energy corresponding to the ground state is E1= 6.0307521978e-18 J\n",
+ "\t\t\t\t\t = 37.6922012363 eV\n",
+ "The energy corresponding to the 1st excited state is E2= 2.41230087912e-17 \n",
+ "\t\t\t\t\t = 150.768804945 eV\n",
+ "The energy corresponding to the 2nd excited state is E3= 5.42767697802e-17 J\n",
+ "\t\t\t\t\t = 339.229811126 eV\n",
+ " There is slight variation in the answer due to round off error\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.25 , Page no:113"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "dx=1E-8; #in m (length of box or uncertainty in the position)\n",
+ "h=6.626E-34; #in J-s (Planck'c constant)\n",
+ "m=9.1E-31; #in Kg (mass of electron)\n",
+ "\n",
+ "#calculate\n",
+ "#From uncertainty principle dx*dp=h and dp=m*dv\n",
+ "#therefore we have\n",
+ "dv=h/(m*dx); #calculation of minimum uncertainty in the velocity\n",
+ "dv1=dv*1E-3; #changing unit from m/s to Km/s\n",
+ "\n",
+ "#result\n",
+ "print\"The minimum uncertainty in the velocity of electron i dv=\",dv,\"m/s\";\n",
+ "print\"\\t\\t\\t\\t\\t\\t =\",round(dv1,2),\"Km/s\";\n",
+ "print \"Note: There is slight variation in the answer due to round off error\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The minimum uncertainty in the velocity of electron i dv= 72813.1868132 m/s\n",
+ "\t\t\t\t\t\t = 72.81 Km/s\n",
+ "Note: There is slight variation in the answer due to round off error\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.26 , Page no:113"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "a=4E-10; #in m (length of the box)\n",
+ "m=9.1E-31; #in Kg (mass of electron)\n",
+ "h=6.626E-34; #in J-s (Planck'c constant)\n",
+ "n1=1; #ground state\n",
+ "e=1.6*1E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "#Since E_n=n^2*h^2/(8*m*a^2) (Energy corresponding to nth quantum state)\n",
+ "E1=n1**2*h**2/(8*m*a**2); #calculation of energy corresponding to the ground state\n",
+ "E11=E1/e; #changing unit from J to eV\n",
+ "\n",
+ "#result\n",
+ "print\"The minimum energy of electron is E1=\",E1,\"J (Note: The answer in the book is wrong due to printing error)\";\n",
+ "print\"\\t\\t\\t\\t =\",round(E11,3),\"eV\";\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The minimum energy of electron is E1= 3.76922012363e-19 J (Note: The answer in the book is wrong due to printing error)\n",
+ "\t\t\t\t = 2.356 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Applied_Physics/Chapter6.ipynb b/Applied_Physics/Chapter6.ipynb new file mode 100755 index 00000000..ff063bb6 --- /dev/null +++ b/Applied_Physics/Chapter6.ipynb @@ -0,0 +1,561 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:0c3d2d354976eaba7e5a0f2e0cf4b1fd72cd55b3826ca7e7cd6acde8afff2cdf"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 6:Electron Theory and Band Theory of Metals"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.1 , Page no:146"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "n=8.5E28; #in 1/m^3 (density of electron)\n",
+ "m_e=9.11E-31; #in Kg (mass of electron)\n",
+ "k=1.38E-23; #in J/K (Boltzmann's constant)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "T=300; #in K (temperature)\n",
+ "p=1.69E-8; #in ohm-m (resistivity)\n",
+ "\n",
+ "#calculate\n",
+ "lambda1=math.sqrt(3*k*m_e*T)/(n*e**2*p); #calculation of mean free path\n",
+ "lambda2=lambda1*1E9; #changing unit from m to nanometer\n",
+ "\n",
+ "#result\n",
+ "print\"The mean free path of electron is =\",lambda1,\"m\";\n",
+ "print\"\\t\\t\\t\\t =\",round(lambda2,2),\"nm\";\n",
+ "#Note: answer in the book is wrong due to printing mistake"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The mean free path of electron is = 2.89250681437e-09 m\n",
+ "\t\t\t\t = 2.89 nm\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.2 , Page no:146"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "k=1.38E-23; #in J/K (Boltzmann's constant)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "P_E=1; #in percentage (probability that a state with an energy 0.5 eV above Fermi energy will be occupied)\n",
+ "E=0.5; #in eV (energy above Fermi level)\n",
+ "\n",
+ "#calculate\n",
+ "P_E=1/100; #changing percentage into ratio\n",
+ "E=E*e; #changing unit from eV to J\n",
+ "#P_E=1/(1+exp((E-E_F)/k*T))\n",
+ "#Rearranging this equation, we get\n",
+ "#T=(E-E_F)/k*log((1/P_E)-1)\n",
+ "#Since E-E_F has been denoted by E therefore\n",
+ "T=E/(k*math.log((1/P_E)-1));\n",
+ "\n",
+ "#result\n",
+ "print\"The temperature is T=\",round(T),\"K\";\n",
+ "# Note: There is slight variation in the answer due to logarithm function"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The temperature is T= 1262.0 K\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.3 , Page no:147"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "n=5.8E28; #in 1/m^3 (density of electron)\n",
+ "m=9.1E-31; #in Kg (mass of electron)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "p=1.54E-8; #in ohm-m (resistivity)\n",
+ "\n",
+ "#calculate\n",
+ "t=m/(n*e**2*p); #calculation of relaxation time\n",
+ "\n",
+ "#result\n",
+ "print\"The relaxation time of conduction electrons is =\",t,\"sec\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The relaxation time of conduction electrons is = 3.97972178683e-14 sec\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.4 , Page no:147"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "n=8.5E28; #in 1/m^3 (density of electron)\n",
+ "m=9.1E-31; #in Kg (mass of electron)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "sigma=6E7; #in 1/ohm-m (conductivity)\n",
+ "E_F=7; #in E=eV (Fermi energy of Copper)\n",
+ "\n",
+ "#calculate\n",
+ "E_F=E_F*e; #changing unit from eV to J\n",
+ "v_F=math.sqrt(2*E_F/m); #calculation of velocity of electrons\n",
+ "#Since sigma=n*e^2*lambda/(2*m*v_F)\n",
+ "#Therefore we have\n",
+ "lambda1=2*m*v_F*sigma/(n*e**2); #calculation of mean free path\n",
+ "lambda2=lambda1*1E10; #changing unit from m to Angstrom\n",
+ "\n",
+ "#result\n",
+ "print\"The velocity of the electrons is v_F=\",v_F,\"m/s (roundoff error)\";\n",
+ "print\"The mean free path traveled by the electrons is=\",round(lambda2),\"Angstrom\";\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The velocity of the electrons is v_F= 1568929.08111 m/s (roundoff error)\n",
+ "The mean free path traveled by the electrons is= 787.0 Angstrom\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.5 , Page no:147"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "n=6.5E28; #in 1/m^3 (density of electron)\n",
+ "m=9.1E-31; #in Kg (mass of electron)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "p=1.43E-8; #in ohm-m (resistivity)\n",
+ "\n",
+ "#calculate\n",
+ "t=m/(n*e**2*p); #calculation of relaxation time\n",
+ "\n",
+ "#result\n",
+ "print\"The relaxation time of conduction electrons is =\",t,\"sec\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The relaxation time of conduction electrons is = 3.8243006993e-14 sec\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6 , Page no:148"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "T=30; #in Celcius (temperature)\n",
+ "k=1.38E-23; #in J/K (Boltzmann's constant)\n",
+ "m_p=1.67E-27; #in Kg (mass of proton)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "T=T+273; #changing temperature from Celcius to Kelvin\n",
+ "KE=(3/2)*k*T; #calculation of average kinetic energy\n",
+ "KE1=KE/e; #changing unit from J to eV\n",
+ "m=1.008*2*m_p; #calculating mass of hydrogen gas molecule\n",
+ "c=math.sqrt(3*k*T/m); #calculation of velocity\n",
+ "\n",
+ "#result\n",
+ "print\"The average kinetic energy of gas ,molecules is KE=\",KE,\"J\";\n",
+ "print\"\\t\\t\\t\\t\\t\\t =\",KE1,\"eV (Answer is wrong in textbook)\";\n",
+ "print\"The velocity of molecules is c=\",round(c,2),\"m/s\";\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The average kinetic energy of gas ,molecules is KE= 6.2721e-21 J\n",
+ "\t\t\t\t\t\t = 0.039200625 eV (Answer is wrong in textbook)\n",
+ "The velocity of molecules is c= 1930.27 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.7 , Page no:148"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "E=10; #in eV (kinetic energy for each electron and proton)\n",
+ "m_e=9.1E-31; #in Kg (mass of electron)\n",
+ "m_p=1.67E-27; #in Kg (mass of proton)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "E=E*e; #changing unit from eV to J\n",
+ "#since E=m*v^2/2\n",
+ "#therefore v=sqrt(2E/m)\n",
+ "v_e=math.sqrt(2*E/m_e); #calculation of kinetic energy of electron\n",
+ "v_p=math.sqrt(2*E/m_p); #calculation of kinetic energy of proton\n",
+ "\n",
+ "#result\n",
+ "print\"The kinetic energy of electron is v_e=\",v_e,\"m/s\";\n",
+ "print\"The kinetic energy of proton is v_p=\",v_p,\"m/s\";\n",
+ "print \"Note: Answer is wrong in textbook\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The kinetic energy of electron is v_e= 1875228.92375 m/s\n",
+ "The kinetic energy of proton is v_p= 43774.0524132 m/s\n",
+ "Note: Answer is wrong in textbook\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.8 , Page no:149"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "I=100; #in A (current in the wire)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "A=10; #in mm^2 (cross-sectional area)\n",
+ "n=8.5E28; #in 1/m^3 (density of electron)\n",
+ "\n",
+ "#calculate\n",
+ "A=A*1E-6; #changing unit from mm^2 to m^2\n",
+ "v_d=I/(n*A*e);\n",
+ "\n",
+ "#result\n",
+ "print\"The drift velocity of free electrons is v_d=\",v_d,\"m/s\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The drift velocity of free electrons is v_d= 0.000735294117647 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.9 , Page no:149"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "I=4;#in A (current in the conductor)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "A=1E-6; #in m^2 (cross-sectional area)\n",
+ "N_A=6.02E23; #in atoms/gram-atom (Avogadro's number)\n",
+ "p=8.9; #in g/cm^3 (density)\n",
+ "M=63.6; #atomic mass of copper\n",
+ "\n",
+ "#calculate\n",
+ "n=N_A*p/M; #Calculation of density of electrons in g/cm^3\n",
+ "n1=n*1E6; #changing unit from g/cm^3 to g/m^3\n",
+ "v_d=I/(n1*A*e);\n",
+ "\n",
+ "#result\n",
+ "print\"The density of copper atoms is n=\",n,\"atoms/m^3\";\n",
+ "print\"\\t\\t\\t =\",n1,\"atoms/m^3\";\n",
+ "print\"The average drift velocity of free electrons is v_d=\",v_d,\"m/s\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The density of copper atoms is n= 8.42421383648e+22 atoms/m^3\n",
+ "\t\t\t = 8.42421383648e+28 atoms/m^3\n",
+ "The average drift velocity of free electrons is v_d= 0.000296763596999 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.10 , Page no:149"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "n=9E28; #in 1/m^3 (density of valence electrons)\n",
+ "sigma=6E7; #in mho/m (conductivity of copper)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "#Since sigma=n*e*mu therefore\n",
+ "mu=sigma/(n*e); #calculation of mobility of electron\n",
+ "\n",
+ "#result\n",
+ "print\"The mobility of electrons is =\",mu,\"m^2/V-s\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The mobility of electrons is = 0.00416666666667 m^2/V-s\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.11 , Page no:150"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "E_F=5.51; #in eV (Fermi energy in Silver)\n",
+ "k=1.38E-23; #in J/K (Boltzmann's constant)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "#part-(a)\n",
+ "Eo=(3/5)*E_F; #calculation of average energy of free electron at 0K\n",
+ "#part-(b)\n",
+ "Eo1=Eo*e; #changing unit from eV to J\n",
+ "#Since for a classical particle E=(3/2)*k*T\n",
+ "#therefroe we have\n",
+ "T=(2/3)*Eo1/k; #calculation of temperature for a classical particle (an ideal gas)\n",
+ "\n",
+ "#result\n",
+ "print\"The average energy of free electron at 0K is Eo=\",Eo,\"eV\";\n",
+ "print\"The temperature at which a classical particle have this much energy is T=\",T,\"K\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The average energy of free electron at 0K is Eo= 3.306 eV\n",
+ "The temperature at which a classical particle have this much energy is T= 25553.6231884 K\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.12 , Page no:150"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "E_F_L=4.7; #in eV (Fermi energy in Lithium)\n",
+ "E_F_M=2.35; #in eV (Fermi energy in a metal)\n",
+ "n_L=4.6E28; #in 1/m^3 (density of electron in Lithium)\n",
+ "\n",
+ "#calculate\n",
+ "#Since n=((2*m/h)^3/2)*E_F^(3/2)*(8*pi/3) and all things except E_F are constant\n",
+ "# Therefore we have n=C*E_F^(3/2) where C is proportionality constant\n",
+ "#n1/n2=(E_F_1/E_F_2)^(3/2)\n",
+ "#Therefore we have\n",
+ "n_M=n_L*(E_F_M/E_F_L); #calculation of electron density for a metal\n",
+ "\n",
+ "#result\n",
+ "print\"The lectron density for a metal is =\",n_M,\"1/m^3\";\n",
+ "print \"Note: Answer in the book is wrong \""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The lectron density for a metal is = 2.3e+28 1/m^3\n",
+ "Note: Answer in the book is wrong \n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Applied_Physics/Chapter7.ipynb b/Applied_Physics/Chapter7.ipynb new file mode 100755 index 00000000..7280033d --- /dev/null +++ b/Applied_Physics/Chapter7.ipynb @@ -0,0 +1,488 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:5117f9d565431f3f25738652845b14f8a6fa036b5c0992b0b2e5d69631937306"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 7:Dielectric Properties"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.1 , Page no:187"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "A=100; #in cm^2 (cross-sectional area)\n",
+ "d=1; #in cm (seperation between plates)\n",
+ "Eo=8.85E-12; #in F/m (absolute permittivity)\n",
+ "V=100; #in V (potential difference)\n",
+ "\n",
+ "#calculate\n",
+ "A1=A*1E-4; #changing unit from cm^2 to m^2\n",
+ "d1=d*1E-2; #changing unit from cm to m\n",
+ "C=Eo*A1/d1; #calculation of capacitance\n",
+ "Q=C*V; #calculation of charge\n",
+ "C1=C*1E12; #changing unit of capacitance from F to pF\n",
+ "\n",
+ "#result\n",
+ "print\"The capacitance of capacitor is C=\",C,\"C\";\n",
+ "print\"\\t\\t\\t\\t =\",C1,\"C\";\n",
+ "print\"The charge on the plates is Q=\",Q,\"C\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The capacitance of capacitor is C= 8.85e-12 C\n",
+ "\t\t\t\t = 8.85 C\n",
+ "The charge on the plates is Q= 8.85e-10 C\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.2 , Page no:187"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "A=650; #in mm^2 (cross-sectional area)\n",
+ "d=4; #in mm (seperation between plates)\n",
+ "Eo=8.85E-12; #in F/m (absolute permittivity)\n",
+ "Er=3.5; #di-electric constant of the material\n",
+ "Q=2E-10; #in C (charge on plates)\n",
+ "\n",
+ "#calculate\n",
+ "A1=A*1E-6; #changing unit from mm^2 to m^2\n",
+ "d1=d*1E-3; #changing unit from mm to m\n",
+ "C=Er*Eo*A1/d1; #calculation of capacitance\n",
+ "V=Q/C; #calculation of charge\n",
+ "C1=C*1E12; #changing unit of capacitance from F to pF\n",
+ "\n",
+ "#result\n",
+ "print\"The capacitance of capacitor is C=\",C,\"C\",\n",
+ "print\"\\n\\t\\t\\t =\",C1,\"pF\";\n",
+ "print\"The resultant voltage across the capacitor is V=\",round(V,3),\"V\";\n",
+ "print \"NOTE: The answer in the textbook is wrong\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The capacitance of capacitor is C= 5.0334375e-12 C \n",
+ "\t\t\t = 5.0334375 pF\n",
+ "The resultant voltage across the capacitor is V= 39.734 V\n",
+ "NOTE: The answer in the textbook is wrong\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.3 , Page no:188"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "N=2.7E25; #in 1/m^3 (density of atoms)\n",
+ "E=1E6; #in V/m (electric field)\n",
+ "Z=2; #atomic number of Helium \n",
+ "Eo=8.85E-12; #in F/m (absolute permittivity)\n",
+ "Er=1.0000684; #(dielectric constant of the material)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "pi=3.14; #value of pi used in the solution\n",
+ "\n",
+ "#calculate\n",
+ "#since alpha=Eo*(Er-1)/N=4*pi*Eo*r_0^3 \n",
+ "#Therefore we have r_0^3=(Er-1)/(4*pi*N)\n",
+ "r_0=((Er-1)/(4*pi*N))**(1/3); #calculation of radius of electron cloud\n",
+ "x=4*pi*Eo*E*r_0/(Z*e); #calculation of dispalcement\n",
+ "\n",
+ "#result\n",
+ "print\"The radius of electron cloud is r_0=\",r_0,\"m\";\n",
+ "print\"The displacement is x=\",x,\"m\";\n",
+ "print \"NOTE: The answer in the textbook is wrong\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The radius of electron cloud is r_0= 5.86454377988e-11 m\n",
+ "The displacement is x= 20371.2258874 m\n",
+ "NOTE: The answer in the textbook is wrong\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.4 , Page no:188"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "K=1.000134; #di-elecrtic constant of the neon gas at NTP\n",
+ "E=90000; #in V/m (electric field)\n",
+ "Eo=8.85E-12; #in C/N-m^2 (absolute premittivity)\n",
+ "N_A=6.023E26; #in atoms/Kg-mole (Avogadro's number)\n",
+ "V=22.4; #in m^3 (volume of gas at NTP\n",
+ "\n",
+ "#calculate\n",
+ "n=N_A/V; #calculaton of density of atoms\n",
+ "#Since P=n*p=(k-1)*Eo*E\n",
+ "#therefore we have\n",
+ "p=(K-1)*Eo*E/n; #calculation of dipole moment induced\n",
+ "alpha=p/E; #calculation of atomic polarisability\n",
+ "\n",
+ "#result\n",
+ "print\"The dipole moment induced in each atom is p=\",p,\"C-m\";\n",
+ "print\"The atomic polarisability of neon is =\",alpha,\"c-m^2/V\";\n",
+ "print \"NOTE: The answer in the textbook is wrong\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The dipole moment induced in each atom is p= 3.96940793625e-36 C-m\n",
+ "The atomic polarisability of neon is = 4.4104532625e-41 c-m^2/V\n",
+ "NOTE: The answer in the textbook is wrong\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.5 , Page no:189"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "Er=3.75; #di-elecrtic constant of sulphur at 27 degree Celcius\n",
+ "gama=1/3; #internal field constant\n",
+ "p=2050; #in Kg/m^3 (density)\n",
+ "M_A=32; #in amu (atomic weight of sulphur)\n",
+ "Eo=8.85E-12; #in F/m (absolute permittivity)\n",
+ "N=6.022E23; #Avogadro's number\n",
+ "\n",
+ "#calculate\n",
+ "#Since ((Er-1)/(Er+2))*(M_A/p)=(N/(3*Eo))*alpha_e\n",
+ "#therefore we have\n",
+ "alpha_e=((Er-1)/(Er+2))*(M_A/p)*(3*Eo/N); #calculation of electronic polarisability of sulphur\n",
+ "\n",
+ "#result\n",
+ "print\"The electronic polarisability of sulphur is =\",alpha_e,\"Fm^2 (roundoff error)\";\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The electronic polarisability of sulphur is = 3.29143088985e-37 Fm^2 (roundoff error)\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.6 , Page no:189"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "Er=1.0000684; #di-elecrtic constant of Helium gas at NTP\n",
+ "Eo=8.85E-12; #in F/m (absolute permittivity)\n",
+ "N=2.7E25; #number of atomsper unit volume\n",
+ "\n",
+ "#calculate\n",
+ "#Since Er-1=(N/Eo)*alpha_e\n",
+ "#therefore we have\n",
+ "alpha_e=Eo*(Er-1)/N; #calculation of electronic polarisability of Helium\n",
+ "\n",
+ "#result\n",
+ "print\"The electronic polarisability of Helium gas is =\",alpha_e,\"Fm^2 (roundoff error)\";\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The electronic polarisability of Helium gas is = 2.242e-41 Fm^2 (roundoff error)\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.7 , Page no:190"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "N=3E28; #in atoms/m^3 (density of atoms)\n",
+ "alpha_e=1E-40; #in F-m^2 (electronic polarisability)\n",
+ "Eo=8.85E-12; #in F/m (absolute permittivity)\n",
+ "\n",
+ "#calculate\n",
+ "#Since (Er-1)/(Er+2)=N*alpha_e/(3*Eo)\n",
+ "#therefore we have\n",
+ "Er=(2*(N*alpha_e/(3*Eo))+1)/(1-(N*alpha_e/(3*Eo)));\n",
+ "#calculation of dielectric constant of the material\n",
+ "\n",
+ "#result\n",
+ "print\"The dielectric constant of the material is Er=\",Er,\"F/m\",\n",
+ "print \"NOTE: The answer in the book is wrong \""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The dielectric constant of the material is Er= 1.3821656051 F/m NOTE: The answer in the book is wrong \n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.8 , Page no:190"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "Er=4; #relative permittivity of sulphur\n",
+ "Eo=8.85E-12; #in F/m (absolute permittivity)\n",
+ "NA=2.08E3; #in Kg/m^3 (density of atoms in sulphur)\n",
+ "\n",
+ "#calculate\n",
+ "#Since ((Er-1)/(Er+2))*(M_A/p)=(N/(3*Eo))*alpha_e\n",
+ "#therefore we have\n",
+ "alpha_e=((Er-1)/(Er+2))*(3*Eo/NA); #calculation of electronic polarisability of sulphur\n",
+ "\n",
+ "#result\n",
+ "print\"The electronic polarisability of sulphur is =\",alpha_e,\"Fm^2\";\n",
+ "print \"NOTE: The answer in the book is wrong and they used the wrong formula \" \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The electronic polarisability of sulphur is = 6.38221153846e-15 Fm^2\n",
+ "NOTE: The answer in the book is wrong and they used the wrong formula \n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.9 , Page no:190"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "alpha1=2.5E-39; #in C^2-m/N (dielectric constant at 300K)\n",
+ "alpha2=2.0E-39; #in C^2-m/N (dielectric constant at 400K)\n",
+ "T1=300; #in K(first temperature)\n",
+ "T2=400; #in K(second temperature)\n",
+ "\n",
+ "#calculate\n",
+ "#since alpha=alpha_d+alpha0 and alpha0=Beta/T\n",
+ "#therefore alpha=alpha_d+(Beta/T)\n",
+ "#since alpha1=alpha_d+(Beta/T1) and alpha2=alpha_d+(Beta/T2)\n",
+ "#therefore alpha1-apha2=Beta*((1/T1)-(1/T2))\n",
+ "#or Beta= (alpha1-apha2)/ ((1/T1)-(1/T2))\n",
+ "Beta= (alpha1-alpha2)/ ((1/T1)-(1/T2)); #calculation of Beta\n",
+ "alpha_d=alpha1-(Beta/T1); #calculation of polarisability due to defromation\n",
+ "alpha0_1=Beta/T1; #calculation of polarisability due to permanent dipole moment at 300K\n",
+ "alpha0_2=Beta/T2; #calculation of polarisability due to permanent dipole moment at 400K\n",
+ "\n",
+ "#result\n",
+ "print\"The polarisability due to permanent dipole moment at 300K is =\",alpha0_1,\"C^2-m/N\";\n",
+ "print\"The polarisability due to permanent dipole moment at 400K is =\",alpha0_2,\"C^2-m/N\";\n",
+ "print\"The polarisability due to deformation of the molecules is =\",alpha_d,\"C^2-m/N\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The polarisability due to permanent dipole moment at 300K is = 2e-39 C^2-m/N\n",
+ "The polarisability due to permanent dipole moment at 400K is = 1.5e-39 C^2-m/N\n",
+ "The polarisability due to deformation of the molecules is = 5e-40 C^2-m/N\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.10 , Page no:191"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "n=1.5; #refractive index\n",
+ "Er=5.6; #dielectric constant\n",
+ "\n",
+ "#calculate\n",
+ "#since (Er-1)/(Er+2)=N*(alpha_e+alpha_i)/(3*E0) Clausius-Mossotti equation\n",
+ "#and (n^2-1)/(n^2+2)=N*alpha_e/(3*E0) \n",
+ "#from above two equations, we get ((n^2-1)/(n^2+2))*((Er+2)/(Er-1))=alpha_e/(alpha_e+alpha_i)\n",
+ "#or alpha_i/ (alpha_e+alpha_i)= 1-((n^2-1)/(n^2+2))*((Er+2)/(Er-1))= (say P)\n",
+ "#where P is fractional ionisational polarisability\n",
+ "P=1-((n**2-1)/(n**2+2))*((Er+2)/(Er-1)); #calculation of fractional ionisational polarisability\n",
+ "P1=P*100; #calculation of percentage of ionisational polarisability\n",
+ "\n",
+ "#result\n",
+ "print\"The percentage of ionisational polarisability is =\",P1,\"percent\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The percentage of ionisational polarisability is = 51.4066496164 percent\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Applied_Physics/Chapter8.ipynb b/Applied_Physics/Chapter8.ipynb new file mode 100755 index 00000000..353fc099 --- /dev/null +++ b/Applied_Physics/Chapter8.ipynb @@ -0,0 +1,323 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:aecad4e644a378cab0352aa29e9045a53c52646739e4f4bd6cdff80f1d0741e6"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 8:Magnetic Properties"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.1 , Page no:236"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "X=-0.5E-5; #magnetic susceptibility of silicon\n",
+ "H=0.9E4; #in A/m (magnetic field intensity)\n",
+ "mu0=4*3.14*1E-7; #in H/m (absolute permeability)\n",
+ "\n",
+ "#calculate\n",
+ "I=X*H; #calculation of intensity of magnetism\n",
+ "B=mu0*H*(1+X); #calculation of magnetic flux density\n",
+ "\n",
+ "#result\n",
+ "print\"The intensity of magnetism is I=\",I,\"A/m\";\n",
+ "print\"The magnetic flux density is B=\",round(B,3),\"Wb/m^2\";\n",
+ "print \"NOTE: The answer in the textbook is wrong\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The intensity of magnetism is I= -0.045 A/m\n",
+ "The magnetic flux density is B= 0.011 Wb/m^2\n",
+ "NOTE: The answer in the textbook is wrong\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.2 , Page no:236"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "r=0.052; #in nm (radius of orbit)\n",
+ "B=1; #in Wb/m^2 (magnetic field of induction)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "m=9.1E-31; #in Kg (mass of electron)\n",
+ "\n",
+ "#calculate\n",
+ "r=0.052*1E-9; #changing unit from nm to m\n",
+ "d_mu=(e**2*r**2*B)/(4*m); #calculation of change in magnetic moment\n",
+ "\n",
+ "#result\n",
+ "print\"The change in magnetic moment is =\",d_mu,\"Am^2\";\n",
+ "print \"NOTE: The answer in the textbook is wrong\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The change in magnetic moment is = 1.90171428571e-29 Am^2\n",
+ "NOTE: The answer in the textbook is wrong\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.3 , Page no:236"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "H=220; #in A/m (magnetic field intensity)\n",
+ "I=3300; #in A/m (intensity of magnetisation)\n",
+ "\n",
+ "#calculate\n",
+ "mu_r=1+(I/H); #calculation of relative permeability\n",
+ "\n",
+ "#result\n",
+ "print\"The relative permeability of a ferromagentic material is =\",mu_r;"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The relative permeability of a ferromagentic material is = 16.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.4 , Page no:236"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "I=3000; #in A/m (intensity of magnetisation)\n",
+ "B=0.005; #in Wb/m^2 (magnetic flus intensity)\n",
+ "pi=3.14; #value of pi used in the solution\n",
+ "mu0=4*pi*1E-7; #in H/m (absolute permeability)\n",
+ "\n",
+ "#calculate\n",
+ "H=(B/mu0)-I; #calculation of magnetic force\n",
+ "mu_r=(I/H)+1; #calculation of relative permeability\n",
+ "\n",
+ "#result\n",
+ "print\"The magnetic force is H=\",H;\n",
+ "print\"The relative permeability is =\",mu_r;"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The magnetic force is H= 980.891719745\n",
+ "The relative permeability is = 4.05844155844\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.5 , Page no:237"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "H=4E3; #in A/m (magnetic field intensity)\n",
+ "N=60; #number of turns\n",
+ "l=12; #in cm (length of solenoid)\n",
+ "\n",
+ "#calculate\n",
+ "n=N/(l*1E-2); #calculation of number of turns per unit metre\n",
+ "#Snice H=n*i;\n",
+ "i=H/n; #calculation of current through the solenoid\n",
+ "\n",
+ "#result\n",
+ "print\"The current through the solenoid is i=\",i,\"A\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The current through the solenoid is i= 8.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.6 , Page no:237"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "l=30; #in cm (length of solenoid)\n",
+ "A=1; #in cm^2 (cross-sectional area)\n",
+ "N=300; #number of turns\n",
+ "i=0.032; #in A (current through the winding)\n",
+ "phi_B=2E-6; #in Wb (magnetic flux)\n",
+ "pi=3.14; #value of pi used in the solution\n",
+ "mu0=4*pi*1E-7; #in H/m (absolute permeability)\n",
+ "\n",
+ "#calculate\n",
+ "l=l*1E-2; #changing unit from cm to m\n",
+ "A=A*1E-4; #changing unit from cm^2 to m^2\n",
+ "B=phi_B/A; #calculation of flux density\n",
+ "H=N*i/l; #calculation of magnetic intensity\n",
+ "mu=B/H; #calcluation of absolute permeability of iron\n",
+ "mu_r=mu/mu0; #calcluation of relative permeability of iron\n",
+ "\n",
+ "#result\n",
+ "print\"The flux density is B=\",B,\"Wb/m^2\";\n",
+ "print\"The magnetic intensity is H=\",H,\"A-turns/m\";\n",
+ "print\"The relative permeability of iron is =\",round(mu_r),\" (roundoff error)\";\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The flux density is B= 0.02 Wb/m^2\n",
+ "The magnetic intensity is H= 32.0 A-turns/m\n",
+ "The relative permeability of iron is = 498.0 (roundoff error)\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.7 , Page no:238"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "A=100; #in m^2 (area of Hysteresis loop)\n",
+ "B=0.01; #in Wb/m^2 (unit space along vertical axis or magnetic flux density)\n",
+ "H=40; #in A/m (unit space along horizontal axis or magnetic fild ntensity)\n",
+ "\n",
+ "#calculate\n",
+ "H_L=A*B*H; #calculation of magnetic intensity\n",
+ "\n",
+ "#result\n",
+ "print\"The Hystersis loss per cycle is =\",round(H_L),\"J/m^2\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Hystersis loss per cycle is = 40.0 J/m^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Applied_Physics/Chapter9.ipynb b/Applied_Physics/Chapter9.ipynb new file mode 100755 index 00000000..f3c26825 --- /dev/null +++ b/Applied_Physics/Chapter9.ipynb @@ -0,0 +1,782 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:eb4f284d8d12c103b674cc454947abe9c7a1a76e2d11836f2841ba55d4adeef7"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 9:Semiconductors"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.2 , Page no:272"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "Eg=0.67; #in eV (Energy band gap)\n",
+ "k=1.38E-23; #in J/K (Boltzmann\u2019s constant)\n",
+ "T1=298; #in K (room temperature)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "K=10; #ratio of number of electrons at different temperature\n",
+ "\n",
+ "#calculate\n",
+ "Eg=Eg*e; #changing unit from eV to Joule\n",
+ "#since ne=Ke*exp(-Eg/(2*k*T))\n",
+ "#and ne/ne1=exp(-Eg/(2*k*T))/exp(-Eg/(2*k*T1)) and ne/ne1=K=10\n",
+ "#therefore we have 10=exp(-Eg/(2*k*T))/exp(-Eg/(2*k*T1))\n",
+ "#re-arranging the equation for T, we get T2=1/((1/T1)-((2*k*log(10))/Eg))\n",
+ "T=1/((1/T1)-((2*k*math.log(10))/Eg)); #calculation of the temperature\n",
+ "\n",
+ "#result\n",
+ "print\"The temperature at which number of electrons in the conduction band of a semiconductor increases by a factor of 10 is T=\",round(T),\"K (roundoff error)\";\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The temperature at which number of electrons in the conduction band of a semiconductor increases by a factor of 10 is T= 362.0 K (roundoff error)\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.3 , Page no:272"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "ni=2.5E13; #in /cm^3 (intrinsic carrier density)\n",
+ "ue=3900; #in cm^2/(V-s) (electron mobilities)\n",
+ "uh=1900; #in cm^2/(V-s) (hole mobilities)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "l=1; #in cm (lenght of the box)\n",
+ "b=1;h=1; #in mm (dimensions of germanium rod )\n",
+ "\n",
+ "#calculate\n",
+ "ni=ni*1E6; #changing unit from 1/cm^3 to 1/m^3\n",
+ "ue=ue*1E-4; #changing unit from cm^2 to m^2\n",
+ "uh=uh*1E-4; #changing unit from cm^2 to m^2\n",
+ "sigma=ni*e*(ue+uh); #calculation of conductivity\n",
+ "rho=1/sigma; #calculation of resistivity\n",
+ "l=l*1E-2; #changing unit from mm to m for length\n",
+ "A=(b*1E-3)*(h*1E-3); #changing unit from mm to m for width and height and calculation of cross-sectional area\n",
+ "R=rho*l/A; #calculation of resistance\n",
+ "\n",
+ "#result\n",
+ "print\"The resistance of intrinsic germanium is R=\",R,\"ohm\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The resistance of intrinsic germanium is R= 4310.34482759 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.4 , Page no:273"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "ne=2.5E19; #in /m^3 (electron density)\n",
+ "nh=2.5E19; #in /m^3 (hole density)\n",
+ "ue=0.36; #in m^2/(V-s) (electron mobilities)\n",
+ "uh=0.17; #in m^2/(V-s) (hole mobilities)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "#since ne=nh=ni, therefore we have \n",
+ "ni=nh;\n",
+ "sigma=ni*e*(ue+uh); #calculation of conductivity\n",
+ "rho=1/sigma; #calculation of resistivity\n",
+ "\n",
+ "#result\n",
+ "print\"The conductivity of germanium is =\",round(sigma,2),\"/ohm-m\";\n",
+ "print\"The resistivity of germanium is =\",round(rho,2),\"ohm-m\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The conductivity of germanium is = 2.12 /ohm-m\n",
+ "The resistivity of germanium is = 0.47 ohm-m\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.5 , Page no:273"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "ni=1.5E16; #in /m^3 (intrinsic carrier density)\n",
+ "ue=0.135; #in m^2/(V-s) (electron mobilities)\n",
+ "uh=0.048; #in m^2/(V-s) (hole mobilities)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "ND=1E23; #in atom/m^3 (doping concentration)\n",
+ "\n",
+ "#calculate\n",
+ "sigma_i=ni*e*(ue+uh); #calculation of intrinsic conductivity\n",
+ "sigma=ND*ue*e; #calculation of conductivity after doping\n",
+ "rho=ni**2/ND; #calculation of equilibrium hole concentration\n",
+ "\n",
+ "#result\n",
+ "print\"The intrinsic conductivity for silicon is =\",sigma_i,\"S\";\n",
+ "print\"The conductivity after doping with phosphorus atoms is =\",sigma,\"S\";\n",
+ "print\"The equilibrium hole concentration is =\",rho,\"/m^3\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The intrinsic conductivity for silicon is = 0.0004392 S\n",
+ "The conductivity after doping with phosphorus atoms is = 2160.0 S\n",
+ "The equilibrium hole concentration is = 2250000000.0 /m^3\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.6 , Page no:274"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "ni=1.5E16; #in /m^3 (intrinsic carrier density)\n",
+ "ue=0.13; #in m^2/(V-s) (electron mobilities)\n",
+ "uh=0.05; #in m^2/(V-s) (hole mobilities)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "ne=5E20; #in /m^3 (concentration of donor type impurity)\n",
+ "nh=5E20; #in /m^3 (concentration of acceptor type impurity)\n",
+ "\n",
+ "#calculate\n",
+ "#part-i\n",
+ "sigma=ni*e*(ue+uh); #calculation of intrinsic conductivity\n",
+ "#part-ii\n",
+ "#since 1 donor atom is in 1E8 Si atoms, hence holes concentration can be neglected\n",
+ "sigma1=ne*e*ue; #calculation of conductivity after doping with donor type impurity\n",
+ "#part-iii\n",
+ "#since 1 acceptor atom is in 1E8 Si atoms, hence electron concentration can be neglected\n",
+ "sigma2=nh*e*uh; #calculation of conductivity after doping with acceptor type impurity\n",
+ "\n",
+ "#result\n",
+ "print\"The intrinsic conductivity for silicon is =\",sigma,\"(ohm-m)^-1\";\n",
+ "print\"The conductivity after doping with donor type impurity is =\",sigma1,\"(ohm-m)^-1\";\n",
+ "print\"The conductivity after doping with acceptor type impurity is =\",sigma2,\"(ohm-m)^-1\";\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The intrinsic conductivity for silicon is = 0.000432 (ohm-m)^-1\n",
+ "The conductivity after doping with donor type impurity is = 10.4 (ohm-m)^-1\n",
+ "The conductivity after doping with acceptor type impurity is = 4.0 (ohm-m)^-1\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.7 , Page no:274"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "ni=1E20; #in /m^3 (intrinsic carrier density)\n",
+ "ND=1E21; #in /m^3 (donor impurity concentration)\n",
+ "\n",
+ "#calculate\n",
+ "nh=ni**2/ND; #calculation of density of hole carriers at room temperature\n",
+ "\n",
+ "#result\n",
+ "print\"The density of hole carriers at room temperature is nh=\",nh,\"/m^3\";\n",
+ "#Note: answer in the book is wrong due to printing mistake"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The density of hole carriers at room temperature is nh= 1e+19 /m^3\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.8 , Page no:275"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "M=72.6; #atomic mass of germanium\n",
+ "P=5400; #in Kg/m^3 (density)\n",
+ "ue=0.4; #in m^2/V-s (mobility of electrons)\n",
+ "uh=0.2; #in m^2/V-s (mobility of holes)\n",
+ "Eg=0.7; #in eV (Band gap)\n",
+ "m=9.1E-31; #in Kg (mass of electron)\n",
+ "k=1.38E-23; #in J/K (Boltzmann\u2019s constant)\n",
+ "T=300; #in K (temperature)\n",
+ "h=6.63E-34; #in J/s (Planck\u2019s constant)\n",
+ "pi=3.14; #value of pi used in the solution\n",
+ "e=1.6E-19; #in C(charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "Eg=Eg*e; #changing unit from eV to J\n",
+ "ni=2*(2*pi*m*k*T/h**2)**(3/2)*math.exp(-Eg/(2*k*T));\n",
+ "sigma=ni*e*(ue+uh);\n",
+ "\n",
+ "#result\n",
+ "print\"The intrinsic carrier density for germanium at 300K is ni=\",ni,\"/m^3\";\n",
+ "print\"The conductivity of germanium is=\",round(sigma,3),\"(ohm-m)^-1\";\n",
+ "print \"NOTE: The answer in the textbook is wrong\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The intrinsic carrier density for germanium at 300K is ni= 3.33408559508e+19 /m^3\n",
+ "The conductivity of germanium is= 3.201 (ohm-m)^-1\n",
+ "NOTE: The answer in the textbook is wrong\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9 , Page no:275"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "rho1=4.5; #in ohm-m (resistivity at 20 degree Celcius)\n",
+ "rho2=2.0; #in ohm-m (resistivity at 32 degree Celcius)\n",
+ "k=1.38E-23; #in J/K (Boltzmann\u2019s constant)\n",
+ "T1=20; T2=32; #in degree Celcius (two temperatures)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "T1=T1+273; #changing unit from degree Celius to K\n",
+ "T2=T2+273; #changing unit from degree Celius to K\n",
+ "#since sigma=e*u*C*T^(3/2)*exp(-Eg/(2*k*T))\n",
+ "#therefore sigma1/sigma2=(T1/T2)^3/2*exp((-Eg/(2*k)*((1/T1)-(1/T2))\n",
+ "#and sigma=1/rho \n",
+ "#therefore we have rho2/rho1=(T1/T2)^3/2*exp((-Eg/(2*k)*((1/T1)-(1/T2))\n",
+ "#re-arranging above equation for Eg, we get Eg=(2*k/((1/T1)-(1/T2)))*((3/2)*log(T1/T2)-log(rho2/rho1))\n",
+ "Eg=(2*k/((1/T1)-(1/T2)))*((3/2)*math.log(T1/T2)-math.log(rho2/rho1));\n",
+ "Eg1=Eg/e;#changing unit from J to eV\n",
+ "\n",
+ "#result\n",
+ "print\"The energy band gap is Eg=\",Eg,\"J\";\n",
+ "print\"\\t\\t\\t =\",round(Eg1,3),\"eV\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The energy band gap is Eg= 1.54302914521e-19 J\n",
+ "\t\t\t = 0.964 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.10 , Page no:276"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "rho=2300; #in ohm-m (resistivity of pure silicon)\n",
+ "ue=0.135; #in m^2/V-s (mobility of electron)\n",
+ "uh=0.048; #in m^2/V-s (mobility of electron)\n",
+ "Nd=1E19; #in /m^3 (doping concentration)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "#since sigma=ni*e*(ue+uh) and sigma=1/rho\n",
+ "#therefore ni=1/(rho*e*(ue+uh))\n",
+ "ni=1/(rho*e*(ue+uh)); #calculation of intrinsic concentration\n",
+ "ne=Nd; #calculation of electron concentration\n",
+ "nh=ni**2/Nd; #calculation of hole concentration\n",
+ "sigma=ne*ue*e+nh*uh*e; #calculation of conductivity\n",
+ "rho=1/sigma; #calculation of resistivity\n",
+ "\n",
+ "#result\n",
+ "print\"The electron concentration is ne=\",ne,\"/m^3\";\n",
+ "print\"The hole concentration is nh=\",nh,\"/m^3\";\n",
+ "print\"The resistivity of the specimen is =\",round(rho,3),\"ohm-m\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The electron concentration is ne= 1e+19 /m^3\n",
+ "The hole concentration is nh= 2.20496745228e+13 /m^3\n",
+ "The resistivity of the specimen is = 4.63 ohm-m\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.11 , Page no:276"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "uh=1900; #in cm^2/V-s (mobility of electron)\n",
+ "Na=2E17; #in /m^3 (acceptor doping concentration)\n",
+ "e=1.6E-19; #in C(charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "uh=uh*1E-4; #changing unit from cm^2/V-s to m^2/V-s\n",
+ "Na=Na*1E6; #changing unit from 1/cm^3 to 1/m^3\n",
+ "nh=Na; #hole concentration \n",
+ "#since sigma=ne*ue*e+nh*uh*e and nh>>ne\n",
+ "#therefore sigma=nh*uh*e\n",
+ "sigma=nh*uh*e; #calculation of conductivity\n",
+ "\n",
+ "#result\n",
+ "print\"The conductivity of p-type Ge crystal is =\",sigma,\"/ohm-m (roundoff error)\";\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The conductivity of p-type Ge crystal is = 6080.0 /ohm-m (roundoff error)\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.12 , Page no:277"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "ue=0.19; #in m^2/V-s (mobility of electron)\n",
+ "T=300; #in K (temperature)\n",
+ "k=1.38E-23; #in J/K (Boltzmann\u2019s constant)\n",
+ "e=1.6E-19; #in C(charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "Dn=ue*k*T/e; #calculation of diffusion co-efficient\n",
+ "\n",
+ "#result\n",
+ "print\"The diffusion co-efficient of electron in silicon is Dn=\",Dn,\"m^2/s\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The diffusion co-efficient of electron in silicon is Dn= 0.00491625 m^2/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.13 , Page no:277"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "Eg=0.4; #in eV (Band gap of semiconductor)\n",
+ "k=1.38E-23; #in J/K (Boltzmann\u2019s constant)\n",
+ "T1=0; #in degree Celcius (first temperature)\n",
+ "T2=50; #in degree Celcius (second temperature)\n",
+ "T3=100; #in degree Celcius (third temperature)\n",
+ "e=1.602E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "T1=T1+273; #changing temperature form Celcius to Kelvin\n",
+ "T2=T2+273; #changing temperature form Celcius to Kelvin\n",
+ "T3=T3+273; #changing temperature form Celcius to Kelvin\n",
+ "Eg=Eg*e; #changing unit from eV to Joule\n",
+ "#Using F_E=1/(1+exp(Eg/2*k*T))\n",
+ "F_E1=1/(1+math.exp(Eg/(2*k*T1))); #calculation of probability of occupation of lowest level at 0 degree Celcius\n",
+ "F_E2=1/(1+math.exp(Eg/(2*k*T2))); #calculation of probability of occupation of lowest level at 50 degree Celcius\n",
+ "F_E3=1/(1+math.exp(Eg/(2*k*T3))); #calculation of probability of occupation of lowest level at 100 degree Celcius\n",
+ "\n",
+ "#result\n",
+ "print\"The probability of occupation of lowest level in conduction band is\";\n",
+ "print\"\\t at 0 degree Celcius, F_E=\",F_E1,\"eV\";\n",
+ "print\"\\t at 50 degree Celcius, F_E=\",F_E2,\"eV\";\n",
+ "print\"\\t at 100 degree Celcius, F_E=\",F_E3,\"eV\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The probability of occupation of lowest level in conduction band is\n",
+ "\t at 0 degree Celcius, F_E= 0.000202505914236 eV\n",
+ "\t at 50 degree Celcius, F_E= 0.000754992968827 eV\n",
+ "\t at 100 degree Celcius, F_E= 0.00197639649915 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.14 , Page no:278"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "Eg=1.2; #in eV (Energy band gap)\n",
+ "k=1.38E-23; #in J/K (Boltzmann\u2019s constant)\n",
+ "T1=600; T2=300; #in K (two temperatures)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "Eg=Eg*e; #changing unit from eV to Joule\n",
+ "#since sigma is proportional to exp(-Eg/(2*k*T))\n",
+ "#therefore ratio=sigma1/sigma2=exp(-Eg/(2*k*((1/T1)-(1/T2))));\n",
+ "ratio= math.exp((-Eg/(2*k))*((1/T1)-(1/T2))); #calculation of ratio of conductivity at 600K and at 300K\n",
+ "\n",
+ "#result\n",
+ "print\"The ratio of conductivity at 600K and at 300K is =\",ratio;"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The ratio of conductivity at 600K and at 300K is = 108467.17792\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.15 , Page no:278"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "ue=0.39; #in m^2/V-s (mobility of electron)\n",
+ "n=5E13; #number of donor atoms\n",
+ "ni=2.4E19; #in atoms/m^3 (intrinsic carrier density)\n",
+ "l=10; #in mm (length of rod)\n",
+ "a=1; #in mm (side of square cross-section)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "l=l*1E-3; #changing unit from mm to m\n",
+ "a=a*1E-3; #changing unit from mm to m\n",
+ "A=a**2; #calculation of cross-section area\n",
+ "Nd=n/(l*A); #calculation of donor concentration\n",
+ "ne=Nd; #calculation of electron density\n",
+ "nh=ni**2/Nd; #calculation of hole density\n",
+ "#since sigma=ne*e*ue+nh*e*ue and since ne>>nh\n",
+ "#therefore sigma=ne*e*ue\n",
+ "sigma=ne*e*ue; #calculation of conductivity\n",
+ "rho=1/sigma; #calculation of resistivity\n",
+ "R=rho*l/A; #calculation of resistance \n",
+ "\n",
+ "#result\n",
+ "print\"The electron density is ne=\",ne,\"/m^3\";\n",
+ "print\"The hole density is nh=\",nh,\"/m^3\";\n",
+ "print\"The conductivity is =\",sigma,\"/ohm-m\";\n",
+ "print\"The resistance is R=\",round(R),\"ohm\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The electron density is ne= 5e+21 /m^3\n",
+ "The hole density is nh= 1.152e+17 /m^3\n",
+ "The conductivity is = 312.0 /ohm-m\n",
+ "The resistance is R= 32.0 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.16 , Page no:279"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "RH=3.66E-4; #in m^3/C (Hall coefficient)\n",
+ "rho=8.93E-3; #in ohm-m (resistivity)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "u=RH/rho; #calculation of mobility\n",
+ "n=1/(RH*e); #calculation of density\n",
+ "\n",
+ "#result\n",
+ "print\"The mobility is u=\",u,\"m^2/(V-s)\";\n",
+ "print\"The density is n=\",n,\"/m^3\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The mobility is u= 0.0409854423292 m^2/(V-s)\n",
+ "The density is n= 1.70765027322e+22 /m^3\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.17 , Page no:279"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "RH=3.66E-4; #in m^3/C (Hall coefficient)\n",
+ "rho=8.93E-3; #in ohm-m (resistivity)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "nh=1/(RH*e); #calculation of density of charge carrier\n",
+ "uh=1/(rho*nh*e); #calculation of mobility of charge carrier\n",
+ "\n",
+ "#result\n",
+ "print\"The density of charge carrier is nh=\",nh,\"/m^3\";\n",
+ "print\"The mobility of charge carrier is uh=\",uh,\"m^2/(V-s)\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The density of charge carrier is nh= 1.70765027322e+22 /m^3\n",
+ "The mobility of charge carrier is uh= 0.0409854423292 m^2/(V-s)\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Applied_Physics/Chapter_01_Bonding_in_Solids_.ipynb b/Applied_Physics/Chapter_01_Bonding_in_Solids_.ipynb new file mode 100755 index 00000000..e7e0fc16 --- /dev/null +++ b/Applied_Physics/Chapter_01_Bonding_in_Solids_.ipynb @@ -0,0 +1,248 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:7b7dfd33b7ab2e69421d0f26bfbfaa20f35bc1084164b445bbbc400215113c08"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 1:Bonding in Solids"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.3 , Page no:15"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "r=2; #in angstrom(distance)\n",
+ "e=1.6E-19; # in C (charge of electron)\n",
+ "E_o= 8.85E-12;# absolute premittivity\n",
+ "\n",
+ "#calculate\n",
+ "r=2*1*10**-10; # since r is in angstrom\n",
+ "V=-e**2/(4*3.14*E_o*r); # calculate potential\n",
+ "V1=V/e; # changing to eV\n",
+ "\n",
+ "#result\n",
+ "print \"\\nThe potential energy is V = \",V,\"J\";\n",
+ "print \"In electron-Volt V = \",round(V1,3),\"eV\"; \n",
+ "print \"Note: the answer in the book is wrong due to calculation mistake\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The potential energy is V = -1.15153477995e-18 J\n",
+ "In electron-Volt V = -7.197 eV\n",
+ "Note: the answer in the book is wrong due to calculation mistake\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.4 , Page no:15"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "#given\n",
+ "r0=0.236; #in nanometer(interionic distance)\n",
+ "e=1.6E-19; # in C (charge of electron)\n",
+ "E_o= 8.85E-12;# absolute premittivity\n",
+ "N=8; # Born constant\n",
+ "IE=5.14;# in eV (ionisation energy of sodium)\n",
+ "EA=3.65;# in eV (electron affinity of Chlorine)\n",
+ "pi=3.14; # value of pi used in the solution\n",
+ "\n",
+ "#calculate\n",
+ "r0=r0*1E-9; # since r is in nanometer\n",
+ "PE=(e**2/(4*pi*E_o*r0))*(1-1/N); # calculate potential energy\n",
+ "PE=PE/e; #changing unit from J to eV\n",
+ "NE=IE-EA;# calculation of Net energy\n",
+ "BE=PE-NE;# calculation of Bond Energy\n",
+ "\n",
+ "#result\n",
+ "print\"The potential energy is PE= \",round(PE,2),\"eV\";\n",
+ "print\"The net energy is NE= \",round(NE,2),\"eV\";\n",
+ "print\"The bond energy is BE= \",round(BE,2),\"eV\";\n",
+ "# Note: (1)-In order to make the answer prcatically feasible and avoid the unusual answer, I have used r_0=0.236 nm instead of 236 nm. because using this value will give very much irrelevant answer.\n",
+ "# Note: (2) There is slight variation in the answer due to round off."
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The potential energy is PE= 5.34 eV\n",
+ "The net energy is NE= 1.49 eV\n",
+ "The bond energy is BE= 3.85 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.5 , Page no:16"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "r_0=.41; #in mm(lattice constant)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "E_o= 8.85E-12; #absolute premittivity\n",
+ "n=0.5; #repulsive exponent value\n",
+ "alpha=1.76; #Madelung constant\n",
+ "pi=3.14; # value of pi used in the solution\n",
+ "\n",
+ "#calculate\n",
+ "r=.41*1E-3; #since r is in mm\n",
+ "Beta=72*pi*E_o*r**4/(alpha*e**2*(n-1)); #calculation compressibility\n",
+ "\n",
+ "#result\n",
+ "print\"The compressibility is\tBeta=\",round(Beta);"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The compressibility is\tBeta= -2.50967916144e+15\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.6 , Page no:16"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "r_0=3.56; #in Angstrom\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "IE=3.89; #in eV (ionisation energy of Cs)\n",
+ "EA=-3.61; #in eV (electron affinity of Cl)\n",
+ "n=10.5; #Born constant\n",
+ "E_o= 8.85E-12; #absolute premittivity\n",
+ "alpha=1.763; #Madelung constant\n",
+ "pi=3.14; #value of pi used in the solution\n",
+ "\n",
+ "#calculate\n",
+ "r_0=r_0*1E-10; #since r is in nanometer\n",
+ "U=-alpha*(e**2/(4*pi*E_o*r_0))*(1-1/n); #calculate potential energy\n",
+ "U=U/e; #changing unit from J to eV\n",
+ "ACE=U+EA+IE; #calculation of atomic cohesive energy\n",
+ "\n",
+ "#result\n",
+ "print\"The ionic cohesive energy is \",round(U),\"eV\";\n",
+ "print\"The atomic cohesive energy is\",round(ACE),\"eV\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The ionic cohesive energy is -6.0 eV\n",
+ "The atomic cohesive energy is -6.0 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.7 , Page no:17"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "r_0=2.81; #in Angstrom\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "n=9; #Born constant\n",
+ "E_o= 8.85E-12; #absolute premittivity\n",
+ "alpha=1.748; #Madelung constant\n",
+ "pi=3.14; #value of pi used in the solution\n",
+ "\n",
+ "#calculate\n",
+ "r_0=r_0*1E-10; #since r is in nanometer\n",
+ "V=-alpha*(e**2/(4*pi*E_o*r_0))*(1-1/n); #calculate potential energy\n",
+ "V=V/e; #changing unit from J to eV\n",
+ "V_1=V/2; #Since only half of the energy contribute per ion to the cohecive energy therfore\n",
+ "\n",
+ "#result\n",
+ "print\"The potential energy is V=\",round(V,2),\"eV\";\n",
+ "print\"The energy contributing per ions to the cohesive energy is \",round(V_1,2),\"eV\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The potential energy is V= -7.96 eV\n",
+ "The energy contributing per ions to the cohesive energy is -3.98 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Applied_Physics/Chapter_01_Bonding_in_Solids__1.ipynb b/Applied_Physics/Chapter_01_Bonding_in_Solids__1.ipynb new file mode 100755 index 00000000..db06e802 --- /dev/null +++ b/Applied_Physics/Chapter_01_Bonding_in_Solids__1.ipynb @@ -0,0 +1,248 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:2ef3d4592a57ed325da4c5f5624afbf334fc1ea3982d155f4e16e40570d72ab9"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 1:Bonding in Solids"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.3 , Page no:15"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "r=2; #in angstrom(distance)\n",
+ "e=1.6E-19; # in C (charge of electron)\n",
+ "E_o= 8.85E-12;# absolute premittivity\n",
+ "\n",
+ "#calculate\n",
+ "r=2*1*10**-10; # since r is in angstrom\n",
+ "V=-e**2/(4*3.14*E_o*r); # calculate potential\n",
+ "V1=V/e; # changing to eV\n",
+ "\n",
+ "#result\n",
+ "print \"\\nThe potential energy is V = \",'%.4E'%V,\"J\";\n",
+ "print \"In electron-Volt V = \",round(V1,3),\"eV\"; \n",
+ "print \"Note: the answer in the book is wrong due to calculation mistake\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The potential energy is V = -1.1515E-18 J\n",
+ "In electron-Volt V = -7.197 eV\n",
+ "Note: the answer in the book is wrong due to calculation mistake\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.4 , Page no:15"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "#given\n",
+ "r0=0.236; #in nanometer(interionic distance)\n",
+ "e=1.6E-19; # in C (charge of electron)\n",
+ "E_o= 8.85E-12;# absolute premittivity\n",
+ "N=8; # Born constant\n",
+ "IE=5.14;# in eV (ionisation energy of sodium)\n",
+ "EA=3.65;# in eV (electron affinity of Chlorine)\n",
+ "pi=3.14; # value of pi used in the solution\n",
+ "\n",
+ "#calculate\n",
+ "r0=r0*1E-9; # since r is in nanometer\n",
+ "PE=(e**2/(4*pi*E_o*r0))*(1-1/N); # calculate potential energy\n",
+ "PE=PE/e; #changing unit from J to eV\n",
+ "NE=IE-EA;# calculation of Net energy\n",
+ "BE=PE-NE;# calculation of Bond Energy\n",
+ "\n",
+ "#result\n",
+ "print\"The potential energy is PE= \",round(PE,2),\"eV\";\n",
+ "print\"The net energy is NE= \",round(NE,2),\"eV\";\n",
+ "print\"The bond energy is BE= \",round(BE,2),\"eV\";\n",
+ "# Note: (1)-In order to make the answer prcatically feasible and avoid the unusual answer, I have used r_0=0.236 nm instead of 236 nm. because using this value will give very much irrelevant answer.\n",
+ "# Note: (2) There is slight variation in the answer due to round off."
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The potential energy is PE= 5.34 eV\n",
+ "The net energy is NE= 1.49 eV\n",
+ "The bond energy is BE= 3.85 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.5 , Page no:16"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "r_0=.41; #in mm(lattice constant)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "E_o= 8.85E-12; #absolute premittivity\n",
+ "n=0.5; #repulsive exponent value\n",
+ "alpha=1.76; #Madelung constant\n",
+ "pi=3.14; # value of pi used in the solution\n",
+ "\n",
+ "#calculate\n",
+ "r=.41*1E-3; #since r is in mm\n",
+ "Beta=72*pi*E_o*r**4/(alpha*e**2*(n-1)); #calculation compressibility\n",
+ "\n",
+ "#result\n",
+ "print\"The compressibility is\tBeta=\",'%.4E'%Beta;"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The compressibility is\tBeta= -2.5097E+15\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.6 , Page no:16"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "r_0=3.56; #in Angstrom\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "IE=3.89; #in eV (ionisation energy of Cs)\n",
+ "EA=-3.61; #in eV (electron affinity of Cl)\n",
+ "n=10.5; #Born constant\n",
+ "E_o= 8.85E-12; #absolute premittivity\n",
+ "alpha=1.763; #Madelung constant\n",
+ "pi=3.14; #value of pi used in the solution\n",
+ "\n",
+ "#calculate\n",
+ "r_0=r_0*1E-10; #since r is in nanometer\n",
+ "U=-alpha*(e**2/(4*pi*E_o*r_0))*(1-1/n); #calculate potential energy\n",
+ "U=U/e; #changing unit from J to eV\n",
+ "ACE=U+EA+IE; #calculation of atomic cohesive energy\n",
+ "\n",
+ "#result\n",
+ "print\"The ionic cohesive energy is \",round(U),\"eV\";\n",
+ "print\"The atomic cohesive energy is\",round(ACE),\"eV\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The ionic cohesive energy is -6.0 eV\n",
+ "The atomic cohesive energy is -6.0 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.7 , Page no:17"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "r_0=2.81; #in Angstrom\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "n=9; #Born constant\n",
+ "E_o= 8.85E-12; #absolute premittivity\n",
+ "alpha=1.748; #Madelung constant\n",
+ "pi=3.14; #value of pi used in the solution\n",
+ "\n",
+ "#calculate\n",
+ "r_0=r_0*1E-10; #since r is in nanometer\n",
+ "V=-alpha*(e**2/(4*pi*E_o*r_0))*(1-1/n); #calculate potential energy\n",
+ "V=V/e; #changing unit from J to eV\n",
+ "V_1=V/2; #Since only half of the energy contribute per ion to the cohecive energy therfore\n",
+ "\n",
+ "#result\n",
+ "print\"The potential energy is V=\",round(V,2),\"eV\";\n",
+ "print\"The energy contributing per ions to the cohesive energy is \",round(V_1,2),\"eV\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The potential energy is V= -7.96 eV\n",
+ "The energy contributing per ions to the cohesive energy is -3.98 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Applied_Physics/Chapter_02_Crystal_Structure.ipynb b/Applied_Physics/Chapter_02_Crystal_Structure.ipynb new file mode 100755 index 00000000..2d0b7a14 --- /dev/null +++ b/Applied_Physics/Chapter_02_Crystal_Structure.ipynb @@ -0,0 +1,291 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:f14517ab3de72640bce8c38f7d46947caa54ae7067bb37477a81849b7b917982"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 2:Crystal Structure"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.1 , Page no:40"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "N=6.02*10**26; #in /Kg-molecule (Avogadro's number)\n",
+ "n=4; #number of molecules per unit cell ofr NaCl\n",
+ "M=58.5; #in Kg/Kg-molecule (molecular weight of NaCl)\n",
+ "p=2189; #in Kg/m^3 (density)\n",
+ "\n",
+ "#calculate\n",
+ "a=pow(((n*M)/(N*p)),0.3333333);\n",
+ "a1=a*10**10; #changing unit to Angstrom\n",
+ "\n",
+ "#result\n",
+ "print\"The lattice constant is a=\",a,\"m\";\n",
+ "print\"\\t\\t\\ta=\",a1,\"Angstrom\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The lattice constant is a= 5.62072226997e-10 m\n",
+ "\t\t\ta= 5.62072226997 Angstrom\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2 , Page no:41"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "N=6.02*10**23; #in /gram-atom (Avogadro's number)\n",
+ "n=4; #number of atom per unit cell for fcc structure\n",
+ "M=63.5; #in gram/gram-atom (atomic weight of Cu)\n",
+ "p=8.96; #in g/cm^3 (density)\n",
+ "\n",
+ "#calculate\n",
+ "a=pow((n*M/(N*p)),0.3333333);\n",
+ "a=a*1E8; #changing unit from cm to Angstrom\n",
+ "d=a/math.sqrt(2); #distance infcc lattice\n",
+ "\n",
+ "#result\n",
+ "print\"lattice constant is a=\",a,\"E-08 cm\";\n",
+ "print\"\\t\\t a=\",round(a,4),\"Angstrom\";\n",
+ "print\"distance between two nearest Cu atoms is d=\",round(d,2),\"Angstrom\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "lattice constant is a= 3.61113576149 E-08 cm\n",
+ "\t\t a= 3.6111 Angstrom\n",
+ "distance between two nearest Cu atoms is d= 2.55 Angstrom\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.3 , Page no:41"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "N=6.02E26; #in /Kg-atom (Avogadro's number)\n",
+ "n=2; #number of molecules per unit cell for bcc lattice\n",
+ "M=55.85; #in Kg/Kg-atom (atomic weight of Iron)\n",
+ "p=7860; #in Kg/m^3 (density)\n",
+ "\n",
+ "#calculate\n",
+ "a=pow((n*M/(N*p)),0.33333);\n",
+ "a1=a*1E10; #changing unit to Angstrom\n",
+ "\n",
+ "#result\n",
+ "print\"The lattice constant is a=\",a,\"m\";\n",
+ "print\"\\t\\t\\ta=\",round(a1,3),\"Angstrom\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The lattice constant is a= 2.86928355016e-10 m\n",
+ "\t\t\ta= 2.869 Angstrom\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.4 , Page no:42"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "N=6.02*10**26; #in /Kg-atom (Avogadro's number)\n",
+ "n=2; #number of molecules per unit cell for bcc lattice\n",
+ "M=6.94; #in Kg/Kg-atom (atomic weight of Iron)\n",
+ "p=530; #in Kg/m^3 (density)\n",
+ "\n",
+ "#calculate\n",
+ "a=pow((n*M/(N*p)),0.33333);\n",
+ "a1=a*1E10; #changing unit to Angstrom\n",
+ "\n",
+ "#result\n",
+ "print\"The lattice constant is a=\",a,\"m\";\n",
+ "print\"\\t\\t\\ta=\",round(a1,3),\"Angstrom\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The lattice constant is a= 3.51776567326e-10 m\n",
+ "\t\t\ta= 3.518 Angstrom\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.5 , Page no:42"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "N=6.02*10**23; #in /gram-molecule (Avogadro's number)\n",
+ "M=58.5; #in gram/gram-molecule (atomic weight of NaCl)\n",
+ "p=2.17; #in g/cm^3 (density)\n",
+ "\n",
+ "#calculate\n",
+ "#since V=M/p\n",
+ "#(1/d)^-3=2N/V=2Np/M\n",
+ "#therefore d= (M/2Np)^-3\n",
+ "d=pow((M/(2*N*p)),0.33333333);\n",
+ "d1=d*1*10**8; #changing unit from cm to Angstrom\n",
+ "\n",
+ "#result\n",
+ "print\"The distance between two adjacent atoms of NaCl is d=\",d,\"m\";\n",
+ "print\"\\t\\t\\t\\t\\t\\t d=\",round(d1,3),\"Angstrom\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The distance between two adjacent atoms of NaCl is d= 2.81853408124e-08 m\n",
+ "\t\t\t\t\t\t d= 2.819 Angstrom\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.6 , Page no:43"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "r_Na=0.98; #in Angstrom (radius of sodium ion)\n",
+ "r_Cl=1.81; #in Angstrom (radius of chloride ion)\n",
+ "M_Na=22.99; #in amu (atomic mass of sodium)\n",
+ "M_Cl=35.45; #in amu (atomic mass of chlorine)\n",
+ "\n",
+ "#calculate\n",
+ "a=2*(r_Na+r_Cl); #lattice parameter\n",
+ "#PF=volume of ions present in the unit cell/volume of unit cell\n",
+ "PF=((4*(4/3)*3.14)*r_Na**3+(4*(4/3)*3.14)*r_Cl**3)/a**3;\n",
+ "#Density=mass of unit cell/volume of unit cell\n",
+ "p=4*(M_Na+M_Cl)*1.66E-27/(a*1E-10)**3;\n",
+ "p1=p*1E-3; #changing unit to gm/cm^-3\n",
+ "\n",
+ "#result\n",
+ "print\"Lattice constant is a=\",round(a,3),\"Angstrom\";\n",
+ "print\"Packing fraction is =\",round(PF,3); \n",
+ "print\"Density is p=\",round(p),\"Kg/m^3\";\n",
+ "print\"Density is p=\",round(p1,2),\"g/cm^3\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Lattice constant is a= 5.58 Angstrom\n",
+ "Packing fraction is = 0.662\n",
+ "Density is p= 2233.0 Kg/m^3\n",
+ "Density is p= 2.23 g/cm^3\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Applied_Physics/Chapter_02_Crystal_Structure_1.ipynb b/Applied_Physics/Chapter_02_Crystal_Structure_1.ipynb new file mode 100755 index 00000000..b6038072 --- /dev/null +++ b/Applied_Physics/Chapter_02_Crystal_Structure_1.ipynb @@ -0,0 +1,291 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:693d1c12519596ec538fd8d806511b7a1fdb44b316b98121270f6b11b5d8d004"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 2:Crystal Structure"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.1 , Page no:40"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "N=6.02*10**26; #in /Kg-molecule (Avogadro's number)\n",
+ "n=4; #number of molecules per unit cell ofr NaCl\n",
+ "M=58.5; #in Kg/Kg-molecule (molecular weight of NaCl)\n",
+ "p=2189; #in Kg/m^3 (density)\n",
+ "\n",
+ "#calculate\n",
+ "a=pow(((n*M)/(N*p)),0.3333333);\n",
+ "a1=a*10**10; #changing unit to Angstrom\n",
+ "\n",
+ "#result\n",
+ "print\"The lattice constant is a=\",'%.2E'%a,\"m\";\n",
+ "print\"\\t\\t\\ta=\",round(a1,2),\"Angstrom\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The lattice constant is a= 5.62E-10 m\n",
+ "\t\t\ta= 5.62 Angstrom\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2 , Page no:41"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "N=6.02*10**23; #in /gram-atom (Avogadro's number)\n",
+ "n=4; #number of atom per unit cell for fcc structure\n",
+ "M=63.5; #in gram/gram-atom (atomic weight of Cu)\n",
+ "p=8.96; #in g/cm^3 (density)\n",
+ "\n",
+ "#calculate\n",
+ "a=pow((n*M/(N*p)),0.3333333);\n",
+ "a1=a*1E8; #changing unit from cm to Angstrom\n",
+ "d=a1/math.sqrt(2); #distance infcc lattice\n",
+ "\n",
+ "#result\n",
+ "print\"lattice constant is a=\",'%.2E'%a,\"cm\";\n",
+ "print\"\\t\\t a=\",round(a1,2),\"Angstrom\";\n",
+ "print\"distance between two nearest Cu atoms is d=\",round(d,2),\"Angstrom\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "lattice constant is a= 3.61E-08 cm\n",
+ "\t\t a= 3.61 Angstrom\n",
+ "distance between two nearest Cu atoms is d= 2.55 Angstrom\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.3 , Page no:41"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "N=6.02E26; #in /Kg-atom (Avogadro's number)\n",
+ "n=2; #number of molecules per unit cell for bcc lattice\n",
+ "M=55.85; #in Kg/Kg-atom (atomic weight of Iron)\n",
+ "p=7860; #in Kg/m^3 (density)\n",
+ "\n",
+ "#calculate\n",
+ "a=pow((n*M/(N*p)),0.33333);\n",
+ "a1=a*1E10; #changing unit to Angstrom\n",
+ "\n",
+ "#result\n",
+ "print\"The lattice constant is a=\",'%.2E'%a,\"m\";\n",
+ "print\"\\t\\t\\ta=\",round(a1,2),\"Angstrom\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The lattice constant is a= 2.87E-10 m\n",
+ "\t\t\ta= 2.87 Angstrom\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.4 , Page no:42"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "N=6.02*10**26; #in /Kg-atom (Avogadro's number)\n",
+ "n=2; #number of molecules per unit cell for bcc lattice\n",
+ "M=6.94; #in Kg/Kg-atom (atomic weight of Iron)\n",
+ "p=530; #in Kg/m^3 (density)\n",
+ "\n",
+ "#calculate\n",
+ "a=pow((n*M/(N*p)),0.33333);\n",
+ "a1=a*1E10; #changing unit to Angstrom\n",
+ "\n",
+ "#result\n",
+ "print\"The lattice constant is a=\",'%.2E'%a,\"m\";\n",
+ "print\"\\t\\t\\ta=\",round(a1,2),\"Angstrom\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The lattice constant is a= 3.52E-10 m\n",
+ "\t\t\ta= 3.52 Angstrom\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.5 , Page no:42"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "N=6.02*10**23; #in /gram-molecule (Avogadro's number)\n",
+ "M=58.5; #in gram/gram-molecule (atomic weight of NaCl)\n",
+ "p=2.17; #in g/cm^3 (density)\n",
+ "\n",
+ "#calculate\n",
+ "#since V=M/p\n",
+ "#(1/d)^-3=2N/V=2Np/M\n",
+ "#therefore d= (M/2Np)^-3\n",
+ "d=pow((M/(2*N*p)),0.33333333);\n",
+ "d1=d*1*10**8; #changing unit from cm to Angstrom\n",
+ "\n",
+ "#result\n",
+ "print\"The distance between two adjacent atoms of NaCl is d=\",'%.2E'%d,\"m\";\n",
+ "print\"\\t\\t\\t\\t\\t\\t d=\",round(d1,2),\"Angstrom\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The distance between two adjacent atoms of NaCl is d= 2.82E-08 m\n",
+ "\t\t\t\t\t\t d= 2.82 Angstrom\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.6 , Page no:43"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "r_Na=0.98; #in Angstrom (radius of sodium ion)\n",
+ "r_Cl=1.81; #in Angstrom (radius of chloride ion)\n",
+ "M_Na=22.99; #in amu (atomic mass of sodium)\n",
+ "M_Cl=35.45; #in amu (atomic mass of chlorine)\n",
+ "\n",
+ "#calculate\n",
+ "a=2*(r_Na+r_Cl); #lattice parameter\n",
+ "#PF=volume of ions present in the unit cell/volume of unit cell\n",
+ "PF=((4*(4/3)*3.14)*r_Na**3+(4*(4/3)*3.14)*r_Cl**3)/a**3;\n",
+ "#Density=mass of unit cell/volume of unit cell\n",
+ "p=4*(M_Na+M_Cl)*1.66E-27/(a*1E-10)**3;\n",
+ "p1=p*1E-3; #changing unit to gm/cm^-3\n",
+ "\n",
+ "#result\n",
+ "print\"Lattice constant is a=\",round(a,3),\"Angstrom\";\n",
+ "print\"Packing fraction is =\",round(PF,3); \n",
+ "print\"Density is p=\",round(p),\"Kg/m^3\";\n",
+ "print\"Density is p=\",round(p1,2),\"g/cm^3\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Lattice constant is a= 5.58 Angstrom\n",
+ "Packing fraction is = 0.662\n",
+ "Density is p= 2233.0 Kg/m^3\n",
+ "Density is p= 2.23 g/cm^3\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Applied_Physics/Chapter_03_Planes_in_Crystals.ipynb b/Applied_Physics/Chapter_03_Planes_in_Crystals.ipynb new file mode 100755 index 00000000..763a53c2 --- /dev/null +++ b/Applied_Physics/Chapter_03_Planes_in_Crystals.ipynb @@ -0,0 +1,144 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:1ec2372be7a9974010a31557c9658c85707025e7a87f82e6ef26be63dbcfb0e7"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 3:Planes in Crystals"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.11 , Page no:61"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "h=3;\n",
+ "k=2;\n",
+ "l=1; #miller indices\n",
+ "a=4.2E-8; #in cm (lattice constant)\n",
+ "\n",
+ "#calculate\n",
+ "d=a/math.sqrt(h**2+k**2+l**2); #calculation for interplanar spacing\n",
+ "d=d*1E8; #changing unit from cm to Angstrom\n",
+ "\n",
+ "#result\n",
+ "print\"The interplanar spacing is d=\",round(d,2),\"cm\";\n",
+ "print\"\\t\\t\\t d=\",round(d,2),\"Angstrom\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The interplanar spacing is d= 1.12 cm\n",
+ "\t\t\t d= 1.12 Angstrom\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.12 , Page no:61"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "h=1;k=1;l=1; #miller indices\n",
+ "a=2.5;b=2.5;c=1.8; #in Angstrom (lattice constants for tetragonal lattice )\n",
+ "\n",
+ "#calculate\n",
+ "d=1/math.sqrt((h/a)**2+(k/b)**2+(l/c)**2); #calculation for interplanar spacing\n",
+ "\n",
+ "#result\n",
+ "print\"The lattice spacing is d=\",round(d,2),\"Angstrom\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The lattice spacing is d= 1.26 Angstrom\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.15 , Page no:63"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "h=1;k=0;l=0; #miller indices\n",
+ "a=2.5; #in Angstrom (lattice constant)\n",
+ "\n",
+ "#calculate\n",
+ "a=a*1E-10; #hence a is in Angstrom\n",
+ "d=a/math.sqrt(h**2+k**2+l**2); #calculation for interplanar spacing\n",
+ "p=d/a**3;\n",
+ "\n",
+ "#result\n",
+ "print\"The density is p=\",round(p),\"lattice points/m^2\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The density is p= 1.6e+19 lattice points/m^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Applied_Physics/Chapter_03_Planes_in_Crystals_1.ipynb b/Applied_Physics/Chapter_03_Planes_in_Crystals_1.ipynb new file mode 100755 index 00000000..92dba19b --- /dev/null +++ b/Applied_Physics/Chapter_03_Planes_in_Crystals_1.ipynb @@ -0,0 +1,144 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:05935aae8ee517fae6abe7a89dceb1ca0d0cc367834bce08b5f427ec49852dcb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 3:Planes in Crystals"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.11 , Page no:61"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "h=3;\n",
+ "k=2;\n",
+ "l=1; #miller indices\n",
+ "a=4.2E-8; #in cm (lattice constant)\n",
+ "\n",
+ "#calculate\n",
+ "d=a/math.sqrt(h**2+k**2+l**2); #calculation for interplanar spacing\n",
+ "d=d*1E8; #changing unit from cm to Angstrom\n",
+ "\n",
+ "#result\n",
+ "print\"The interplanar spacing is d=\",round(d,2),\"cm\";\n",
+ "print\"\\t\\t\\t d=\",round(d,2),\"Angstrom\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The interplanar spacing is d= 1.12 cm\n",
+ "\t\t\t d= 1.12 Angstrom\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.12 , Page no:61"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "h=1;k=1;l=1; #miller indices\n",
+ "a=2.5;b=2.5;c=1.8; #in Angstrom (lattice constants for tetragonal lattice )\n",
+ "\n",
+ "#calculate\n",
+ "d=1/math.sqrt((h/a)**2+(k/b)**2+(l/c)**2); #calculation for interplanar spacing\n",
+ "\n",
+ "#result\n",
+ "print\"The lattice spacing is d=\",round(d,2),\"Angstrom\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The lattice spacing is d= 1.26 Angstrom\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.15 , Page no:63"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "h=1;k=0;l=0; #miller indices\n",
+ "a=2.5; #in Angstrom (lattice constant)\n",
+ "\n",
+ "#calculate\n",
+ "a=a*1E-10; #hence a is in Angstrom\n",
+ "d=a/math.sqrt(h**2+k**2+l**2); #calculation for interplanar spacing\n",
+ "p=d/a**3;\n",
+ "\n",
+ "#result\n",
+ "print\"The density is p=\",round(p),\"lattice points/m^2\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The density is p= 1.6e+19 lattice points/m^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Applied_Physics/Chapter_04_Crystal_Diffraction.ipynb b/Applied_Physics/Chapter_04_Crystal_Diffraction.ipynb new file mode 100755 index 00000000..6c9c83eb --- /dev/null +++ b/Applied_Physics/Chapter_04_Crystal_Diffraction.ipynb @@ -0,0 +1,494 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:177e44357417b2cfb0b476ada5b36a934225ade51d50763b09586ba8679f08fc"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 4:Crystal Diffraction"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.1 , Page no:75"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "lambda1=2.6; #in Angstrom (wavelength)\n",
+ "theta=20; #in Degree (angle)\n",
+ "n=2;\n",
+ "\n",
+ "#calculate\n",
+ "lambda1=lambda1*1E-10; #since lambda is in Angstrom\n",
+ "#Since 2dsin(theta)=n(lambda)\n",
+ "#therefore d=n(lambda)/2sin(theta)\n",
+ "d=n*lambda1/(2*math.sin(math.radians(theta)));\n",
+ "\n",
+ "#result\n",
+ "print\"The spacing constant is d=\",d,\"m\";\n",
+ "print\"\\t\\t\\td=\",d*10**10,\"Angstrom\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The spacing constant is d= 7.60189144042e-10 m\n",
+ "\t\t\td= 7.60189144042 Angstrom\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.2 , Page no:75"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "h=1;k=1;l=0; #miller indices\n",
+ "a=0.26; #in nanometer (lattice constant)\n",
+ "lambda1=0.065; #in nanometer (wavelength)\n",
+ "n=2; #order\n",
+ "\n",
+ "#calculate\n",
+ "d=a/math.sqrt(h**2+k**2+l**2); #calculation of interlattice spacing\n",
+ "#Since 2dsin(theta)=n(lambda)\n",
+ "#therefore we have\n",
+ "theta=math.asin(n*lambda1/(2*d));\n",
+ "theta1=theta*180/3.14;\n",
+ "\n",
+ "#result\n",
+ "print\"The glancing angle is =\",round(theta1,2),\"degree\";\n",
+ "print \"Note: there is slight variation in the answer due to round off error\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The glancing angle is = 20.72 degree\n",
+ "Note: there is slight variation in the answer due to round off error\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.3 , Page no:75"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "d=3.04E-10; #in mm (spacing constant)\n",
+ "lambda1=0.79; #in Angstrom (wavelength)\n",
+ "n=3; #order\n",
+ "\n",
+ "#calculate\n",
+ "#Since 2dsin(theta)=n(lambda)\n",
+ "#therefore we have\n",
+ "lambda2=lambda1*1E-10; #since lambda is in angstrom\n",
+ "theta=math.asin(n*lambda2/(2*d));\n",
+ "theta1=theta*180/3.14;\n",
+ "\n",
+ "#result\n",
+ "print\"The glancing angle is\",round(theta1,3),\"degree\";\n",
+ "print \"Note: In question the value of d=3.04E-9 cm but in solution is using d=3.04E-10 m. So I have used d=3.04E-10 cm as used in the solution\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The glancing angle is 22.954 degree\n",
+ "Note: In question the value of d=3.04E-9 cm but in solution is using d=3.04E-10 m. So I have used d=3.04E-10 cm as used in the solution\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4 , Page no:76"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "d=0.282; #in nanometer (spacing constant)\n",
+ "n=1; #order\n",
+ "theta=8.35; #in degree (glancing angle)\n",
+ "\n",
+ "#calculate\n",
+ "d=d*1E-9; #since d is in nanometer\n",
+ "#Since 2dsin(theta)=n(lambda)\n",
+ "#therefore we have\n",
+ "lambda1=2*d*math.sin(math.radians(theta))/n; \n",
+ "lambda_1=lambda1*1E10; #changing unit from m to Angstrom\n",
+ "theta_1=90; #in degree (for maximum order theta=90)\n",
+ "n_max=2*d*math.sin(math.radians(theta_1))/lambda1; #calculation of maximum order. \n",
+ "\n",
+ "#result\n",
+ "print\"The wavelength =\",lambda1,\"m\";\n",
+ "print\"\\t\\t=\" ,round(lambda_1,3),\"Angstrom\";\n",
+ "print\"The maximum order possible is n=\",round(n_max);\n",
+ "print \"Note: In question value of theta=8 degree and 35 minutes but solution uses theta=8.35 degree.So I am using theta=8.35 degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The wavelength = 8.19038939236e-11 m\n",
+ "\t\t= 0.819 Angstrom\n",
+ "The maximum order possible is n= 7.0\n",
+ "Note: In question value of theta=8 degree and 35 minutes but solution uses theta=8.35 degree.So I am using theta=8.35 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.5 , Page no:76"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "theta=6; #in degree (glancing angle)\n",
+ "p=2170; #in Kg/m^3 (density)\n",
+ "M=58.46; #Molecular weight of NaCl\n",
+ "N=6.02E26; #in Kg-molecule (Avogadro's number)\n",
+ "n=1; #order\n",
+ "XU=1E-12; #since 1X.U.= 1E-12m\n",
+ "\n",
+ "#calculate\n",
+ "d=(M/(2*N*p))**(1/3); #calclation of lattice constant\n",
+ "#Since 2dsin(theta)=n(lambda)\n",
+ "#therefore we have\n",
+ "lambda1=2*d*math.sin(math.radians(theta))/n; #calculation of wavelength\n",
+ "lambda2=lambda1/XU;\n",
+ "\n",
+ "#result\n",
+ "print\"The spacing constant is d=\",d,\"m\";\n",
+ "print\"The wavelength is =\",lambda1,\"m\";\n",
+ "print\"\\t\\t =\",lambda2,\"X.U\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The spacing constant is d= 2.81789104396e-10 m\n",
+ "The wavelength is = 5.89099640962e-11 m\n",
+ "\t\t = 58.9099640962 X.U\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.6 , Page no:77"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "h=1;k=1;l=1; #miller indices\n",
+ "a=5.63; #in Angstrom (lattice constant)\n",
+ "theta=27.5; #in degree (Glancing angle)\n",
+ "n=1; #order\n",
+ "H=6.625E-34; #in J-s (Plank's constant)\n",
+ "c=3E8; #in m/s (velocity of light)\n",
+ "e=1.6E-19; #charge of electron\n",
+ "\n",
+ "#calculate\n",
+ "d=a/math.sqrt(h**2+k**2+l**2); #calculation for interplanar spacing\n",
+ "#Since 2dsin(theta)=n(lambda)\n",
+ "#therefore we have\n",
+ "lambda1=2*d*math.sin(math.radians(theta))/n; #calculation for wavelength\n",
+ "E=H*c/(lambda1*1E-10); #calculation of Energy\n",
+ "E1=E/e; #changing unit from J to eV\n",
+ "\n",
+ "#result\n",
+ "print\"The lattice spacing is d=\",round(d,2),\"Angstrom\";\n",
+ "print\"The wavelength is =\",round(lambda1),\"Angstrom\";\n",
+ "print\"The energy of X-rays is E=\",E,\"J\";\n",
+ "print\"\\t\\t\\tE=\",E1,\"eV\";\n",
+ "print \"Note: c=3E8 m/s but in solution c=3E10 m/s \""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The lattice spacing is d= 3.25 Angstrom\n",
+ "The wavelength is = 3.0 Angstrom\n",
+ "The energy of X-rays is E= 6.62100284311e-16 J\n",
+ "\t\t\tE= 4138.12677695 eV\n",
+ "Note: c=3E8 m/s but in solution c=3E10 m/s \n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.7 , Page no:77"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "V=344; #in V (accelerating voltage)\n",
+ "theta=60; #in degree (glancing angle)\n",
+ "m=9.1E-31; #in Kg (mass of electron)\n",
+ "h=6.625e-34; #in J-s (Plank's constant)\n",
+ "n=1; #order\n",
+ "e=1.6E-19; #charge on electron\n",
+ "\n",
+ "#calculate\n",
+ "#Since K=m*v^2/2=e*V\n",
+ "#therefore v=sqrt(2*e*V/m)\n",
+ "#since lambda=h/(m*v)\n",
+ "#therefore we have lambda=h/sqrt(2*m*e*V)\n",
+ "lambda1=h/math.sqrt(2*m*e*V); #calculation of lambda\n",
+ "lambda2=lambda1*1E10; #changing unit from m to Angstrom\n",
+ "#Since 2dsin(theta)=n(lambda)\n",
+ "#therefore we have\n",
+ "d=n*lambda2/(2*math.sin(math.radians(theta)));\n",
+ "\n",
+ "#result\n",
+ "print\"The wavelength is =\",lambda1,\"m\";\n",
+ "print\"\\t\\t =\",round(lambda2,2),\"Angstrom\";\n",
+ "print\"The interplanar spacing is d=\",round(d,2),\"Angstrom\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The wavelength is = 6.61928340764e-11 m\n",
+ "\t\t = 0.66 Angstrom\n",
+ "The interplanar spacing is d= 0.38 Angstrom\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.8 , Page no:78"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "K=0.02; #in eV (kinetic energy)\n",
+ "d=2.0; #in Angstrom (Bragg's spacing)\n",
+ "m=1.00898; #in amu (mass of neutron)\n",
+ "amu=1.66E-27; #in Kg (1amu=1.66E-27 Kg)\n",
+ "h=6.625e-34; #in J-s (Plank's constant)\n",
+ "n=1; #order\n",
+ "e=1.6E-19; #charge on electron\n",
+ "\n",
+ "#calculate\n",
+ "#Since K=m*v^2/2\n",
+ "#therefore v=sqrt(2*K/m)\n",
+ "#since lambda=h/(m*v)\n",
+ "#therefore we have lambda=h/sqrt(2*m*K)\n",
+ "m=m*amu; #changing unit from amu to Kg\n",
+ "K=K*e; #changing unit to J from eV\n",
+ "lambda1=h/math.sqrt(2*m*K); #calculation of lambda\n",
+ "lambda2=lambda1*1E10; #changing unit from m to Angstrom\n",
+ "theta=math.asin(n*lambda2/(2*d)); #calculation of angle of first order diffraction maximum\n",
+ "theta1=theta*180/3.14;\n",
+ "#result\n",
+ "print\"The wavelength is =\",lambda1,\"m\";\n",
+ "print\"\\t \\t =\",round(lambda2),\"Angstrom\";\n",
+ "print\"The angle of first order diffraction maximum is \",round(theta1),\"Degree\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The wavelength is = 2.02348771817e-10 m\n",
+ "\t \t = 2.0 Angstrom\n",
+ "The angle of first order diffraction maximum is 30.0 Degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.9 , Page no:79"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "lambda1=0.586; #in Angstrom (wavelength of X-rays)\n",
+ "n1=1; n2=2; n3=3; #orders of diffraction\n",
+ "theta1=5+(58/60); #in degree (Glancing angle for first order of diffraction)\n",
+ "theta2=12+(01/60); #in degree (Glancing angle for second order of diffraction)\n",
+ "theta3=18+(12/60); #in degree (Glancing angle for third order of diffraction)\n",
+ "\n",
+ "#calculate\n",
+ "K1=math.sin(math.radians(theta1));\n",
+ "K2=math.sin(math.radians(theta2));\n",
+ "K3=math.sin(math.radians(theta3));\n",
+ "#Since 2dsin(theta)=n(lambda)\n",
+ "#therefore we have\n",
+ "d1=n1*lambda1/(2*K1);\n",
+ "d2=n2*lambda1/(2*K2);\n",
+ "d3=n3*lambda1/(2*K3);\n",
+ "d1=d1*1E-10; #changing unit from Angstrom to m\n",
+ "d2=d2*1E-10; #changing unit from Angstrom to m\n",
+ "d3=d3*1E-10; #changing unit from Angstrom to m\n",
+ "d=(d1+d2+d3)/3;\n",
+ "\n",
+ "#result\n",
+ "print\"The value of sine of different angle of diffraction is \\nK1=\",K1;\n",
+ "print\"K2=\",K2;\n",
+ "print\"K3=\",K3;\n",
+ "#Taking the ratios of K1:K2:K3\n",
+ "#We get K1:K2:K3=1:2:3\n",
+ "#Therefore we have\n",
+ "print\"Or we have K1:K2:K3=1:2:3\";\n",
+ "print\"Hence these angles of incidence are for Ist, 2nd and 3rd order reflections respectively\";\n",
+ "print\"The spacing constants are \\nd1=\",d1;\n",
+ "print\"d2=\",d2;\n",
+ "print\"d3=\",d3;\n",
+ "print\"The mean value of crystal spacing is d=\",d,\"m\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of sine of different angle of diffraction is \n",
+ "K1= 0.103949856225\n",
+ "K2= 0.208196213621\n",
+ "K3= 0.312334918512\n",
+ "Or we have K1:K2:K3=1:2:3\n",
+ "Hence these angles of incidence are for Ist, 2nd and 3rd order reflections respectively\n",
+ "The spacing constants are \n",
+ "d1= 2.81866671719e-10\n",
+ "d2= 2.81465253286e-10\n",
+ "d3= 2.81428667722e-10\n",
+ "The mean value of crystal spacing is d= 2.81586864243e-10 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Applied_Physics/Chapter_04_Crystal_Diffraction_1.ipynb b/Applied_Physics/Chapter_04_Crystal_Diffraction_1.ipynb new file mode 100755 index 00000000..fef74a69 --- /dev/null +++ b/Applied_Physics/Chapter_04_Crystal_Diffraction_1.ipynb @@ -0,0 +1,503 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:513926db1524f2764781365e8a81c6fa3119b7bc0f032a42233e49f5c3b6140a"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 4:Crystal Diffraction"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.1 , Page no:75"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "lambda1=2.6; #in Angstrom (wavelength)\n",
+ "theta=20; #in Degree (angle)\n",
+ "n=2;\n",
+ "\n",
+ "#calculate\n",
+ "lambda1=lambda1*1E-10; #since lambda is in Angstrom\n",
+ "#Since 2dsin(theta)=n(lambda)\n",
+ "#therefore d=n(lambda)/2sin(theta)\n",
+ "d=n*lambda1/(2*math.sin(math.radians(theta)));\n",
+ "\n",
+ "#result\n",
+ "print\"The spacing constant is d=\",'%.2E'%d,\"m\";\n",
+ "print\"\\t\\t\\td=\",round(d*10**10,3),\"Angstrom\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The spacing constant is d= 7.60E-10 m\n",
+ "\t\t\td= 7.602 Angstrom\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.2 , Page no:75"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "h=1;k=1;l=0; #miller indices\n",
+ "a=0.26; #in nanometer (lattice constant)\n",
+ "lambda1=0.065; #in nanometer (wavelength)\n",
+ "n=2; #order\n",
+ "\n",
+ "#calculate\n",
+ "d=a/math.sqrt(h**2+k**2+l**2); #calculation of interlattice spacing\n",
+ "#Since 2dsin(theta)=n(lambda)\n",
+ "#therefore we have\n",
+ "theta=math.asin(n*lambda1/(2*d));\n",
+ "theta1=theta*180/3.14;\n",
+ "\n",
+ "#result\n",
+ "print\"The glancing angle is =\",round(theta1,2),\"degree\";\n",
+ "print \"Note: there is slight variation in the answer due to round off error\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The glancing angle is = 20.72 degree\n",
+ "Note: there is slight variation in the answer due to round off error\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.3 , Page no:75"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "d=3.04E-10; #in mm (spacing constant)\n",
+ "lambda1=0.79; #in Angstrom (wavelength)\n",
+ "n=3; #order\n",
+ "\n",
+ "#calculate\n",
+ "#Since 2dsin(theta)=n(lambda)\n",
+ "#therefore we have\n",
+ "lambda2=lambda1*1E-10; #since lambda is in angstrom\n",
+ "theta=math.asin(n*lambda2/(2*d));\n",
+ "theta1=theta*180/3.14;\n",
+ "\n",
+ "#result\n",
+ "print\"The glancing angle is\",round(theta1,3),\"degree\";\n",
+ "print \"Note: In question the value of d=3.04E-9 cm but in solution is using d=3.04E-10 m. So I have used d=3.04E-10 cm as used in the solution\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The glancing angle is 22.954 degree\n",
+ "Note: In question the value of d=3.04E-9 cm but in solution is using d=3.04E-10 m. So I have used d=3.04E-10 cm as used in the solution\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4 , Page no:76"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "d=0.282; #in nanometer (spacing constant)\n",
+ "n=1; #order\n",
+ "theta=8.35; #in degree (glancing angle)\n",
+ "\n",
+ "#calculate\n",
+ "d=d*1E-9; #since d is in nanometer\n",
+ "#Since 2dsin(theta)=n(lambda)\n",
+ "#therefore we have\n",
+ "lambda1=2*d*math.sin(math.radians(theta))/n; \n",
+ "lambda_1=lambda1*1E10; #changing unit from m to Angstrom\n",
+ "theta_1=90; #in degree (for maximum order theta=90)\n",
+ "n_max=2*d*math.sin(math.radians(theta_1))/lambda1; #calculation of maximum order. \n",
+ "\n",
+ "#result\n",
+ "print\"The wavelength =\",'%.2E'%lambda1,\"m\";\n",
+ "print\"\\t\\t=\" ,round(lambda_1,3),\"Angstrom\";\n",
+ "print\"The maximum order possible is n=\",round(n_max);\n",
+ "print \"Note: In question value of theta=8 degree and 35 minutes but solution uses theta=8.35 degree.So I am using theta=8.35 degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The wavelength = 8.19E-11 m\n",
+ "\t\t= 0.819 Angstrom\n",
+ "The maximum order possible is n= 7.0\n",
+ "Note: In question value of theta=8 degree and 35 minutes but solution uses theta=8.35 degree.So I am using theta=8.35 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.5 , Page no:76"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "theta=6; #in degree (glancing angle)\n",
+ "p=2170; #in Kg/m^3 (density)\n",
+ "M=58.46; #Molecular weight of NaCl\n",
+ "N=6.02E26; #in Kg-molecule (Avogadro's number)\n",
+ "n=1; #order\n",
+ "XU=1E-12; #since 1X.U.= 1E-12m\n",
+ "\n",
+ "#calculate\n",
+ "d=(M/(2*N*p))**(1/3); #calclation of lattice constant\n",
+ "#Since 2dsin(theta)=n(lambda)\n",
+ "#therefore we have\n",
+ "lambda1=2*d*math.sin(math.radians(theta))/n; #calculation of wavelength\n",
+ "lambda2=lambda1/XU;\n",
+ "\n",
+ "#result\n",
+ "print\"The spacing constant is d=\",'%.2E'%d,\"m\";\n",
+ "print\"The wavelength is =\",'%.2E'%lambda1,\"m\";\n",
+ "print\"\\t\\t =\",round(lambda2,1),\"X.U\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The spacing constant is d= 2.82E-10 m\n",
+ "The wavelength is = 5.89E-11 m\n",
+ "\t\t = 58.9 X.U\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.6 , Page no:77"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "h=1;k=1;l=1; #miller indices\n",
+ "a=5.63; #in Angstrom (lattice constant)\n",
+ "theta=27.5; #in degree (Glancing angle)\n",
+ "n=1; #order\n",
+ "H=6.625E-34; #in J-s (Plank's constant)\n",
+ "c=3E8; #in m/s (velocity of light)\n",
+ "e=1.6E-19; #charge of electron\n",
+ "\n",
+ "#calculate\n",
+ "d=a/math.sqrt(h**2+k**2+l**2); #calculation for interplanar spacing\n",
+ "#Since 2dsin(theta)=n(lambda)\n",
+ "#therefore we have\n",
+ "lambda1=2*d*math.sin(math.radians(theta))/n; #calculation for wavelength\n",
+ "E=H*c/(lambda1*1E-10); #calculation of Energy\n",
+ "E1=E/e; #changing unit from J to eV\n",
+ "\n",
+ "#result\n",
+ "print\"The lattice spacing is d=\",round(d,2),\"Angstrom\";\n",
+ "print\"The wavelength is =\",round(lambda1),\"Angstrom\";\n",
+ "print\"The energy of X-rays is E=\",'%.2E'%E,\"J\";\n",
+ "print\"\\t\\t\\tE=\",'%.3E'%E1,\"eV\";\n",
+ "print \"Note: c=3E8 m/s but in solution c=3E10 m/s \""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The lattice spacing is d= 3.25 Angstrom\n",
+ "The wavelength is = 3.0 Angstrom\n",
+ "The energy of X-rays is E= 6.62E-16 J\n",
+ "\t\t\tE= 4.138E+03 eV\n",
+ "Note: c=3E8 m/s but in solution c=3E10 m/s \n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.7 , Page no:77"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "V=344; #in V (accelerating voltage)\n",
+ "theta=60; #in degree (glancing angle)\n",
+ "m=9.1E-31; #in Kg (mass of electron)\n",
+ "h=6.625e-34; #in J-s (Plank's constant)\n",
+ "n=1; #order\n",
+ "e=1.6E-19; #charge on electron\n",
+ "\n",
+ "#calculate\n",
+ "#Since K=m*v^2/2=e*V\n",
+ "#therefore v=sqrt(2*e*V/m)\n",
+ "#since lambda=h/(m*v)\n",
+ "#therefore we have lambda=h/sqrt(2*m*e*V)\n",
+ "lambda1=h/math.sqrt(2*m*e*V); #calculation of lambda\n",
+ "lambda2=lambda1*1E10; #changing unit from m to Angstrom\n",
+ "#Since 2dsin(theta)=n(lambda)\n",
+ "#therefore we have\n",
+ "d=n*lambda2/(2*math.sin(math.radians(theta)));\n",
+ "\n",
+ "#result\n",
+ "print\"The wavelength is =\",'%.2E'%lambda1,\"m\";\n",
+ "print\"\\t\\t =\",round(lambda2,2),\"Angstrom\";\n",
+ "print\"The interplanar spacing is d=\",round(d,2),\"Angstrom\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The wavelength is = 6.62E-11 m\n",
+ "\t\t = 0.66 Angstrom\n",
+ "The interplanar spacing is d= 0.38 Angstrom\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.8 , Page no:78"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "K=0.02; #in eV (kinetic energy)\n",
+ "d=2.0; #in Angstrom (Bragg's spacing)\n",
+ "m=1.00898; #in amu (mass of neutron)\n",
+ "amu=1.66E-27; #in Kg (1amu=1.66E-27 Kg)\n",
+ "h=6.625e-34; #in J-s (Plank's constant)\n",
+ "n=1; #order\n",
+ "e=1.6E-19; #charge on electron\n",
+ "\n",
+ "#calculate\n",
+ "#Since K=m*v^2/2\n",
+ "#therefore v=sqrt(2*K/m)\n",
+ "#since lambda=h/(m*v)\n",
+ "#therefore we have lambda=h/sqrt(2*m*K)\n",
+ "m=m*amu; #changing unit from amu to Kg\n",
+ "K=K*e; #changing unit to J from eV\n",
+ "lambda1=h/math.sqrt(2*m*K); #calculation of lambda\n",
+ "lambda2=lambda1*1E10; #changing unit from m to Angstrom\n",
+ "theta=math.asin(n*lambda2/(2*d)); #calculation of angle of first order diffraction maximum\n",
+ "theta1=theta*180/3.14;\n",
+ "#result\n",
+ "print\"The wavelength is =\",'%.2E'%lambda1,\"m\";\n",
+ "print\"\\t \\t =\",round(lambda2),\"Angstrom\";\n",
+ "print\"The angle of first order diffraction maximum is \",round(theta1),\"Degree\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The wavelength is = 2.02E-10 m\n",
+ "\t \t = 2.0 Angstrom\n",
+ "The angle of first order diffraction maximum is 30.0 Degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.9 , Page no:79"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "lambda1=0.586; #in Angstrom (wavelength of X-rays)\n",
+ "n1=1; n2=2; n3=3; #orders of diffraction\n",
+ "theta1=5+(58/60); #in degree (Glancing angle for first order of diffraction)\n",
+ "theta2=12+(01/60); #in degree (Glancing angle for second order of diffraction)\n",
+ "theta3=18+(12/60); #in degree (Glancing angle for third order of diffraction)\n",
+ "\n",
+ "#calculate\n",
+ "K1=math.sin(math.radians(theta1));\n",
+ "K2=math.sin(math.radians(theta2));\n",
+ "K3=math.sin(math.radians(theta3));\n",
+ "#Since 2dsin(theta)=n(lambda)\n",
+ "#therefore we have\n",
+ "d1=n1*lambda1/(2*K1);\n",
+ "d2=n2*lambda1/(2*K2);\n",
+ "d3=n3*lambda1/(2*K3);\n",
+ "d1=d1*1E-10; #changing unit from Angstrom to m\n",
+ "d2=d2*1E-10; #changing unit from Angstrom to m\n",
+ "d3=d3*1E-10; #changing unit from Angstrom to m\n",
+ "d=(d1+d2+d3)/3;\n",
+ "\n",
+ "#result\n",
+ "print\"The value of sine of different angle of diffraction is \\nK1=\",round(K1,4);\n",
+ "print\"K2=\",round(K2,4);\n",
+ "print\"K3=\",round(K3,4);\n",
+ "#Taking the ratios of K1:K2:K3\n",
+ "#We get K1:K2:K3=1:2:3\n",
+ "#Therefore we have\n",
+ "print\"Or we have K1:K2:K3=1:2:3\";\n",
+ "print\"Hence these angles of incidence are for Ist, 2nd and 3rd order reflections respectively\";\n",
+ "print\"The spacing constants are \\nd1=\",'%.2E'%d1;\n",
+ "print\"d2=\",'%.3E'%d2;\n",
+ "print\"d3=\",'%.3E'%d3;\n",
+ "print\"The mean value of crystal spacing is d=\",'%.3E'%d,\"m\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of sine of different angle of diffraction is \n",
+ "K1= 0.1039\n",
+ "K2= 0.2082\n",
+ "K3= 0.3123\n",
+ "Or we have K1:K2:K3=1:2:3\n",
+ "Hence these angles of incidence are for Ist, 2nd and 3rd order reflections respectively\n",
+ "The spacing constants are \n",
+ "d1= 2.82E-10\n",
+ "d2= 2.815E-10\n",
+ "d3= 2.814E-10\n",
+ "The mean value of crystal spacing is d= 2.816E-10 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [],
+ "language": "python",
+ "metadata": {},
+ "outputs": [],
+ "prompt_number": 9
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Applied_Physics/Chapter_05_Principles_of_Quantum_Mechanics.ipynb b/Applied_Physics/Chapter_05_Principles_of_Quantum_Mechanics.ipynb new file mode 100755 index 00000000..cf16b683 --- /dev/null +++ b/Applied_Physics/Chapter_05_Principles_of_Quantum_Mechanics.ipynb @@ -0,0 +1,1258 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:17ded29a54411f524d1901cc530f956cb4706c517e993f5acba4014789eca0bd"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 5:Principles of Quantum Mechanics"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.1 , Page no:102"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "lambda1=1; #in Angstrom (wavelength)\n",
+ "m=1.67E-27; #in Kg (mass of neutron)\n",
+ "h=6.625E-34; #in J-s (Planck's constant)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "lambda2=lambda1*1E-10; #since lambda is in Angstrom\n",
+ "#Since lambda=h/(m*v)\n",
+ "#Therefore we have\n",
+ "v=h/(m*lambda2); #calculation of velocity\n",
+ "K=m*v**2/2; #calculation of kinetic energy\n",
+ "K1=K/e; #changing unit fro J to eV\n",
+ "\n",
+ "#result\n",
+ "print\"The velocity is v=\",v,\"m/s\";\n",
+ "print\"The kinetic energy is K=\",K,\"J\";\n",
+ "print\"\\t\\t =\",K1,\"eV\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The velocity is v= 3967.06586826 m/s\n",
+ "The kinetic energy is K= 1.31409056886e-20 J\n",
+ "\t\t = 0.0821306605539 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.2 , Page no:103"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "K=50; #in eV (Kinetic energy)\n",
+ "m0=9.1E-31; #in Kg (mass of electron)\n",
+ "h=6.625E-34; #in J-s (Planck's constant)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "K=K*e; #changing unit from eV to J\n",
+ "#Since K=m*v^2/2\n",
+ "#Therefore v=sqrt(2*K/m)\n",
+ "#Since lambda=h/(m*v)\n",
+ "#Therefore we have\n",
+ "lambda1=h/math.sqrt(2*m0*K); #calculation of wavelength\n",
+ "lambda2=lambda1*1E10; #changing unit from m to Angstrom\n",
+ "\n",
+ "#result\n",
+ "print\"The wavelength is =\",lambda1,\"m\";\n",
+ "print\"\\t\\t =\",round(lambda2,3),\"Angstrom\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The wavelength is = 1.73622176082e-10 m\n",
+ "\t\t = 1.736 Angstrom\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.3 , Page no:104"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "E=2000; #in eV (Kinetic energy)\n",
+ "m=9.1E-31; #in Kg (mass of electron)\n",
+ "h=6.625E-34; #in J-s (Planck's constant)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "E=E*e; #changing unit from eV to J\n",
+ "#Since E=m*v^2/2\n",
+ "#Therefore v=sqrt(2*E/m)\n",
+ "#Since lambda=h/(m*v)\n",
+ "#Therefore we have\n",
+ "lambda1=h/math.sqrt(2*m*E); #calculation of wavelength\n",
+ "lambda2=lambda1*1E9; #changing unit from m to nanometer\n",
+ "\n",
+ "#result\n",
+ "print\"The wavelength is =\",lambda1,\"m\";\n",
+ "print\"\\t\\t =\",round(lambda2,4),\"nm\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The wavelength is = 2.74520764367e-11 m\n",
+ "\t\t = 0.0275 nm\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.4 , Page no:104"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "m_e=9.1E-31; #in Kg (mass of electron)\n",
+ "m_n=1.676E-27; #in Kg (mass of neutron)\n",
+ "h=6.625E-34; #in J-s (Planck's constant)\n",
+ "c=3E8; #in m/s (velocity of light)\n",
+ "\n",
+ "#calculate\n",
+ "E_e=m_e*c**2; #rest mass energy of electron\n",
+ "E_n=2*E_e; #given (kinetic energy of neutron)\n",
+ "#Since K=m*v^2/2\n",
+ "#Therefore v=sqrt(2*K/m)\n",
+ "#Since lambda=h/(m*v)\n",
+ "#Therefore we have\n",
+ "lambda1=h/math.sqrt(2*m_n*E_n); #calculation of wavelength\n",
+ "lambda2=lambda1*1E10; #changing unit from m to Angstrom\n",
+ "\n",
+ "#result\n",
+ "print\"The wavelength is =\",lambda1,\"m\";\n",
+ "print\"\\t\\t =\",lambda2,\"Angstrom\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The wavelength is = 2.82733278331e-14 m\n",
+ "\t\t = 0.000282733278331 Angstrom\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5 , Page no:104"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "V=1600; #in V (Potential)\n",
+ "\n",
+ "#calculate\n",
+ "lambda1=12.27/math.sqrt(V); #calculation of wavelength in Angstrom\n",
+ "\n",
+ "#result\n",
+ "print\"The wavelength is =\",round(lambda1,3),\"Angstrom\";\n",
+ "print \"Note : The answer in the book is wrong \""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The wavelength is = 0.307 Angstrom\n",
+ "Note : The answer in the book is wrong \n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.6 , Page no:105"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "E=1000; #in eV (Kinetic energy of photon)\n",
+ "K=1000; #in eV (Kinetic energy of electron)\n",
+ "m0=9.1E-31; #in Kg (mass of electron)\n",
+ "h=6.6E-34; #in J-s (Planck's constant)\n",
+ "c=3E8; #in m/s (velocity of light)\n",
+ "e=1.6E-19; #in C (charge on electron)\n",
+ "\n",
+ "#calculate\n",
+ "E=E*e; #changing unit from eV to J\n",
+ "lambda_p=h*c/E; #For photon E=hc/lambda\n",
+ "lambda_p1=lambda_p*1E10; #changing unit from m to Angstrom\n",
+ "#Since K=m*v^2/2\n",
+ "#Therefore v=sqrt(2*K/m)\n",
+ "#Since lambda=h/(m*v)\n",
+ "#Therefore we have\n",
+ "K=K*e; #changing unit from eV to J\n",
+ "lambda_e=h/math.sqrt(2*m0*K); #calculation of wavelength\n",
+ "lambda_e1=lambda_e*1E10; #changing unit from m to Angstrom\n",
+ "\n",
+ "#result\n",
+ "print\"For photon,the wavelength is =\",lambda_p,\"m\";\n",
+ "print\"\\t\\t\\t =\",round(lambda_p1,1),\"Angstrom\";\n",
+ "print\"For electron,the wavelength is =\",lambda_e,\"m\";\n",
+ "print\"\\t\\t\\t =\",round(lambda_e1,2),\"Angstrom\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "For photon,the wavelength is = 1.2375e-09 m\n",
+ "\t\t\t = 12.4 Angstrom\n",
+ "For electron,the wavelength is = 3.86765965524e-11 m\n",
+ "\t\t\t = 0.39 Angstrom\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.7 , Page no:105"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "lambda1=1.66*10**-10; #in m (wavelength)\n",
+ "m=9.1*10**-31; #in Kg (mass of electron)\n",
+ "h=6.626*10**-34; #in J-s (Planck's constant)\n",
+ "e=1.6E-19; #in C (charge on electron)\n",
+ "\n",
+ "#calculate\n",
+ "#Since lambda=h/(m*v)\n",
+ "#Therefore we have\n",
+ "v=h/(m*lambda1); #calculation of velocity\n",
+ "K=m*v**2/2; #calculation of kinetic energy\n",
+ "K1=K/e; #changing unit from J to eV\n",
+ "\n",
+ "#result\n",
+ "print\"The velocity of electron is v=\",v,\"m/s\";\n",
+ "print\"The kinetic energy is K=\",K,\"J\";\n",
+ "print\"\\t\\t =\",round(K1,3),\"eV\";\n",
+ "print \"Note: The answer in the book for kinetic energy is wrong \""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The velocity of electron is v= 4386336.55501 m/s\n",
+ "The kinetic energy is K= 8.75417651009e-18 J\n",
+ "\t\t = 54.714 eV\n",
+ "Note: The answer in the book for kinetic energy is wrong \n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.8 , Page no:106"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "T=400; #in K (temperature)\n",
+ "m=6.7E-27; #in Kg (mass of He-atom)\n",
+ "h=6.6E-34; #in J-s (Planck's constant)\n",
+ "k=1.376E-23; #in J/degree (Boltzmann constant)\n",
+ "\n",
+ "#calculate\n",
+ "#Since lambda=h/(m*v)\n",
+ "#E=mv^2/2;\n",
+ "#Therefore lambda=h/sqrt(2*m*E)\n",
+ "#E=kT\n",
+ "#Therefore lambda=h/sqrt(2*m*k*T)\n",
+ "lambda1=h/math.sqrt(2*m*k*T)\n",
+ "lambda2=lambda1*1E10; #changing unit from m to Angstrom\n",
+ "\n",
+ "#result\n",
+ "print\"The de-Broglie wavelength of He-atom is =\",lambda1,\"m\";\n",
+ "print\"\\t\\t\\t\\t =\",round(lambda2,4),\"Angstrom\"; "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The de-Broglie wavelength of He-atom is = 7.6851495611e-11 m\n",
+ "\t\t\t\t = 0.7685 Angstrom\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.9 , Page no:106"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "m_e=9.1E-31; #in Kg (mass of electron)\n",
+ "m_p=1.6E-27; #in Kg (mass of proton)\n",
+ "h=6.626E-34; #in J-s (Planck's constant)\n",
+ "c=3E8; #in m/s (velocity of light)\n",
+ "\n",
+ "#calculate\n",
+ "E=m_e*c**2; #in J (rest energy of electron)\n",
+ "#Since lambda=h/(m*v)\n",
+ "#E=mv^2/2;\n",
+ "#herefore lambda=h/sqrt(2*m*E)\n",
+ "#Also E=m_e*c^2;\n",
+ "#therefore lambda=h/sqrt(2*m_p*m_e*c^2)\n",
+ "lambda1=h/math.sqrt(2*m_p*m_e*c**2); #calculation of wavelength\n",
+ "lambda2=lambda1*1*10**10; #changing unit from m to Angstrom\n",
+ "\n",
+ "#result\n",
+ "print\"The de-Broglie wavelength of proton is =\",lambda1,\"m\";\n",
+ "print\"\\t\\t\\t\\t =\",lambda2,\"Angstrom\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The de-Broglie wavelength of proton is = 4.0929316435e-14 m\n",
+ "\t\t\t\t = 0.00040929316435 Angstrom\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.10 , Page no:106"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "V=10000; #in V (Potential)\n",
+ "\n",
+ "#calculate\n",
+ "lambda1=12.27/math.sqrt(V); #calculation of wavelength in Angstrom\n",
+ "\n",
+ "#result\n",
+ "print\"The wavelength is =\",round(lambda1,3),\"Angstrom\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The wavelength is = 0.123 Angstrom\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.11 , Page no:107"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "V=100; #in V (Potential)\n",
+ "n=1; #order of diffraction\n",
+ "d=2.15; #in Angstrom (lattice spacing)\n",
+ "\n",
+ "#calculate\n",
+ "lambda1=12.27/math.sqrt(V); #calculation of wavelength in Angstrom\n",
+ "#Since 2*d*sind(theta)=n*lambda\n",
+ "#therefore we have\n",
+ "theta=math.asin(n*lambda1/(2*d)); #calculation of glancing angle\n",
+ "theta1=theta*180/3.14;\n",
+ "\n",
+ "#result\n",
+ "print\"The wavelength is =\",round(lambda1,3),\"Angstrom\";\n",
+ "print\"The glancing angle is =\",round(theta1,1),\"degree\";\n",
+ "print \"Note: In question V=100 eV but the solution is using V=100V \"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The wavelength is = 1.227 Angstrom\n",
+ "The glancing angle is = 16.6 degree\n",
+ "Note: In question V=100 eV but the solution is using V=100V \n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.12 , Page no:107"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "V=344; #in V (Potential)\n",
+ "n=1; #order of diffraction\n",
+ "theta=60; #in degree (glancing angle)\n",
+ "\n",
+ "#calculate\n",
+ "lambda1=12.27/math.sqrt(V); #calculation of wavelength in Angstrom\n",
+ "#Since 2*d*sind(theta)=n*lambda\n",
+ "#therefore we have\n",
+ "d=n*lambda1/(2*math.sin(math.radians(theta))); #calculation of spacing constant\n",
+ "\n",
+ "#result\n",
+ "print\"The wavelength is =\",round(lambda1,3),\"Angstrom\";\n",
+ "print\"The spacing of the crystal is =\",round(d,4),\"Angstrom\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The wavelength is = 0.662 Angstrom\n",
+ "The spacing of the crystal is = 0.3819 Angstrom\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.13 , Page no:107"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "r=0.53*10**-10; #in m (radius of first Bohr orbit)\n",
+ "h=6.6*10**-34; #in J-s (Planck's constant)\n",
+ "m=9.1*10**-31; #in Kg (mass of electron)\n",
+ "n=1; #First Bohr orbit\n",
+ "pi=3.14; #value of pi used in the solution\n",
+ "\n",
+ "#calculate\n",
+ "#Since 2*pi*r=n*lambda and lambda=h/(m*v)\n",
+ "#Threfore we have v=h*n/(2*pi*r*m)\n",
+ "v=h*n/(2*pi*r*m); #calculation of velocity\n",
+ "\n",
+ "#result\n",
+ "print\"The velocity of electron is =\",v,\"m/s\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The velocity of electron is = 2179049.16859 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.14 , Page no:108"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "dx=0.2; #in Angstrom (uncertainty in the position)\n",
+ "h=6.6E-34; #in J-s (Planck's constant)\n",
+ "m0=9.1E-31; #in Kg (mass of electron)\n",
+ "pi=3.14; #value of pi used in the solution\n",
+ "\n",
+ "#calculate\n",
+ "dx=dx*1E-10; #since dx is in Angstrom\n",
+ "#Since dx*dp=h/4*pi (uncertainty relation)\n",
+ "dp=h/(4*pi*dx); #calculation of uncertainty in the momentum\n",
+ "#since dp=m*dv\n",
+ "dv=dp/m0; #calculation of uncertainty in the velocity\n",
+ "\n",
+ "#result\n",
+ "print\"The uncertainty in the momentum is dp=\",dp,\"Kg-m/\";\n",
+ "print\"The uncertainty in the velocity is dv=\",dv,\"m/s\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The uncertainty in the momentum is dp= 2.62738853503e-24 Kg-m/\n",
+ "The uncertainty in the velocity is dv= 2887240.14839 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.15 , Page no:108"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "m_e=9.1E-31; #in Kg (mass of electron)\n",
+ "m_p=1.67E-27; #in Kg (mass of proton)\n",
+ "dx_p=1; #in nanometer (uncertainty in position of electron)\n",
+ "dx_n=1; #in nanometer (uncertainty in position of proton)\n",
+ "\n",
+ "#calculate\n",
+ "#since dp=h/(4*pi*dx)\n",
+ "#since h/(4*pi) is constant and dx is same for electron and proton \n",
+ "#therefor both electron and proton have same uncertainty in the momentum\n",
+ "#since dv=dp/m and dp is same for both\n",
+ "#therefore dv_e/dv_p=m_p/m_e\n",
+ "#therefore\n",
+ "K=m_p/m_e; #ratio of uncertainty in the velocity of electron and proton\n",
+ "\n",
+ "#result\n",
+ "print\"The ratio of uncertainty in the velocity of electron to that of proton is =\",round(K);"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The ratio of uncertainty in the velocity of electron to that of proton is = 1835.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.16 , Page no:108"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "dx=5*10**-15; #in m (radius of nucleus or uncertainty in the position)\n",
+ "h=6.6*10**-34; #in J-s (Planck's constant)\n",
+ "m=1.67E-27; #in Kg (mass of proton)\n",
+ "pi=3.14; #value of pi used in the solution\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "#Since dx*dp=h/4*pi (uncertainty relation)\n",
+ "dp=h/(4*pi*dx); #calculation of uncertainty in the momentum\n",
+ "p=dp; #minimum value of momentum to calculate mimimum kinetic energy\n",
+ "K=(p**2)/(2*m); #calculation of minimum kinetic energy of proton\n",
+ "K1=K/e; #changing unit from J to eV\n",
+ "K2=K1/1E6; #changing unit from eV to MeV\n",
+ "\n",
+ "#result\n",
+ "print\"The minimum uncertainty in the momentum of proton is dp =\",dp,\"Kg-m/s\";\n",
+ "print\"The minimum kinetic energy of proton is K=\",K,\"J\";\n",
+ "print\"\\t\\t\\t\\t \\t =\",K1,\"eV\";\n",
+ "print\"\\t\\t\\t\\t\\t =\",K2,\"MeV\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The minimum uncertainty in the momentum of proton is dp = 1.05095541401e-20 Kg-m/s\n",
+ "The minimum kinetic energy of proton is K= 3.30690803067e-14 J\n",
+ "\t\t\t\t \t = 206681.751917 eV\n",
+ "\t\t\t\t\t = 0.206681751917 MeV\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.17 , Page no:109"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "K=1; #in KeV (kinetic energy of electron)\n",
+ "dx=1; #in Angstrom (uncertainty in the position)\n",
+ "h=6.63E-34; #in J-s (Planck's constant)\n",
+ "m=9.1E-31; #in Kg (mass of electron)\n",
+ "pi=3.14; #value of pi used in the solution\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "dx=dx*1E-10; #since dx is in Angstrom\n",
+ "#Since dx*dp=h/4*pi (uncertainty relation)\n",
+ "dp=h/(4*pi*dx); #calculation of uncertainty in the momentum\n",
+ "K=K*1E3*1.6E-19; #changing unit from KeV to J\n",
+ "p=math.sqrt(2*m*K); #calculation of momentum \n",
+ "poc=(dp/p)*100; #calculation of percentage of uncertainty\n",
+ "\n",
+ "#result\n",
+ "print\"The uncertainty in the momentum of electron is dp=\",dp,\"Kg-m/s\";\n",
+ "print\"The momentum of electron is p=\",p,\"Kg-m/s\";\n",
+ "print\"The percentage of uncertainty in the momentum is =\",round(poc,1);"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The uncertainty in the momentum of electron is dp= 5.27866242038e-25 Kg-m/s\n",
+ "The momentum of electron is p= 1.70645832062e-23 Kg-m/s\n",
+ "The percentage of uncertainty in the momentum is = 3.1\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.18 , Page no:109"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "v=6.6E4; #m/s (speed of electron)\n",
+ "poc=0.01; #percentage of uncertainty\n",
+ "h=6.63E-34; #in J-s (Planck's constant)\n",
+ "m=9E-31; #in Kg (mass of electron)\n",
+ "pi=3.14; #value of pi used in the solution\n",
+ "\n",
+ "#calculate\n",
+ "p=m*v; #calculation of momentum\n",
+ "dp=(poc/100)*p; #calculation of uncertainty in the momentum\n",
+ "#Since dx*dp=h/4*pi (uncertainty relation)\n",
+ "dx=h/(4*pi*dp); #calculation of uncertainty in the position\n",
+ "\n",
+ "#result\n",
+ "print\"The momentum of electron is p=\",p,\"Kg-m/s\";\n",
+ "print\"The uncertainty in the momentum of electron is dp=\",dp,\"Kg-m/s\";\n",
+ "print\"The uncertainty in the position of electron is dx=\",dx,\"Kg-m/s\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The momentum of electron is p= 5.94e-26 Kg-m/s\n",
+ "The uncertainty in the momentum of electron is dp= 5.94e-30 Kg-m/s\n",
+ "The uncertainty in the position of electron is dx= 8.88663707135e-06 Kg-m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.19 , Page no:111"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "lambda1=1; #in Angstrom (wavelength)\n",
+ "pi=3.14; #value of pi used in the solution\n",
+ "dlambda=1E-6; #uncertainty in wavelength\n",
+ "\n",
+ "#calculate\n",
+ "lambda2=lambda1*1E-10; #sinc lambda is in Angstrom\n",
+ "#By uncertainty principle, dx*dp>=h/(4*pi) --(1)\n",
+ "#since p=h/lambda -----(2)\n",
+ "#Or p*lambda=h \n",
+ "#diffrentiting this equation\n",
+ "#p*dlambda+lambda*dp=0\n",
+ "#dp=-p*dlambda/lambda ----(3)\n",
+ "#from (2) and (3) dp=-h*dlambda/lambda^2 ---(4)\n",
+ "#from (1) and(4) dx*dlambda>=lambda^2/4*pi\n",
+ "#Or dx=lambda^2/(4*pi*dlambda)\n",
+ "dx=lambda2**2/(4*pi*dlambda); #calculation of uncertainty in the position\n",
+ "\n",
+ "#result\n",
+ "print\"The uncertainty in the position of X-ray photon is dx=\",dx,\"m\";\n",
+ "print \"Note: wavelength accuracy is given as 1 in 1E8 but in book solution has used 1 in 1E6 \"\n",
+ "print \" ANSWEER IS WRONG IN THE TEXTBOOK\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The uncertainty in the position of X-ray photon is dx= 7.96178343949e-16 m\n",
+ "Note: wavelength accuracy is given as 1 in 1E8 but in book solution has used 1 in 1E6 \n",
+ " ANSWEER IS WRONG IN THE TEXTBOOK\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.20 , Page no:111"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "dt=1*10**-8; #in sec (average life time)\n",
+ "pi=3.14; #value of pi used in the solution\n",
+ "\n",
+ "#calculate\n",
+ "#Since dE*dt>=h/(4*pi) (uncertainty relation for energy)\n",
+ "#and E=h*v v is the frequency\n",
+ "#therefore we have dv>=1/(4*pi*dt)\n",
+ "dv=1/(4*pi*dt); #calculation of minimum uncertainty in the frequency\n",
+ "\n",
+ "#result\n",
+ "print\"The minimum uncertainty in the frequency of the photon is dv=\",dv,\"sec^-1\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The minimum uncertainty in the frequency of the photon is dv= 7961783.43949 sec^-1\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.21 , Page no:111"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "dt=1E-12; #in sec (average life time)\n",
+ "h=6.63E-34; #in J-s (Planck'c constant)\n",
+ "pi=3.14; #value of pi used in the solution\n",
+ "e=1.6*1E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "#Since dE*dt>=h/(4*pi) (uncertainty relation for energy)\n",
+ "dE=h/(4*pi*dt); #calculation of minimum uncertainty in the energy\n",
+ "dE1=dE/e; #changing unit from J to eV\n",
+ "\n",
+ "#result\n",
+ "print\"The uncertainty in the energy of the photon is dE=\",dE,\"J\";\n",
+ "print\"\\t\\t\\t\\t\\t =\",dE1,\"eV\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The uncertainty in the energy of the photon is dE= 5.27866242038e-23 J\n",
+ "\t\t\t\t\t = 0.000329916401274 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.22 , Page no:111"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "dT=2.5E-14; #in sec (average life time)\n",
+ "h=6.63E-34; #in J-s (Planck'c constant)\n",
+ "pi=3.14; #value of pi used in the solution\n",
+ "e=1.6*1E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "#Since dE*dt>=h/(4*pi) (uncertainty relation for energy)\n",
+ "dE=h/(4*pi*dT); #calculation of minimum uncertainty in the energy\n",
+ "dE1=dE/e; #changing unit from J to eV\n",
+ "\n",
+ "#result\n",
+ "print\"The uncertainty in the energy of the photon is dE=\",dE,\"J\";\n",
+ "print\"\\t\\t\\t\\t\\t =\",dE1,\"eV\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The uncertainty in the energy of the photon is dE= 2.11146496815e-21 J\n",
+ "\t\t\t\t\t = 0.013196656051 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.23 , Page no:112"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "a=2; #in Angstrom (length of the box)\n",
+ "m=9.1E-31; #in Kg (mass of electron)\n",
+ "h=6.626E-34; #in J-s (Planck'c constant)\n",
+ "n2=2; n4=4; #two quantum states\n",
+ "e=1.6*1E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "a=a*1E-10; #since a is in Angstrom \n",
+ "#Since E_n=n^2*h^2/(8*m*a^2) (Energy corresponding to nth quantum state)\n",
+ "E2=n2**2*h**2/(8*m*a**2); #calculation of energy corresponding to the 2nd quantum state\n",
+ "E21=E2/e; #changing unit from J to eV\n",
+ "E4=n4**2*h**2/(8*m*a**2); #calculation of energy corresponding to the 4nd quantum state\n",
+ "E41=E4/e; #changing unit from J to eV\n",
+ "\n",
+ "#result\n",
+ "print\"The energy corresponding to the 2nd quantum state is E2=\",E2,\"J\";\n",
+ "print\"\\t\\t\\t\\t\\t\\t =\",E21,\"eV\";\n",
+ "print\"The energy corresponding to the 4nd quantum state is E4=\",E4,\"J\";\n",
+ "print\"\\t\\t\\t\\t\\t\\t =\",E41,\"eV\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The energy corresponding to the 2nd quantum state is E2= 6.0307521978e-18 J\n",
+ "\t\t\t\t\t\t = 37.6922012363 eV\n",
+ "The energy corresponding to the 4nd quantum state is E4= 2.41230087912e-17 J\n",
+ "\t\t\t\t\t\t = 150.768804945 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.24 , Page no:113"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "a=1E-10; #in m (width of the well)\n",
+ "m=9.1E-31; #in Kg (mass of electron)\n",
+ "h=6.626E-34; #in J-s (Planck'c constant)\n",
+ "n1=1; n2=2; n3=3; #ground and first two excited states\n",
+ "e=1.6*1E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "#Since E_n=n^2*h^2/(8*m*a^2) (Energy corresponding to nth quantum state)\n",
+ "E1=n1**2*h**2/(8*m*a**2); #calculation of energy corresponding to the Ground state\n",
+ "E11=E1/e; #changing unit from J to eV\n",
+ "E2=n2**2*h**2/(8*m*a**2); #calculation of energy corresponding to the 1st excited state\n",
+ "E21=E2/e; #changing unit from J to eV\n",
+ "E3=n3**2*h**2/(8*m*a**2); #calculation of energy corresponding to the 2nd excited state\n",
+ "E31=E3/e; #changing unit from J to eV\n",
+ "\n",
+ "#result\n",
+ "print\"The energy corresponding to the ground state is E1=\",E1,\"J\";\n",
+ "print\"\\t\\t\\t\\t\\t =\",E11,\"eV\";\n",
+ "print\"The energy corresponding to the 1st excited state is E2=\",E2,\"\";\n",
+ "print\"\\t\\t\\t\\t\\t =\",E21,\"eV\";\n",
+ "print\"The energy corresponding to the 2nd excited state is E3=\",E3,\"J\";\n",
+ "print\"\\t\\t\\t\\t\\t =\",E31,\"eV\";\n",
+ "print \" There is slight variation in the answer due to round off error\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The energy corresponding to the ground state is E1= 6.0307521978e-18 J\n",
+ "\t\t\t\t\t = 37.6922012363 eV\n",
+ "The energy corresponding to the 1st excited state is E2= 2.41230087912e-17 \n",
+ "\t\t\t\t\t = 150.768804945 eV\n",
+ "The energy corresponding to the 2nd excited state is E3= 5.42767697802e-17 J\n",
+ "\t\t\t\t\t = 339.229811126 eV\n",
+ " There is slight variation in the answer due to round off error\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.25 , Page no:113"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "dx=1E-8; #in m (length of box or uncertainty in the position)\n",
+ "h=6.626E-34; #in J-s (Planck'c constant)\n",
+ "m=9.1E-31; #in Kg (mass of electron)\n",
+ "\n",
+ "#calculate\n",
+ "#From uncertainty principle dx*dp=h and dp=m*dv\n",
+ "#therefore we have\n",
+ "dv=h/(m*dx); #calculation of minimum uncertainty in the velocity\n",
+ "dv1=dv*1E-3; #changing unit from m/s to Km/s\n",
+ "\n",
+ "#result\n",
+ "print\"The minimum uncertainty in the velocity of electron i dv=\",dv,\"m/s\";\n",
+ "print\"\\t\\t\\t\\t\\t\\t =\",round(dv1,2),\"Km/s\";\n",
+ "print \"Note: There is slight variation in the answer due to round off error\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The minimum uncertainty in the velocity of electron i dv= 72813.1868132 m/s\n",
+ "\t\t\t\t\t\t = 72.81 Km/s\n",
+ "Note: There is slight variation in the answer due to round off error\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.26 , Page no:113"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "a=4E-10; #in m (length of the box)\n",
+ "m=9.1E-31; #in Kg (mass of electron)\n",
+ "h=6.626E-34; #in J-s (Planck'c constant)\n",
+ "n1=1; #ground state\n",
+ "e=1.6*1E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "#Since E_n=n^2*h^2/(8*m*a^2) (Energy corresponding to nth quantum state)\n",
+ "E1=n1**2*h**2/(8*m*a**2); #calculation of energy corresponding to the ground state\n",
+ "E11=E1/e; #changing unit from J to eV\n",
+ "\n",
+ "#result\n",
+ "print\"The minimum energy of electron is E1=\",E1,\"J (Note: The answer in the book is wrong due to printing error)\";\n",
+ "print\"\\t\\t\\t\\t =\",round(E11,3),\"eV\";\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The minimum energy of electron is E1= 3.76922012363e-19 J (Note: The answer in the book is wrong due to printing error)\n",
+ "\t\t\t\t = 2.356 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Applied_Physics/Chapter_05_Principles_of_Quantum_Mechanics_1.ipynb b/Applied_Physics/Chapter_05_Principles_of_Quantum_Mechanics_1.ipynb new file mode 100755 index 00000000..66041c45 --- /dev/null +++ b/Applied_Physics/Chapter_05_Principles_of_Quantum_Mechanics_1.ipynb @@ -0,0 +1,1258 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:43369901ed597bc7502d54a31a362249abbf2812c557cb432bc86adf72bf30d7"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 5:Principles of Quantum Mechanics"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.1 , Page no:102"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "lambda1=1; #in Angstrom (wavelength)\n",
+ "m=1.67E-27; #in Kg (mass of neutron)\n",
+ "h=6.625E-34; #in J-s (Planck's constant)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "lambda2=lambda1*1E-10; #since lambda is in Angstrom\n",
+ "#Since lambda=h/(m*v)\n",
+ "#Therefore we have\n",
+ "v=h/(m*lambda2); #calculation of velocity\n",
+ "K=m*v**2/2; #calculation of kinetic energy\n",
+ "K1=K/e; #changing unit fro J to eV\n",
+ "\n",
+ "#result\n",
+ "print\"The velocity is v=\",round(v,3),\"m/s\";\n",
+ "print\"The kinetic energy is K=\",'%.3E'%K,\"J\";\n",
+ "print\"\\t\\t =\",'%.3E'%K1,\"eV\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The velocity is v= 3967.066 m/s\n",
+ "The kinetic energy is K= 1.314E-20 J\n",
+ "\t\t = 8.213E-02 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.2 , Page no:103"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "K=50; #in eV (Kinetic energy)\n",
+ "m0=9.1E-31; #in Kg (mass of electron)\n",
+ "h=6.625E-34; #in J-s (Planck's constant)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "K=K*e; #changing unit from eV to J\n",
+ "#Since K=m*v^2/2\n",
+ "#Therefore v=sqrt(2*K/m)\n",
+ "#Since lambda=h/(m*v)\n",
+ "#Therefore we have\n",
+ "lambda1=h/math.sqrt(2*m0*K); #calculation of wavelength\n",
+ "lambda2=lambda1*1E10; #changing unit from m to Angstrom\n",
+ "\n",
+ "#result\n",
+ "print\"The wavelength is =\",'%.3E'%lambda1,\"m\";\n",
+ "print\"\\t\\t =\",round(lambda2,3),\"Angstrom\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The wavelength is = 1.736E-10 m\n",
+ "\t\t = 1.736 Angstrom\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.3 , Page no:104"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "E=2000; #in eV (Kinetic energy)\n",
+ "m=9.1E-31; #in Kg (mass of electron)\n",
+ "h=6.625E-34; #in J-s (Planck's constant)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "E=E*e; #changing unit from eV to J\n",
+ "#Since E=m*v^2/2\n",
+ "#Therefore v=sqrt(2*E/m)\n",
+ "#Since lambda=h/(m*v)\n",
+ "#Therefore we have\n",
+ "lambda1=h/math.sqrt(2*m*E); #calculation of wavelength\n",
+ "lambda2=lambda1*1E9; #changing unit from m to nanometer\n",
+ "\n",
+ "#result\n",
+ "print\"The wavelength is =\",'%.3E'%lambda1,\"m\";\n",
+ "print\"\\t\\t =\",round(lambda2,4),\"nm\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The wavelength is = 2.745E-11 m\n",
+ "\t\t = 0.0275 nm\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.4 , Page no:104"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "m_e=9.1E-31; #in Kg (mass of electron)\n",
+ "m_n=1.676E-27; #in Kg (mass of neutron)\n",
+ "h=6.625E-34; #in J-s (Planck's constant)\n",
+ "c=3E8; #in m/s (velocity of light)\n",
+ "\n",
+ "#calculate\n",
+ "E_e=m_e*c**2; #rest mass energy of electron\n",
+ "E_n=2*E_e; #given (kinetic energy of neutron)\n",
+ "#Since K=m*v^2/2\n",
+ "#Therefore v=sqrt(2*K/m)\n",
+ "#Since lambda=h/(m*v)\n",
+ "#Therefore we have\n",
+ "lambda1=h/math.sqrt(2*m_n*E_n); #calculation of wavelength\n",
+ "lambda2=lambda1*1E10; #changing unit from m to Angstrom\n",
+ "\n",
+ "#result\n",
+ "print\"The wavelength is =\",'%.3E'%lambda1,\"m\";\n",
+ "print\"\\t\\t =\",'%.3E'%lambda2,\"Angstrom\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The wavelength is = 2.827E-14 m\n",
+ "\t\t = 2.827E-04 Angstrom\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5 , Page no:104"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "V=1600; #in V (Potential)\n",
+ "\n",
+ "#calculate\n",
+ "lambda1=12.27/math.sqrt(V); #calculation of wavelength in Angstrom\n",
+ "\n",
+ "#result\n",
+ "print\"The wavelength is =\",round(lambda1,3),\"Angstrom\";\n",
+ "print \"Note : The answer in the book is wrong \""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The wavelength is = 0.307 Angstrom\n",
+ "Note : The answer in the book is wrong \n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.6 , Page no:105"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "E=1000; #in eV (Kinetic energy of photon)\n",
+ "K=1000; #in eV (Kinetic energy of electron)\n",
+ "m0=9.1E-31; #in Kg (mass of electron)\n",
+ "h=6.6E-34; #in J-s (Planck's constant)\n",
+ "c=3E8; #in m/s (velocity of light)\n",
+ "e=1.6E-19; #in C (charge on electron)\n",
+ "\n",
+ "#calculate\n",
+ "E=E*e; #changing unit from eV to J\n",
+ "lambda_p=h*c/E; #For photon E=hc/lambda\n",
+ "lambda_p1=lambda_p*1E10; #changing unit from m to Angstrom\n",
+ "#Since K=m*v^2/2\n",
+ "#Therefore v=sqrt(2*K/m)\n",
+ "#Since lambda=h/(m*v)\n",
+ "#Therefore we have\n",
+ "K=K*e; #changing unit from eV to J\n",
+ "lambda_e=h/math.sqrt(2*m0*K); #calculation of wavelength\n",
+ "lambda_e1=lambda_e*1E10; #changing unit from m to Angstrom\n",
+ "\n",
+ "#result\n",
+ "print\"For photon,the wavelength is =\",'%.3E'%lambda_p,\"m\";\n",
+ "print\"\\t\\t\\t =\",round(lambda_p1,1),\"Angstrom\";\n",
+ "print\"For electron,the wavelength is =\",'%.3E'%lambda_e,\"m\";\n",
+ "print\"\\t\\t\\t =\",round(lambda_e1,2),\"Angstrom\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "For photon,the wavelength is = 1.238E-09 m\n",
+ "\t\t\t = 12.4 Angstrom\n",
+ "For electron,the wavelength is = 3.868E-11 m\n",
+ "\t\t\t = 0.39 Angstrom\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.7 , Page no:105"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "lambda1=1.66*10**-10; #in m (wavelength)\n",
+ "m=9.1*10**-31; #in Kg (mass of electron)\n",
+ "h=6.626*10**-34; #in J-s (Planck's constant)\n",
+ "e=1.6E-19; #in C (charge on electron)\n",
+ "\n",
+ "#calculate\n",
+ "#Since lambda=h/(m*v)\n",
+ "#Therefore we have\n",
+ "v=h/(m*lambda1); #calculation of velocity\n",
+ "K=m*v**2/2; #calculation of kinetic energy\n",
+ "K1=K/e; #changing unit from J to eV\n",
+ "\n",
+ "#result\n",
+ "print\"The velocity of electron is v=\",'%.2E'%v,\"m/s\";\n",
+ "print\"The kinetic energy is K=\",'%.2E'%K,\"J\";\n",
+ "print\"\\t\\t =\",round(K1,3),\"eV\";\n",
+ "print \"Note: The answer in the book for kinetic energy is wrong \""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The velocity of electron is v= 4.39E+06 m/s\n",
+ "The kinetic energy is K= 8.75E-18 J\n",
+ "\t\t = 54.714 eV\n",
+ "Note: The answer in the book for kinetic energy is wrong \n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.8 , Page no:106"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "T=400; #in K (temperature)\n",
+ "m=6.7E-27; #in Kg (mass of He-atom)\n",
+ "h=6.6E-34; #in J-s (Planck's constant)\n",
+ "k=1.376E-23; #in J/degree (Boltzmann constant)\n",
+ "\n",
+ "#calculate\n",
+ "#Since lambda=h/(m*v)\n",
+ "#E=mv^2/2;\n",
+ "#Therefore lambda=h/sqrt(2*m*E)\n",
+ "#E=kT\n",
+ "#Therefore lambda=h/sqrt(2*m*k*T)\n",
+ "lambda1=h/math.sqrt(2*m*k*T)\n",
+ "lambda2=lambda1*1E10; #changing unit from m to Angstrom\n",
+ "\n",
+ "#result\n",
+ "print\"The de-Broglie wavelength of He-atom is =\",'%.2E'%lambda1,\"m\";\n",
+ "print\"\\t\\t\\t\\t =\",round(lambda2,4),\"Angstrom\"; "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The de-Broglie wavelength of He-atom is = 7.69E-11 m\n",
+ "\t\t\t\t = 0.7685 Angstrom\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.9 , Page no:106"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "m_e=9.1E-31; #in Kg (mass of electron)\n",
+ "m_p=1.6E-27; #in Kg (mass of proton)\n",
+ "h=6.626E-34; #in J-s (Planck's constant)\n",
+ "c=3E8; #in m/s (velocity of light)\n",
+ "\n",
+ "#calculate\n",
+ "E=m_e*c**2; #in J (rest energy of electron)\n",
+ "#Since lambda=h/(m*v)\n",
+ "#E=mv^2/2;\n",
+ "#herefore lambda=h/sqrt(2*m*E)\n",
+ "#Also E=m_e*c^2;\n",
+ "#therefore lambda=h/sqrt(2*m_p*m_e*c^2)\n",
+ "lambda1=h/math.sqrt(2*m_p*m_e*c**2); #calculation of wavelength\n",
+ "lambda2=lambda1*1*10**10; #changing unit from m to Angstrom\n",
+ "\n",
+ "#result\n",
+ "print\"The de-Broglie wavelength of proton is =\",'%.2E'%lambda1,\"m\";\n",
+ "print\"\\t\\t\\t\\t =\",'%.2E'%lambda2,\"Angstrom\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The de-Broglie wavelength of proton is = 4.09E-14 m\n",
+ "\t\t\t\t = 4.09E-04 Angstrom\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.10 , Page no:106"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "V=10000; #in V (Potential)\n",
+ "\n",
+ "#calculate\n",
+ "lambda1=12.27/math.sqrt(V); #calculation of wavelength in Angstrom\n",
+ "\n",
+ "#result\n",
+ "print\"The wavelength is =\",round(lambda1,3),\"Angstrom\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The wavelength is = 0.123 Angstrom\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.11 , Page no:107"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "V=100; #in V (Potential)\n",
+ "n=1; #order of diffraction\n",
+ "d=2.15; #in Angstrom (lattice spacing)\n",
+ "\n",
+ "#calculate\n",
+ "lambda1=12.27/math.sqrt(V); #calculation of wavelength in Angstrom\n",
+ "#Since 2*d*sind(theta)=n*lambda\n",
+ "#therefore we have\n",
+ "theta=math.asin(n*lambda1/(2*d)); #calculation of glancing angle\n",
+ "theta1=theta*180/3.14;\n",
+ "\n",
+ "#result\n",
+ "print\"The wavelength is =\",round(lambda1,3),\"Angstrom\";\n",
+ "print\"The glancing angle is =\",round(theta1,1),\"degree\";\n",
+ "print \"Note: In question V=100 eV but the solution is using V=100V \"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The wavelength is = 1.227 Angstrom\n",
+ "The glancing angle is = 16.6 degree\n",
+ "Note: In question V=100 eV but the solution is using V=100V \n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.12 , Page no:107"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "V=344; #in V (Potential)\n",
+ "n=1; #order of diffraction\n",
+ "theta=60; #in degree (glancing angle)\n",
+ "\n",
+ "#calculate\n",
+ "lambda1=12.27/math.sqrt(V); #calculation of wavelength in Angstrom\n",
+ "#Since 2*d*sind(theta)=n*lambda\n",
+ "#therefore we have\n",
+ "d=n*lambda1/(2*math.sin(math.radians(theta))); #calculation of spacing constant\n",
+ "\n",
+ "#result\n",
+ "print\"The wavelength is =\",round(lambda1,3),\"Angstrom\";\n",
+ "print\"The spacing of the crystal is =\",round(d,4),\"Angstrom\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The wavelength is = 0.662 Angstrom\n",
+ "The spacing of the crystal is = 0.3819 Angstrom\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.13 , Page no:107"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "r=0.53*10**-10; #in m (radius of first Bohr orbit)\n",
+ "h=6.6*10**-34; #in J-s (Planck's constant)\n",
+ "m=9.1*10**-31; #in Kg (mass of electron)\n",
+ "n=1; #First Bohr orbit\n",
+ "pi=3.14; #value of pi used in the solution\n",
+ "\n",
+ "#calculate\n",
+ "#Since 2*pi*r=n*lambda and lambda=h/(m*v)\n",
+ "#Threfore we have v=h*n/(2*pi*r*m)\n",
+ "v=h*n/(2*pi*r*m); #calculation of velocity\n",
+ "\n",
+ "#result\n",
+ "print\"The velocity of electron is =\",'%.3E'%v,\"m/s\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The velocity of electron is = 2.179E+06 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.14 , Page no:108"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "dx=0.2; #in Angstrom (uncertainty in the position)\n",
+ "h=6.6E-34; #in J-s (Planck's constant)\n",
+ "m0=9.1E-31; #in Kg (mass of electron)\n",
+ "pi=3.14; #value of pi used in the solution\n",
+ "\n",
+ "#calculate\n",
+ "dx=dx*1E-10; #since dx is in Angstrom\n",
+ "#Since dx*dp=h/4*pi (uncertainty relation)\n",
+ "dp=h/(4*pi*dx); #calculation of uncertainty in the momentum\n",
+ "#since dp=m*dv\n",
+ "dv=dp/m0; #calculation of uncertainty in the velocity\n",
+ "\n",
+ "#result\n",
+ "print\"The uncertainty in the momentum is dp=\",'%.3E'%dp,\"Kg-m/\";\n",
+ "print\"The uncertainty in the velocity is dv=\",'%.3E'%dv,\"m/s\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The uncertainty in the momentum is dp= 2.627E-24 Kg-m/\n",
+ "The uncertainty in the velocity is dv= 2.887E+06 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.15 , Page no:108"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "m_e=9.1E-31; #in Kg (mass of electron)\n",
+ "m_p=1.67E-27; #in Kg (mass of proton)\n",
+ "dx_p=1; #in nanometer (uncertainty in position of electron)\n",
+ "dx_n=1; #in nanometer (uncertainty in position of proton)\n",
+ "\n",
+ "#calculate\n",
+ "#since dp=h/(4*pi*dx)\n",
+ "#since h/(4*pi) is constant and dx is same for electron and proton \n",
+ "#therefor both electron and proton have same uncertainty in the momentum\n",
+ "#since dv=dp/m and dp is same for both\n",
+ "#therefore dv_e/dv_p=m_p/m_e\n",
+ "#therefore\n",
+ "K=m_p/m_e; #ratio of uncertainty in the velocity of electron and proton\n",
+ "\n",
+ "#result\n",
+ "print\"The ratio of uncertainty in the velocity of electron to that of proton is =\",round(K);"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The ratio of uncertainty in the velocity of electron to that of proton is = 1835.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.16 , Page no:108"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "dx=5*10**-15; #in m (radius of nucleus or uncertainty in the position)\n",
+ "h=6.6*10**-34; #in J-s (Planck's constant)\n",
+ "m=1.67E-27; #in Kg (mass of proton)\n",
+ "pi=3.14; #value of pi used in the solution\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "#Since dx*dp=h/4*pi (uncertainty relation)\n",
+ "dp=h/(4*pi*dx); #calculation of uncertainty in the momentum\n",
+ "p=dp; #minimum value of momentum to calculate mimimum kinetic energy\n",
+ "K=(p**2)/(2*m); #calculation of minimum kinetic energy of proton\n",
+ "K1=K/e; #changing unit from J to eV\n",
+ "K2=K1/1E6; #changing unit from eV to MeV\n",
+ "\n",
+ "#result\n",
+ "print\"The minimum uncertainty in the momentum of proton is dp =\",'%.3E'%dp,\"Kg-m/s\";\n",
+ "print\"The minimum kinetic energy of proton is K=\",'%.3E'%K,\"J\";\n",
+ "print\"\\t\\t\\t\\t \\t =\",'%.3E'%K1,\"eV\";\n",
+ "print\"\\t\\t\\t\\t\\t =\",round(K2,1),\"MeV\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The minimum uncertainty in the momentum of proton is dp = 1.051E-20 Kg-m/s\n",
+ "The minimum kinetic energy of proton is K= 3.307E-14 J\n",
+ "\t\t\t\t \t = 2.067E+05 eV\n",
+ "\t\t\t\t\t = 0.2 MeV\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.17 , Page no:109"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "K=1; #in KeV (kinetic energy of electron)\n",
+ "dx=1; #in Angstrom (uncertainty in the position)\n",
+ "h=6.63E-34; #in J-s (Planck's constant)\n",
+ "m=9.1E-31; #in Kg (mass of electron)\n",
+ "pi=3.14; #value of pi used in the solution\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "dx=dx*1E-10; #since dx is in Angstrom\n",
+ "#Since dx*dp=h/4*pi (uncertainty relation)\n",
+ "dp=h/(4*pi*dx); #calculation of uncertainty in the momentum\n",
+ "K=K*1E3*1.6E-19; #changing unit from KeV to J\n",
+ "p=math.sqrt(2*m*K); #calculation of momentum \n",
+ "poc=(dp/p)*100; #calculation of percentage of uncertainty\n",
+ "\n",
+ "#result\n",
+ "print\"The uncertainty in the momentum of electron is dp=\",'%.3E'%dp,\"Kg-m/s\";\n",
+ "print\"The momentum of electron is p=\",'%.3E'%p,\"Kg-m/s\";\n",
+ "print\"The percentage of uncertainty in the momentum is =\",round(poc,1);"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The uncertainty in the momentum of electron is dp= 5.279E-25 Kg-m/s\n",
+ "The momentum of electron is p= 1.706E-23 Kg-m/s\n",
+ "The percentage of uncertainty in the momentum is = 3.1\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.18 , Page no:109"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "v=6.6E4; #m/s (speed of electron)\n",
+ "poc=0.01; #percentage of uncertainty\n",
+ "h=6.63E-34; #in J-s (Planck's constant)\n",
+ "m=9E-31; #in Kg (mass of electron)\n",
+ "pi=3.14; #value of pi used in the solution\n",
+ "\n",
+ "#calculate\n",
+ "p=m*v; #calculation of momentum\n",
+ "dp=(poc/100)*p; #calculation of uncertainty in the momentum\n",
+ "#Since dx*dp=h/4*pi (uncertainty relation)\n",
+ "dx=h/(4*pi*dp); #calculation of uncertainty in the position\n",
+ "\n",
+ "#result\n",
+ "print\"The momentum of electron is p=\",'%.3E'%p,\"Kg-m/s\";\n",
+ "print\"The uncertainty in the momentum of electron is dp=\",'%.3E'%dp,\"Kg-m/s\";\n",
+ "print\"The uncertainty in the position of electron is dx=\",'%.3E'%dx,\"Kg-m/s\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The momentum of electron is p= 5.940E-26 Kg-m/s\n",
+ "The uncertainty in the momentum of electron is dp= 5.940E-30 Kg-m/s\n",
+ "The uncertainty in the position of electron is dx= 8.887E-06 Kg-m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.19 , Page no:111"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "lambda1=1; #in Angstrom (wavelength)\n",
+ "pi=3.14; #value of pi used in the solution\n",
+ "dlambda=1E-6; #uncertainty in wavelength\n",
+ "\n",
+ "#calculate\n",
+ "lambda2=lambda1*1E-10; #sinc lambda is in Angstrom\n",
+ "#By uncertainty principle, dx*dp>=h/(4*pi) --(1)\n",
+ "#since p=h/lambda -----(2)\n",
+ "#Or p*lambda=h \n",
+ "#diffrentiting this equation\n",
+ "#p*dlambda+lambda*dp=0\n",
+ "#dp=-p*dlambda/lambda ----(3)\n",
+ "#from (2) and (3) dp=-h*dlambda/lambda^2 ---(4)\n",
+ "#from (1) and(4) dx*dlambda>=lambda^2/4*pi\n",
+ "#Or dx=lambda^2/(4*pi*dlambda)\n",
+ "dx=lambda2**2/(4*pi*dlambda); #calculation of uncertainty in the position\n",
+ "\n",
+ "#result\n",
+ "print\"The uncertainty in the position of X-ray photon is dx=\",'%.3E'%dx,\"m\";\n",
+ "print \"Note: wavelength accuracy is given as 1 in 1E8 but in book solution has used 1 in 1E6 \"\n",
+ "print \" ANSWEER IS WRONG IN THE TEXTBOOK\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The uncertainty in the position of X-ray photon is dx= 7.962E-16 m\n",
+ "Note: wavelength accuracy is given as 1 in 1E8 but in book solution has used 1 in 1E6 \n",
+ " ANSWEER IS WRONG IN THE TEXTBOOK\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.20 , Page no:111"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "dt=1*10**-8; #in sec (average life time)\n",
+ "pi=3.14; #value of pi used in the solution\n",
+ "\n",
+ "#calculate\n",
+ "#Since dE*dt>=h/(4*pi) (uncertainty relation for energy)\n",
+ "#and E=h*v v is the frequency\n",
+ "#therefore we have dv>=1/(4*pi*dt)\n",
+ "dv=1/(4*pi*dt); #calculation of minimum uncertainty in the frequency\n",
+ "\n",
+ "#result\n",
+ "print\"The minimum uncertainty in the frequency of the photon is dv=\",'%.3E'%dv,\"sec^-1\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The minimum uncertainty in the frequency of the photon is dv= 7.962E+06 sec^-1\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.21 , Page no:111"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "dt=1E-12; #in sec (average life time)\n",
+ "h=6.63E-34; #in J-s (Planck'c constant)\n",
+ "pi=3.14; #value of pi used in the solution\n",
+ "e=1.6*1E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "#Since dE*dt>=h/(4*pi) (uncertainty relation for energy)\n",
+ "dE=h/(4*pi*dt); #calculation of minimum uncertainty in the energy\n",
+ "dE1=dE/e; #changing unit from J to eV\n",
+ "\n",
+ "#result\n",
+ "print\"The uncertainty in the energy of the photon is dE=\",'%.3E'%dE,\"J\";\n",
+ "print\"\\t\\t\\t\\t\\t =\",'%.3E'%dE1,\"eV\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The uncertainty in the energy of the photon is dE= 5.279E-23 J\n",
+ "\t\t\t\t\t = 3.299E-04 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.22 , Page no:111"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "dT=2.5E-14; #in sec (average life time)\n",
+ "h=6.63E-34; #in J-s (Planck'c constant)\n",
+ "pi=3.14; #value of pi used in the solution\n",
+ "e=1.6*1E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "#Since dE*dt>=h/(4*pi) (uncertainty relation for energy)\n",
+ "dE=h/(4*pi*dT); #calculation of minimum uncertainty in the energy\n",
+ "dE1=dE/e; #changing unit from J to eV\n",
+ "\n",
+ "#result\n",
+ "print\"The uncertainty in the energy of the photon is dE=\",'%.3E'%dE,\"J\";\n",
+ "print\"\\t\\t\\t\\t\\t =\",'%.3E'%dE1,\"eV\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The uncertainty in the energy of the photon is dE= 2.111E-21 J\n",
+ "\t\t\t\t\t = 1.320E-02 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.23 , Page no:112"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "a=2; #in Angstrom (length of the box)\n",
+ "m=9.1E-31; #in Kg (mass of electron)\n",
+ "h=6.626E-34; #in J-s (Planck'c constant)\n",
+ "n2=2; n4=4; #two quantum states\n",
+ "e=1.6*1E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "a=a*1E-10; #since a is in Angstrom \n",
+ "#Since E_n=n^2*h^2/(8*m*a^2) (Energy corresponding to nth quantum state)\n",
+ "E2=n2**2*h**2/(8*m*a**2); #calculation of energy corresponding to the 2nd quantum state\n",
+ "E21=E2/e; #changing unit from J to eV\n",
+ "E4=n4**2*h**2/(8*m*a**2); #calculation of energy corresponding to the 4nd quantum state\n",
+ "E41=E4/e; #changing unit from J to eV\n",
+ "\n",
+ "#result\n",
+ "print\"The energy corresponding to the 2nd quantum state is E2=\",'%.3E'%E2,\"J\";\n",
+ "print\"\\t\\t\\t\\t\\t\\t =\",'%.3E'%E21,\"eV\";\n",
+ "print\"The energy corresponding to the 4nd quantum state is E4=\",'%.3E'%E4,\"J\";\n",
+ "print\"\\t\\t\\t\\t\\t\\t =\",'%.3E'%E41,\"eV\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The energy corresponding to the 2nd quantum state is E2= 6.031E-18 J\n",
+ "\t\t\t\t\t\t = 3.769E+01 eV\n",
+ "The energy corresponding to the 4nd quantum state is E4= 2.412E-17 J\n",
+ "\t\t\t\t\t\t = 1.508E+02 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.24 , Page no:113"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "a=1E-10; #in m (width of the well)\n",
+ "m=9.1E-31; #in Kg (mass of electron)\n",
+ "h=6.626E-34; #in J-s (Planck'c constant)\n",
+ "n1=1; n2=2; n3=3; #ground and first two excited states\n",
+ "e=1.6*1E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "#Since E_n=n^2*h^2/(8*m*a^2) (Energy corresponding to nth quantum state)\n",
+ "E1=n1**2*h**2/(8*m*a**2); #calculation of energy corresponding to the Ground state\n",
+ "E11=E1/e; #changing unit from J to eV\n",
+ "E2=n2**2*h**2/(8*m*a**2); #calculation of energy corresponding to the 1st excited state\n",
+ "E21=E2/e; #changing unit from J to eV\n",
+ "E3=n3**2*h**2/(8*m*a**2); #calculation of energy corresponding to the 2nd excited state\n",
+ "E31=E3/e; #changing unit from J to eV\n",
+ "\n",
+ "#result\n",
+ "print\"The energy corresponding to the ground state is E1=\",'%.3E'%E1,\"J\";\n",
+ "print\"\\t\\t\\t\\t\\t =\",'%.3E'%E11,\"eV\";\n",
+ "print\"The energy corresponding to the 1st excited state is E2=\",'%.3E'%E2,\"\";\n",
+ "print\"\\t\\t\\t\\t\\t =\",'%.3E'%E21,\"eV\";\n",
+ "print\"The energy corresponding to the 2nd excited state is E3=\",'%.3E'%E3,\"J\";\n",
+ "print\"\\t\\t\\t\\t\\t =\",'%.3E'%E31,\"eV\";\n",
+ "print \" There is slight variation in the answer due to round off error\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The energy corresponding to the ground state is E1= 6.031E-18 J\n",
+ "\t\t\t\t\t = 3.769E+01 eV\n",
+ "The energy corresponding to the 1st excited state is E2= 2.412E-17 \n",
+ "\t\t\t\t\t = 1.508E+02 eV\n",
+ "The energy corresponding to the 2nd excited state is E3= 5.428E-17 J\n",
+ "\t\t\t\t\t = 3.392E+02 eV\n",
+ " There is slight variation in the answer due to round off error\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.25 , Page no:113"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "dx=1E-8; #in m (length of box or uncertainty in the position)\n",
+ "h=6.626E-34; #in J-s (Planck'c constant)\n",
+ "m=9.1E-31; #in Kg (mass of electron)\n",
+ "\n",
+ "#calculate\n",
+ "#From uncertainty principle dx*dp=h and dp=m*dv\n",
+ "#therefore we have\n",
+ "dv=h/(m*dx); #calculation of minimum uncertainty in the velocity\n",
+ "dv1=dv*1E-3; #changing unit from m/s to Km/s\n",
+ "\n",
+ "#result\n",
+ "print\"The minimum uncertainty in the velocity of electron i dv=\",'%.3E'%dv,\"m/s\";\n",
+ "print\"\\t\\t\\t\\t\\t\\t =\",round(dv1,2),\"Km/s\";\n",
+ "print \"Note: There is slight variation in the answer due to round off error\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The minimum uncertainty in the velocity of electron i dv= 7.281E+04 m/s\n",
+ "\t\t\t\t\t\t = 72.81 Km/s\n",
+ "Note: There is slight variation in the answer due to round off error\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.26 , Page no:113"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "a=4E-10; #in m (length of the box)\n",
+ "m=9.1E-31; #in Kg (mass of electron)\n",
+ "h=6.626E-34; #in J-s (Planck'c constant)\n",
+ "n1=1; #ground state\n",
+ "e=1.6*1E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "#Since E_n=n^2*h^2/(8*m*a^2) (Energy corresponding to nth quantum state)\n",
+ "E1=n1**2*h**2/(8*m*a**2); #calculation of energy corresponding to the ground state\n",
+ "E11=E1/e; #changing unit from J to eV\n",
+ "\n",
+ "#result\n",
+ "print\"The minimum energy of electron is E1=\",'%.3E'%E1,\"J (Note: The answer in the book is wrong due to printing error)\";\n",
+ "print\"\\t\\t\\t\\t =\",round(E11,3),\"eV\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The minimum energy of electron is E1= 3.769E-19 J (Note: The answer in the book is wrong due to printing error)\n",
+ "\t\t\t\t = 2.356 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Applied_Physics/Chapter_06_Electron_Theory_and_Band_Theory_of_Metals.ipynb b/Applied_Physics/Chapter_06_Electron_Theory_and_Band_Theory_of_Metals.ipynb new file mode 100755 index 00000000..ff063bb6 --- /dev/null +++ b/Applied_Physics/Chapter_06_Electron_Theory_and_Band_Theory_of_Metals.ipynb @@ -0,0 +1,561 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:0c3d2d354976eaba7e5a0f2e0cf4b1fd72cd55b3826ca7e7cd6acde8afff2cdf"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 6:Electron Theory and Band Theory of Metals"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.1 , Page no:146"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "n=8.5E28; #in 1/m^3 (density of electron)\n",
+ "m_e=9.11E-31; #in Kg (mass of electron)\n",
+ "k=1.38E-23; #in J/K (Boltzmann's constant)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "T=300; #in K (temperature)\n",
+ "p=1.69E-8; #in ohm-m (resistivity)\n",
+ "\n",
+ "#calculate\n",
+ "lambda1=math.sqrt(3*k*m_e*T)/(n*e**2*p); #calculation of mean free path\n",
+ "lambda2=lambda1*1E9; #changing unit from m to nanometer\n",
+ "\n",
+ "#result\n",
+ "print\"The mean free path of electron is =\",lambda1,\"m\";\n",
+ "print\"\\t\\t\\t\\t =\",round(lambda2,2),\"nm\";\n",
+ "#Note: answer in the book is wrong due to printing mistake"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The mean free path of electron is = 2.89250681437e-09 m\n",
+ "\t\t\t\t = 2.89 nm\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.2 , Page no:146"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "k=1.38E-23; #in J/K (Boltzmann's constant)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "P_E=1; #in percentage (probability that a state with an energy 0.5 eV above Fermi energy will be occupied)\n",
+ "E=0.5; #in eV (energy above Fermi level)\n",
+ "\n",
+ "#calculate\n",
+ "P_E=1/100; #changing percentage into ratio\n",
+ "E=E*e; #changing unit from eV to J\n",
+ "#P_E=1/(1+exp((E-E_F)/k*T))\n",
+ "#Rearranging this equation, we get\n",
+ "#T=(E-E_F)/k*log((1/P_E)-1)\n",
+ "#Since E-E_F has been denoted by E therefore\n",
+ "T=E/(k*math.log((1/P_E)-1));\n",
+ "\n",
+ "#result\n",
+ "print\"The temperature is T=\",round(T),\"K\";\n",
+ "# Note: There is slight variation in the answer due to logarithm function"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The temperature is T= 1262.0 K\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.3 , Page no:147"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "n=5.8E28; #in 1/m^3 (density of electron)\n",
+ "m=9.1E-31; #in Kg (mass of electron)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "p=1.54E-8; #in ohm-m (resistivity)\n",
+ "\n",
+ "#calculate\n",
+ "t=m/(n*e**2*p); #calculation of relaxation time\n",
+ "\n",
+ "#result\n",
+ "print\"The relaxation time of conduction electrons is =\",t,\"sec\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The relaxation time of conduction electrons is = 3.97972178683e-14 sec\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.4 , Page no:147"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "n=8.5E28; #in 1/m^3 (density of electron)\n",
+ "m=9.1E-31; #in Kg (mass of electron)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "sigma=6E7; #in 1/ohm-m (conductivity)\n",
+ "E_F=7; #in E=eV (Fermi energy of Copper)\n",
+ "\n",
+ "#calculate\n",
+ "E_F=E_F*e; #changing unit from eV to J\n",
+ "v_F=math.sqrt(2*E_F/m); #calculation of velocity of electrons\n",
+ "#Since sigma=n*e^2*lambda/(2*m*v_F)\n",
+ "#Therefore we have\n",
+ "lambda1=2*m*v_F*sigma/(n*e**2); #calculation of mean free path\n",
+ "lambda2=lambda1*1E10; #changing unit from m to Angstrom\n",
+ "\n",
+ "#result\n",
+ "print\"The velocity of the electrons is v_F=\",v_F,\"m/s (roundoff error)\";\n",
+ "print\"The mean free path traveled by the electrons is=\",round(lambda2),\"Angstrom\";\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The velocity of the electrons is v_F= 1568929.08111 m/s (roundoff error)\n",
+ "The mean free path traveled by the electrons is= 787.0 Angstrom\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.5 , Page no:147"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "n=6.5E28; #in 1/m^3 (density of electron)\n",
+ "m=9.1E-31; #in Kg (mass of electron)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "p=1.43E-8; #in ohm-m (resistivity)\n",
+ "\n",
+ "#calculate\n",
+ "t=m/(n*e**2*p); #calculation of relaxation time\n",
+ "\n",
+ "#result\n",
+ "print\"The relaxation time of conduction electrons is =\",t,\"sec\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The relaxation time of conduction electrons is = 3.8243006993e-14 sec\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6 , Page no:148"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "T=30; #in Celcius (temperature)\n",
+ "k=1.38E-23; #in J/K (Boltzmann's constant)\n",
+ "m_p=1.67E-27; #in Kg (mass of proton)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "T=T+273; #changing temperature from Celcius to Kelvin\n",
+ "KE=(3/2)*k*T; #calculation of average kinetic energy\n",
+ "KE1=KE/e; #changing unit from J to eV\n",
+ "m=1.008*2*m_p; #calculating mass of hydrogen gas molecule\n",
+ "c=math.sqrt(3*k*T/m); #calculation of velocity\n",
+ "\n",
+ "#result\n",
+ "print\"The average kinetic energy of gas ,molecules is KE=\",KE,\"J\";\n",
+ "print\"\\t\\t\\t\\t\\t\\t =\",KE1,\"eV (Answer is wrong in textbook)\";\n",
+ "print\"The velocity of molecules is c=\",round(c,2),\"m/s\";\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The average kinetic energy of gas ,molecules is KE= 6.2721e-21 J\n",
+ "\t\t\t\t\t\t = 0.039200625 eV (Answer is wrong in textbook)\n",
+ "The velocity of molecules is c= 1930.27 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.7 , Page no:148"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "E=10; #in eV (kinetic energy for each electron and proton)\n",
+ "m_e=9.1E-31; #in Kg (mass of electron)\n",
+ "m_p=1.67E-27; #in Kg (mass of proton)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "E=E*e; #changing unit from eV to J\n",
+ "#since E=m*v^2/2\n",
+ "#therefore v=sqrt(2E/m)\n",
+ "v_e=math.sqrt(2*E/m_e); #calculation of kinetic energy of electron\n",
+ "v_p=math.sqrt(2*E/m_p); #calculation of kinetic energy of proton\n",
+ "\n",
+ "#result\n",
+ "print\"The kinetic energy of electron is v_e=\",v_e,\"m/s\";\n",
+ "print\"The kinetic energy of proton is v_p=\",v_p,\"m/s\";\n",
+ "print \"Note: Answer is wrong in textbook\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The kinetic energy of electron is v_e= 1875228.92375 m/s\n",
+ "The kinetic energy of proton is v_p= 43774.0524132 m/s\n",
+ "Note: Answer is wrong in textbook\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.8 , Page no:149"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "I=100; #in A (current in the wire)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "A=10; #in mm^2 (cross-sectional area)\n",
+ "n=8.5E28; #in 1/m^3 (density of electron)\n",
+ "\n",
+ "#calculate\n",
+ "A=A*1E-6; #changing unit from mm^2 to m^2\n",
+ "v_d=I/(n*A*e);\n",
+ "\n",
+ "#result\n",
+ "print\"The drift velocity of free electrons is v_d=\",v_d,\"m/s\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The drift velocity of free electrons is v_d= 0.000735294117647 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.9 , Page no:149"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "I=4;#in A (current in the conductor)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "A=1E-6; #in m^2 (cross-sectional area)\n",
+ "N_A=6.02E23; #in atoms/gram-atom (Avogadro's number)\n",
+ "p=8.9; #in g/cm^3 (density)\n",
+ "M=63.6; #atomic mass of copper\n",
+ "\n",
+ "#calculate\n",
+ "n=N_A*p/M; #Calculation of density of electrons in g/cm^3\n",
+ "n1=n*1E6; #changing unit from g/cm^3 to g/m^3\n",
+ "v_d=I/(n1*A*e);\n",
+ "\n",
+ "#result\n",
+ "print\"The density of copper atoms is n=\",n,\"atoms/m^3\";\n",
+ "print\"\\t\\t\\t =\",n1,\"atoms/m^3\";\n",
+ "print\"The average drift velocity of free electrons is v_d=\",v_d,\"m/s\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The density of copper atoms is n= 8.42421383648e+22 atoms/m^3\n",
+ "\t\t\t = 8.42421383648e+28 atoms/m^3\n",
+ "The average drift velocity of free electrons is v_d= 0.000296763596999 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.10 , Page no:149"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "n=9E28; #in 1/m^3 (density of valence electrons)\n",
+ "sigma=6E7; #in mho/m (conductivity of copper)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "#Since sigma=n*e*mu therefore\n",
+ "mu=sigma/(n*e); #calculation of mobility of electron\n",
+ "\n",
+ "#result\n",
+ "print\"The mobility of electrons is =\",mu,\"m^2/V-s\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The mobility of electrons is = 0.00416666666667 m^2/V-s\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.11 , Page no:150"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "E_F=5.51; #in eV (Fermi energy in Silver)\n",
+ "k=1.38E-23; #in J/K (Boltzmann's constant)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "#part-(a)\n",
+ "Eo=(3/5)*E_F; #calculation of average energy of free electron at 0K\n",
+ "#part-(b)\n",
+ "Eo1=Eo*e; #changing unit from eV to J\n",
+ "#Since for a classical particle E=(3/2)*k*T\n",
+ "#therefroe we have\n",
+ "T=(2/3)*Eo1/k; #calculation of temperature for a classical particle (an ideal gas)\n",
+ "\n",
+ "#result\n",
+ "print\"The average energy of free electron at 0K is Eo=\",Eo,\"eV\";\n",
+ "print\"The temperature at which a classical particle have this much energy is T=\",T,\"K\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The average energy of free electron at 0K is Eo= 3.306 eV\n",
+ "The temperature at which a classical particle have this much energy is T= 25553.6231884 K\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.12 , Page no:150"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "E_F_L=4.7; #in eV (Fermi energy in Lithium)\n",
+ "E_F_M=2.35; #in eV (Fermi energy in a metal)\n",
+ "n_L=4.6E28; #in 1/m^3 (density of electron in Lithium)\n",
+ "\n",
+ "#calculate\n",
+ "#Since n=((2*m/h)^3/2)*E_F^(3/2)*(8*pi/3) and all things except E_F are constant\n",
+ "# Therefore we have n=C*E_F^(3/2) where C is proportionality constant\n",
+ "#n1/n2=(E_F_1/E_F_2)^(3/2)\n",
+ "#Therefore we have\n",
+ "n_M=n_L*(E_F_M/E_F_L); #calculation of electron density for a metal\n",
+ "\n",
+ "#result\n",
+ "print\"The lectron density for a metal is =\",n_M,\"1/m^3\";\n",
+ "print \"Note: Answer in the book is wrong \""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The lectron density for a metal is = 2.3e+28 1/m^3\n",
+ "Note: Answer in the book is wrong \n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Applied_Physics/Chapter_06_Electron_Theory_and_Band_Theory_of_Metals_1.ipynb b/Applied_Physics/Chapter_06_Electron_Theory_and_Band_Theory_of_Metals_1.ipynb new file mode 100755 index 00000000..d52f3a45 --- /dev/null +++ b/Applied_Physics/Chapter_06_Electron_Theory_and_Band_Theory_of_Metals_1.ipynb @@ -0,0 +1,561 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:3c7e08d83e902e2949e4fdcd2bc98d527fe589fabb9636cea78c35c42abb8038"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 6:Electron Theory and Band Theory of Metals"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.1 , Page no:146"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "n=8.5E28; #in 1/m^3 (density of electron)\n",
+ "m_e=9.11E-31; #in Kg (mass of electron)\n",
+ "k=1.38E-23; #in J/K (Boltzmann's constant)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "T=300; #in K (temperature)\n",
+ "p=1.69E-8; #in ohm-m (resistivity)\n",
+ "\n",
+ "#calculate\n",
+ "lambda1=math.sqrt(3*k*m_e*T)/(n*e**2*p); #calculation of mean free path\n",
+ "lambda2=lambda1*1E9; #changing unit from m to nanometer\n",
+ "\n",
+ "#result\n",
+ "print\"The mean free path of electron is =\",'%.3E'%lambda1,\"m\";\n",
+ "print\"\\t\\t\\t\\t =\",round(lambda2,2),\"nm\";\n",
+ "#Note: answer in the book is wrong due to printing mistake"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The mean free path of electron is = 2.893E-09 m\n",
+ "\t\t\t\t = 2.89 nm\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.2 , Page no:146"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "k=1.38E-23; #in J/K (Boltzmann's constant)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "P_E=1; #in percentage (probability that a state with an energy 0.5 eV above Fermi energy will be occupied)\n",
+ "E=0.5; #in eV (energy above Fermi level)\n",
+ "\n",
+ "#calculate\n",
+ "P_E=1/100; #changing percentage into ratio\n",
+ "E=E*e; #changing unit from eV to J\n",
+ "#P_E=1/(1+exp((E-E_F)/k*T))\n",
+ "#Rearranging this equation, we get\n",
+ "#T=(E-E_F)/k*log((1/P_E)-1)\n",
+ "#Since E-E_F has been denoted by E therefore\n",
+ "T=E/(k*math.log((1/P_E)-1));\n",
+ "\n",
+ "#result\n",
+ "print\"The temperature is T=\",round(T),\"K\";\n",
+ "# Note: There is slight variation in the answer due to logarithm function"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The temperature is T= 1262.0 K\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.3 , Page no:147"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "n=5.8E28; #in 1/m^3 (density of electron)\n",
+ "m=9.1E-31; #in Kg (mass of electron)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "p=1.54E-8; #in ohm-m (resistivity)\n",
+ "\n",
+ "#calculate\n",
+ "t=m/(n*e**2*p); #calculation of relaxation time\n",
+ "\n",
+ "#result\n",
+ "print\"The relaxation time of conduction electrons is =\",'%.3E'%t,\"sec\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The relaxation time of conduction electrons is = 3.980E-14 sec\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.4 , Page no:147"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "n=8.5E28; #in 1/m^3 (density of electron)\n",
+ "m=9.1E-31; #in Kg (mass of electron)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "sigma=6E7; #in 1/ohm-m (conductivity)\n",
+ "E_F=7; #in E=eV (Fermi energy of Copper)\n",
+ "\n",
+ "#calculate\n",
+ "E_F=E_F*e; #changing unit from eV to J\n",
+ "v_F=math.sqrt(2*E_F/m); #calculation of velocity of electrons\n",
+ "#Since sigma=n*e^2*lambda/(2*m*v_F)\n",
+ "#Therefore we have\n",
+ "lambda1=2*m*v_F*sigma/(n*e**2); #calculation of mean free path\n",
+ "lambda2=lambda1*1E10; #changing unit from m to Angstrom\n",
+ "\n",
+ "#result\n",
+ "print\"The velocity of the electrons is v_F=\",'%.3E'%v_F,\"m/s (roundoff error)\";\n",
+ "print\"The mean free path traveled by the electrons is=\",round(lambda2),\"Angstrom\";\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The velocity of the electrons is v_F= 1.569E+06 m/s (roundoff error)\n",
+ "The mean free path traveled by the electrons is= 787.0 Angstrom\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.5 , Page no:147"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "n=6.5E28; #in 1/m^3 (density of electron)\n",
+ "m=9.1E-31; #in Kg (mass of electron)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "p=1.43E-8; #in ohm-m (resistivity)\n",
+ "\n",
+ "#calculate\n",
+ "t=m/(n*e**2*p); #calculation of relaxation time\n",
+ "\n",
+ "#result\n",
+ "print\"The relaxation time of conduction electrons is =\",'%.3E'%t,\"sec\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The relaxation time of conduction electrons is = 3.824E-14 sec\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6 , Page no:148"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "T=30; #in Celcius (temperature)\n",
+ "k=1.38E-23; #in J/K (Boltzmann's constant)\n",
+ "m_p=1.67E-27; #in Kg (mass of proton)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "T=T+273; #changing temperature from Celcius to Kelvin\n",
+ "KE=(3/2)*k*T; #calculation of average kinetic energy\n",
+ "KE1=KE/e; #changing unit from J to eV\n",
+ "m=1.008*2*m_p; #calculating mass of hydrogen gas molecule\n",
+ "c=math.sqrt(3*k*T/m); #calculation of velocity\n",
+ "\n",
+ "#result\n",
+ "print\"The average kinetic energy of gas ,molecules is KE=\",'%.3E'%KE,\"J\";\n",
+ "print\"\\t\\t\\t\\t\\t\\t =\",round(KE1,4),\"eV (Answer is wrong in textbook)\";\n",
+ "print\"The velocity of molecules is c=\",round(c,2),\"m/s\";\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The average kinetic energy of gas ,molecules is KE= 6.272E-21 J\n",
+ "\t\t\t\t\t\t = 0.0392 eV (Answer is wrong in textbook)\n",
+ "The velocity of molecules is c= 1930.27 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.7 , Page no:148"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "E=10; #in eV (kinetic energy for each electron and proton)\n",
+ "m_e=9.1E-31; #in Kg (mass of electron)\n",
+ "m_p=1.67E-27; #in Kg (mass of proton)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "E=E*e; #changing unit from eV to J\n",
+ "#since E=m*v^2/2\n",
+ "#therefore v=sqrt(2E/m)\n",
+ "v_e=math.sqrt(2*E/m_e); #calculation of kinetic energy of electron\n",
+ "v_p=math.sqrt(2*E/m_p); #calculation of kinetic energy of proton\n",
+ "\n",
+ "#result\n",
+ "print\"The kinetic energy of electron is v_e=\",'%.3E'%v_e,\"m/s\";\n",
+ "print\"The kinetic energy of proton is v_p=\",'%.3E'%v_p,\"m/s\";\n",
+ "print \"Note: Answer is wrong in textbook\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The kinetic energy of electron is v_e= 1.875E+06 m/s\n",
+ "The kinetic energy of proton is v_p= 4.377E+04 m/s\n",
+ "Note: Answer is wrong in textbook\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.8 , Page no:149"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "I=100; #in A (current in the wire)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "A=10; #in mm^2 (cross-sectional area)\n",
+ "n=8.5E28; #in 1/m^3 (density of electron)\n",
+ "\n",
+ "#calculate\n",
+ "A=A*1E-6; #changing unit from mm^2 to m^2\n",
+ "v_d=I/(n*A*e);\n",
+ "\n",
+ "#result\n",
+ "print\"The drift velocity of free electrons is v_d=\",'%.3E'%v_d,\"m/s\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The drift velocity of free electrons is v_d= 7.353E-04 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.9 , Page no:149"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "I=4;#in A (current in the conductor)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "A=1E-6; #in m^2 (cross-sectional area)\n",
+ "N_A=6.02E23; #in atoms/gram-atom (Avogadro's number)\n",
+ "p=8.9; #in g/cm^3 (density)\n",
+ "M=63.6; #atomic mass of copper\n",
+ "\n",
+ "#calculate\n",
+ "n=N_A*p/M; #Calculation of density of electrons in g/cm^3\n",
+ "n1=n*1E6; #changing unit from g/cm^3 to g/m^3\n",
+ "v_d=I/(n1*A*e);\n",
+ "\n",
+ "#result\n",
+ "print\"The density of copper atoms is n=\",'%.3E'%n,\"atoms/m^3\";\n",
+ "print\"\\t\\t\\t =\",'%.3E'%n1,\"atoms/m^3\";\n",
+ "print\"The average drift velocity of free electrons is v_d=\",'%.3E'%v_d,\"m/s\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The density of copper atoms is n= 8.424E+22 atoms/m^3\n",
+ "\t\t\t = 8.424E+28 atoms/m^3\n",
+ "The average drift velocity of free electrons is v_d= 2.968E-04 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.10 , Page no:149"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "n=9E28; #in 1/m^3 (density of valence electrons)\n",
+ "sigma=6E7; #in mho/m (conductivity of copper)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "#Since sigma=n*e*mu therefore\n",
+ "mu=sigma/(n*e); #calculation of mobility of electron\n",
+ "\n",
+ "#result\n",
+ "print\"The mobility of electrons is =\",'%.3E'%mu,\"m^2/V-s\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The mobility of electrons is = 4.167E-03 m^2/V-s\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.11 , Page no:150"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "E_F=5.51; #in eV (Fermi energy in Silver)\n",
+ "k=1.38E-23; #in J/K (Boltzmann's constant)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "#part-(a)\n",
+ "Eo=(3/5)*E_F; #calculation of average energy of free electron at 0K\n",
+ "#part-(b)\n",
+ "Eo1=Eo*e; #changing unit from eV to J\n",
+ "#Since for a classical particle E=(3/2)*k*T\n",
+ "#therefroe we have\n",
+ "T=(2/3)*Eo1/k; #calculation of temperature for a classical particle (an ideal gas)\n",
+ "\n",
+ "#result\n",
+ "print\"The average energy of free electron at 0K is Eo=\",Eo,\"eV\";\n",
+ "print\"The temperature at which a classical particle have this much energy is T=\",'%.3E'%T,\"K\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The average energy of free electron at 0K is Eo= 3.306 eV\n",
+ "The temperature at which a classical particle have this much energy is T= 2.555E+04 K\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.12 , Page no:150"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "E_F_L=4.7; #in eV (Fermi energy in Lithium)\n",
+ "E_F_M=2.35; #in eV (Fermi energy in a metal)\n",
+ "n_L=4.6E28; #in 1/m^3 (density of electron in Lithium)\n",
+ "\n",
+ "#calculate\n",
+ "#Since n=((2*m/h)^3/2)*E_F^(3/2)*(8*pi/3) and all things except E_F are constant\n",
+ "# Therefore we have n=C*E_F^(3/2) where C is proportionality constant\n",
+ "#n1/n2=(E_F_1/E_F_2)^(3/2)\n",
+ "#Therefore we have\n",
+ "n_M=n_L*(E_F_M/E_F_L); #calculation of electron density for a metal\n",
+ "\n",
+ "#result\n",
+ "print\"The lectron density for a metal is =\",n_M,\"1/m^3\";\n",
+ "print \"Note: Answer in the book is wrong \""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The lectron density for a metal is = 2.3e+28 1/m^3\n",
+ "Note: Answer in the book is wrong \n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Applied_Physics/Chapter_07_Dielectric_Properties.ipynb b/Applied_Physics/Chapter_07_Dielectric_Properties.ipynb new file mode 100755 index 00000000..7280033d --- /dev/null +++ b/Applied_Physics/Chapter_07_Dielectric_Properties.ipynb @@ -0,0 +1,488 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:5117f9d565431f3f25738652845b14f8a6fa036b5c0992b0b2e5d69631937306"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 7:Dielectric Properties"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.1 , Page no:187"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "A=100; #in cm^2 (cross-sectional area)\n",
+ "d=1; #in cm (seperation between plates)\n",
+ "Eo=8.85E-12; #in F/m (absolute permittivity)\n",
+ "V=100; #in V (potential difference)\n",
+ "\n",
+ "#calculate\n",
+ "A1=A*1E-4; #changing unit from cm^2 to m^2\n",
+ "d1=d*1E-2; #changing unit from cm to m\n",
+ "C=Eo*A1/d1; #calculation of capacitance\n",
+ "Q=C*V; #calculation of charge\n",
+ "C1=C*1E12; #changing unit of capacitance from F to pF\n",
+ "\n",
+ "#result\n",
+ "print\"The capacitance of capacitor is C=\",C,\"C\";\n",
+ "print\"\\t\\t\\t\\t =\",C1,\"C\";\n",
+ "print\"The charge on the plates is Q=\",Q,\"C\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The capacitance of capacitor is C= 8.85e-12 C\n",
+ "\t\t\t\t = 8.85 C\n",
+ "The charge on the plates is Q= 8.85e-10 C\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.2 , Page no:187"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "A=650; #in mm^2 (cross-sectional area)\n",
+ "d=4; #in mm (seperation between plates)\n",
+ "Eo=8.85E-12; #in F/m (absolute permittivity)\n",
+ "Er=3.5; #di-electric constant of the material\n",
+ "Q=2E-10; #in C (charge on plates)\n",
+ "\n",
+ "#calculate\n",
+ "A1=A*1E-6; #changing unit from mm^2 to m^2\n",
+ "d1=d*1E-3; #changing unit from mm to m\n",
+ "C=Er*Eo*A1/d1; #calculation of capacitance\n",
+ "V=Q/C; #calculation of charge\n",
+ "C1=C*1E12; #changing unit of capacitance from F to pF\n",
+ "\n",
+ "#result\n",
+ "print\"The capacitance of capacitor is C=\",C,\"C\",\n",
+ "print\"\\n\\t\\t\\t =\",C1,\"pF\";\n",
+ "print\"The resultant voltage across the capacitor is V=\",round(V,3),\"V\";\n",
+ "print \"NOTE: The answer in the textbook is wrong\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The capacitance of capacitor is C= 5.0334375e-12 C \n",
+ "\t\t\t = 5.0334375 pF\n",
+ "The resultant voltage across the capacitor is V= 39.734 V\n",
+ "NOTE: The answer in the textbook is wrong\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.3 , Page no:188"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "N=2.7E25; #in 1/m^3 (density of atoms)\n",
+ "E=1E6; #in V/m (electric field)\n",
+ "Z=2; #atomic number of Helium \n",
+ "Eo=8.85E-12; #in F/m (absolute permittivity)\n",
+ "Er=1.0000684; #(dielectric constant of the material)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "pi=3.14; #value of pi used in the solution\n",
+ "\n",
+ "#calculate\n",
+ "#since alpha=Eo*(Er-1)/N=4*pi*Eo*r_0^3 \n",
+ "#Therefore we have r_0^3=(Er-1)/(4*pi*N)\n",
+ "r_0=((Er-1)/(4*pi*N))**(1/3); #calculation of radius of electron cloud\n",
+ "x=4*pi*Eo*E*r_0/(Z*e); #calculation of dispalcement\n",
+ "\n",
+ "#result\n",
+ "print\"The radius of electron cloud is r_0=\",r_0,\"m\";\n",
+ "print\"The displacement is x=\",x,\"m\";\n",
+ "print \"NOTE: The answer in the textbook is wrong\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The radius of electron cloud is r_0= 5.86454377988e-11 m\n",
+ "The displacement is x= 20371.2258874 m\n",
+ "NOTE: The answer in the textbook is wrong\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.4 , Page no:188"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "K=1.000134; #di-elecrtic constant of the neon gas at NTP\n",
+ "E=90000; #in V/m (electric field)\n",
+ "Eo=8.85E-12; #in C/N-m^2 (absolute premittivity)\n",
+ "N_A=6.023E26; #in atoms/Kg-mole (Avogadro's number)\n",
+ "V=22.4; #in m^3 (volume of gas at NTP\n",
+ "\n",
+ "#calculate\n",
+ "n=N_A/V; #calculaton of density of atoms\n",
+ "#Since P=n*p=(k-1)*Eo*E\n",
+ "#therefore we have\n",
+ "p=(K-1)*Eo*E/n; #calculation of dipole moment induced\n",
+ "alpha=p/E; #calculation of atomic polarisability\n",
+ "\n",
+ "#result\n",
+ "print\"The dipole moment induced in each atom is p=\",p,\"C-m\";\n",
+ "print\"The atomic polarisability of neon is =\",alpha,\"c-m^2/V\";\n",
+ "print \"NOTE: The answer in the textbook is wrong\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The dipole moment induced in each atom is p= 3.96940793625e-36 C-m\n",
+ "The atomic polarisability of neon is = 4.4104532625e-41 c-m^2/V\n",
+ "NOTE: The answer in the textbook is wrong\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.5 , Page no:189"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "Er=3.75; #di-elecrtic constant of sulphur at 27 degree Celcius\n",
+ "gama=1/3; #internal field constant\n",
+ "p=2050; #in Kg/m^3 (density)\n",
+ "M_A=32; #in amu (atomic weight of sulphur)\n",
+ "Eo=8.85E-12; #in F/m (absolute permittivity)\n",
+ "N=6.022E23; #Avogadro's number\n",
+ "\n",
+ "#calculate\n",
+ "#Since ((Er-1)/(Er+2))*(M_A/p)=(N/(3*Eo))*alpha_e\n",
+ "#therefore we have\n",
+ "alpha_e=((Er-1)/(Er+2))*(M_A/p)*(3*Eo/N); #calculation of electronic polarisability of sulphur\n",
+ "\n",
+ "#result\n",
+ "print\"The electronic polarisability of sulphur is =\",alpha_e,\"Fm^2 (roundoff error)\";\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The electronic polarisability of sulphur is = 3.29143088985e-37 Fm^2 (roundoff error)\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.6 , Page no:189"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "Er=1.0000684; #di-elecrtic constant of Helium gas at NTP\n",
+ "Eo=8.85E-12; #in F/m (absolute permittivity)\n",
+ "N=2.7E25; #number of atomsper unit volume\n",
+ "\n",
+ "#calculate\n",
+ "#Since Er-1=(N/Eo)*alpha_e\n",
+ "#therefore we have\n",
+ "alpha_e=Eo*(Er-1)/N; #calculation of electronic polarisability of Helium\n",
+ "\n",
+ "#result\n",
+ "print\"The electronic polarisability of Helium gas is =\",alpha_e,\"Fm^2 (roundoff error)\";\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The electronic polarisability of Helium gas is = 2.242e-41 Fm^2 (roundoff error)\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.7 , Page no:190"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "N=3E28; #in atoms/m^3 (density of atoms)\n",
+ "alpha_e=1E-40; #in F-m^2 (electronic polarisability)\n",
+ "Eo=8.85E-12; #in F/m (absolute permittivity)\n",
+ "\n",
+ "#calculate\n",
+ "#Since (Er-1)/(Er+2)=N*alpha_e/(3*Eo)\n",
+ "#therefore we have\n",
+ "Er=(2*(N*alpha_e/(3*Eo))+1)/(1-(N*alpha_e/(3*Eo)));\n",
+ "#calculation of dielectric constant of the material\n",
+ "\n",
+ "#result\n",
+ "print\"The dielectric constant of the material is Er=\",Er,\"F/m\",\n",
+ "print \"NOTE: The answer in the book is wrong \""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The dielectric constant of the material is Er= 1.3821656051 F/m NOTE: The answer in the book is wrong \n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.8 , Page no:190"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "Er=4; #relative permittivity of sulphur\n",
+ "Eo=8.85E-12; #in F/m (absolute permittivity)\n",
+ "NA=2.08E3; #in Kg/m^3 (density of atoms in sulphur)\n",
+ "\n",
+ "#calculate\n",
+ "#Since ((Er-1)/(Er+2))*(M_A/p)=(N/(3*Eo))*alpha_e\n",
+ "#therefore we have\n",
+ "alpha_e=((Er-1)/(Er+2))*(3*Eo/NA); #calculation of electronic polarisability of sulphur\n",
+ "\n",
+ "#result\n",
+ "print\"The electronic polarisability of sulphur is =\",alpha_e,\"Fm^2\";\n",
+ "print \"NOTE: The answer in the book is wrong and they used the wrong formula \" \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The electronic polarisability of sulphur is = 6.38221153846e-15 Fm^2\n",
+ "NOTE: The answer in the book is wrong and they used the wrong formula \n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.9 , Page no:190"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "alpha1=2.5E-39; #in C^2-m/N (dielectric constant at 300K)\n",
+ "alpha2=2.0E-39; #in C^2-m/N (dielectric constant at 400K)\n",
+ "T1=300; #in K(first temperature)\n",
+ "T2=400; #in K(second temperature)\n",
+ "\n",
+ "#calculate\n",
+ "#since alpha=alpha_d+alpha0 and alpha0=Beta/T\n",
+ "#therefore alpha=alpha_d+(Beta/T)\n",
+ "#since alpha1=alpha_d+(Beta/T1) and alpha2=alpha_d+(Beta/T2)\n",
+ "#therefore alpha1-apha2=Beta*((1/T1)-(1/T2))\n",
+ "#or Beta= (alpha1-apha2)/ ((1/T1)-(1/T2))\n",
+ "Beta= (alpha1-alpha2)/ ((1/T1)-(1/T2)); #calculation of Beta\n",
+ "alpha_d=alpha1-(Beta/T1); #calculation of polarisability due to defromation\n",
+ "alpha0_1=Beta/T1; #calculation of polarisability due to permanent dipole moment at 300K\n",
+ "alpha0_2=Beta/T2; #calculation of polarisability due to permanent dipole moment at 400K\n",
+ "\n",
+ "#result\n",
+ "print\"The polarisability due to permanent dipole moment at 300K is =\",alpha0_1,\"C^2-m/N\";\n",
+ "print\"The polarisability due to permanent dipole moment at 400K is =\",alpha0_2,\"C^2-m/N\";\n",
+ "print\"The polarisability due to deformation of the molecules is =\",alpha_d,\"C^2-m/N\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The polarisability due to permanent dipole moment at 300K is = 2e-39 C^2-m/N\n",
+ "The polarisability due to permanent dipole moment at 400K is = 1.5e-39 C^2-m/N\n",
+ "The polarisability due to deformation of the molecules is = 5e-40 C^2-m/N\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.10 , Page no:191"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "n=1.5; #refractive index\n",
+ "Er=5.6; #dielectric constant\n",
+ "\n",
+ "#calculate\n",
+ "#since (Er-1)/(Er+2)=N*(alpha_e+alpha_i)/(3*E0) Clausius-Mossotti equation\n",
+ "#and (n^2-1)/(n^2+2)=N*alpha_e/(3*E0) \n",
+ "#from above two equations, we get ((n^2-1)/(n^2+2))*((Er+2)/(Er-1))=alpha_e/(alpha_e+alpha_i)\n",
+ "#or alpha_i/ (alpha_e+alpha_i)= 1-((n^2-1)/(n^2+2))*((Er+2)/(Er-1))= (say P)\n",
+ "#where P is fractional ionisational polarisability\n",
+ "P=1-((n**2-1)/(n**2+2))*((Er+2)/(Er-1)); #calculation of fractional ionisational polarisability\n",
+ "P1=P*100; #calculation of percentage of ionisational polarisability\n",
+ "\n",
+ "#result\n",
+ "print\"The percentage of ionisational polarisability is =\",P1,\"percent\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The percentage of ionisational polarisability is = 51.4066496164 percent\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Applied_Physics/Chapter_07_Dielectric_Properties_1.ipynb b/Applied_Physics/Chapter_07_Dielectric_Properties_1.ipynb new file mode 100755 index 00000000..85a277ef --- /dev/null +++ b/Applied_Physics/Chapter_07_Dielectric_Properties_1.ipynb @@ -0,0 +1,488 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:d363aa85787801a09c4a4d5d3e8dc3b5551836ae64a0a0c025744f99f230a4b4"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 7:Dielectric Properties"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.1 , Page no:187"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "A=100; #in cm^2 (cross-sectional area)\n",
+ "d=1; #in cm (seperation between plates)\n",
+ "Eo=8.85E-12; #in F/m (absolute permittivity)\n",
+ "V=100; #in V (potential difference)\n",
+ "\n",
+ "#calculate\n",
+ "A1=A*1E-4; #changing unit from cm^2 to m^2\n",
+ "d1=d*1E-2; #changing unit from cm to m\n",
+ "C=Eo*A1/d1; #calculation of capacitance\n",
+ "Q=C*V; #calculation of charge\n",
+ "C1=C*1E12; #changing unit of capacitance from F to pF\n",
+ "\n",
+ "#result\n",
+ "print\"The capacitance of capacitor is C=\",C,\"C\";\n",
+ "print\"\\t\\t\\t\\t =\",C1,\"C\";\n",
+ "print\"The charge on the plates is Q=\",Q,\"C\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The capacitance of capacitor is C= 8.85e-12 C\n",
+ "\t\t\t\t = 8.85 C\n",
+ "The charge on the plates is Q= 8.85e-10 C\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.2 , Page no:187"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "A=650; #in mm^2 (cross-sectional area)\n",
+ "d=4; #in mm (seperation between plates)\n",
+ "Eo=8.85E-12; #in F/m (absolute permittivity)\n",
+ "Er=3.5; #di-electric constant of the material\n",
+ "Q=2E-10; #in C (charge on plates)\n",
+ "\n",
+ "#calculate\n",
+ "A1=A*1E-6; #changing unit from mm^2 to m^2\n",
+ "d1=d*1E-3; #changing unit from mm to m\n",
+ "C=Er*Eo*A1/d1; #calculation of capacitance\n",
+ "V=Q/C; #calculation of charge\n",
+ "C1=C*1E12; #changing unit of capacitance from F to pF\n",
+ "\n",
+ "#result\n",
+ "print\"The capacitance of capacitor is C=\",'%.3E'%C,\"C\",\n",
+ "print\"\\n\\t\\t\\t =\",round(C1,2),\"pF\";\n",
+ "print\"The resultant voltage across the capacitor is V=\",round(V,2),\"V\";\n",
+ "print \"NOTE: The answer in the textbook is wrong\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The capacitance of capacitor is C= 5.033E-12 C \n",
+ "\t\t\t = 5.03 pF\n",
+ "The resultant voltage across the capacitor is V= 39.73 V\n",
+ "NOTE: The answer in the textbook is wrong\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.3 , Page no:188"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "N=2.7E25; #in 1/m^3 (density of atoms)\n",
+ "E=1E6; #in V/m (electric field)\n",
+ "Z=2; #atomic number of Helium \n",
+ "Eo=8.85E-12; #in F/m (absolute permittivity)\n",
+ "Er=1.0000684; #(dielectric constant of the material)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "pi=3.14; #value of pi used in the solution\n",
+ "\n",
+ "#calculate\n",
+ "#since alpha=Eo*(Er-1)/N=4*pi*Eo*r_0^3 \n",
+ "#Therefore we have r_0^3=(Er-1)/(4*pi*N)\n",
+ "r_0=((Er-1)/(4*pi*N))**(1/3); #calculation of radius of electron cloud\n",
+ "x=4*pi*Eo*E*r_0/(Z*e); #calculation of dispalcement\n",
+ "\n",
+ "#result\n",
+ "print\"The radius of electron cloud is r_0=\",'%.3E'%r_0,\"m\";\n",
+ "print\"The displacement is x=\",'%.3E'%x,\"m\";\n",
+ "print \"NOTE: The answer in the textbook is wrong\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The radius of electron cloud is r_0= 5.865E-11 m\n",
+ "The displacement is x= 2.037E+04 m\n",
+ "NOTE: The answer in the textbook is wrong\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.4 , Page no:188"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "K=1.000134; #di-elecrtic constant of the neon gas at NTP\n",
+ "E=90000; #in V/m (electric field)\n",
+ "Eo=8.85E-12; #in C/N-m^2 (absolute premittivity)\n",
+ "N_A=6.023E26; #in atoms/Kg-mole (Avogadro's number)\n",
+ "V=22.4; #in m^3 (volume of gas at NTP\n",
+ "\n",
+ "#calculate\n",
+ "n=N_A/V; #calculaton of density of atoms\n",
+ "#Since P=n*p=(k-1)*Eo*E\n",
+ "#therefore we have\n",
+ "p=(K-1)*Eo*E/n; #calculation of dipole moment induced\n",
+ "alpha=p/E; #calculation of atomic polarisability\n",
+ "\n",
+ "#result\n",
+ "print\"The dipole moment induced in each atom is p=\",'%.3E'%p,\"C-m\";\n",
+ "print\"The atomic polarisability of neon is =\",'%.3E'%alpha,\"c-m^2/V\";\n",
+ "print \"NOTE: The answer in the textbook is wrong\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The dipole moment induced in each atom is p= 3.969E-36 C-m\n",
+ "The atomic polarisability of neon is = 4.410E-41 c-m^2/V\n",
+ "NOTE: The answer in the textbook is wrong\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.5 , Page no:189"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "Er=3.75; #di-elecrtic constant of sulphur at 27 degree Celcius\n",
+ "gama=1/3; #internal field constant\n",
+ "p=2050; #in Kg/m^3 (density)\n",
+ "M_A=32; #in amu (atomic weight of sulphur)\n",
+ "Eo=8.85E-12; #in F/m (absolute permittivity)\n",
+ "N=6.022E23; #Avogadro's number\n",
+ "\n",
+ "#calculate\n",
+ "#Since ((Er-1)/(Er+2))*(M_A/p)=(N/(3*Eo))*alpha_e\n",
+ "#therefore we have\n",
+ "alpha_e=((Er-1)/(Er+2))*(M_A/p)*(3*Eo/N); #calculation of electronic polarisability of sulphur\n",
+ "\n",
+ "#result\n",
+ "print\"The electronic polarisability of sulphur is =\",'%.3E'%alpha_e,\"Fm^2 (roundoff error)\";\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The electronic polarisability of sulphur is = 3.291E-37 Fm^2 (roundoff error)\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.6 , Page no:189"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "Er=1.0000684; #di-elecrtic constant of Helium gas at NTP\n",
+ "Eo=8.85E-12; #in F/m (absolute permittivity)\n",
+ "N=2.7E25; #number of atomsper unit volume\n",
+ "\n",
+ "#calculate\n",
+ "#Since Er-1=(N/Eo)*alpha_e\n",
+ "#therefore we have\n",
+ "alpha_e=Eo*(Er-1)/N; #calculation of electronic polarisability of Helium\n",
+ "\n",
+ "#result\n",
+ "print\"The electronic polarisability of Helium gas is =\",'%.3E'%alpha_e,\"Fm^2 (roundoff error)\";\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The electronic polarisability of Helium gas is = 2.242E-41 Fm^2 (roundoff error)\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.7 , Page no:190"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "N=3E28; #in atoms/m^3 (density of atoms)\n",
+ "alpha_e=1E-40; #in F-m^2 (electronic polarisability)\n",
+ "Eo=8.85E-12; #in F/m (absolute permittivity)\n",
+ "\n",
+ "#calculate\n",
+ "#Since (Er-1)/(Er+2)=N*alpha_e/(3*Eo)\n",
+ "#therefore we have\n",
+ "Er=(2*(N*alpha_e/(3*Eo))+1)/(1-(N*alpha_e/(3*Eo)));\n",
+ "#calculation of dielectric constant of the material\n",
+ "\n",
+ "#result\n",
+ "print\"The dielectric constant of the material is Er=\",'%.3E'%Er,\"F/m\",\n",
+ "print \"NOTE: The answer in the book is wrong \""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The dielectric constant of the material is Er= 1.382E+00 F/m NOTE: The answer in the book is wrong \n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.8 , Page no:190"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "Er=4; #relative permittivity of sulphur\n",
+ "Eo=8.85E-12; #in F/m (absolute permittivity)\n",
+ "NA=2.08E3; #in Kg/m^3 (density of atoms in sulphur)\n",
+ "\n",
+ "#calculate\n",
+ "#Since ((Er-1)/(Er+2))*(M_A/p)=(N/(3*Eo))*alpha_e\n",
+ "#therefore we have\n",
+ "alpha_e=((Er-1)/(Er+2))*(3*Eo/NA); #calculation of electronic polarisability of sulphur\n",
+ "\n",
+ "#result\n",
+ "print\"The electronic polarisability of sulphur is =\",'%.3E'%alpha_e,\"Fm^2\";\n",
+ "print \"NOTE: The answer in the book is wrong and they used the wrong formula \" \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The electronic polarisability of sulphur is = 6.382E-15 Fm^2\n",
+ "NOTE: The answer in the book is wrong and they used the wrong formula \n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.9 , Page no:190"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "alpha1=2.5E-39; #in C^2-m/N (dielectric constant at 300K)\n",
+ "alpha2=2.0E-39; #in C^2-m/N (dielectric constant at 400K)\n",
+ "T1=300; #in K(first temperature)\n",
+ "T2=400; #in K(second temperature)\n",
+ "\n",
+ "#calculate\n",
+ "#since alpha=alpha_d+alpha0 and alpha0=Beta/T\n",
+ "#therefore alpha=alpha_d+(Beta/T)\n",
+ "#since alpha1=alpha_d+(Beta/T1) and alpha2=alpha_d+(Beta/T2)\n",
+ "#therefore alpha1-apha2=Beta*((1/T1)-(1/T2))\n",
+ "#or Beta= (alpha1-apha2)/ ((1/T1)-(1/T2))\n",
+ "Beta= (alpha1-alpha2)/ ((1/T1)-(1/T2)); #calculation of Beta\n",
+ "alpha_d=alpha1-(Beta/T1); #calculation of polarisability due to defromation\n",
+ "alpha0_1=Beta/T1; #calculation of polarisability due to permanent dipole moment at 300K\n",
+ "alpha0_2=Beta/T2; #calculation of polarisability due to permanent dipole moment at 400K\n",
+ "\n",
+ "#result\n",
+ "print\"The polarisability due to permanent dipole moment at 300K is =\",'%.3E'%alpha0_1,\"C^2-m/N\";\n",
+ "print\"The polarisability due to permanent dipole moment at 400K is =\",'%.3E'%alpha0_2,\"C^2-m/N\";\n",
+ "print\"The polarisability due to deformation of the molecules is =\",'%.3E'%alpha_d,\"C^2-m/N\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The polarisability due to permanent dipole moment at 300K is = 2.000E-39 C^2-m/N\n",
+ "The polarisability due to permanent dipole moment at 400K is = 1.500E-39 C^2-m/N\n",
+ "The polarisability due to deformation of the molecules is = 5.000E-40 C^2-m/N\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.10 , Page no:191"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "n=1.5; #refractive index\n",
+ "Er=5.6; #dielectric constant\n",
+ "\n",
+ "#calculate\n",
+ "#since (Er-1)/(Er+2)=N*(alpha_e+alpha_i)/(3*E0) Clausius-Mossotti equation\n",
+ "#and (n^2-1)/(n^2+2)=N*alpha_e/(3*E0) \n",
+ "#from above two equations, we get ((n^2-1)/(n^2+2))*((Er+2)/(Er-1))=alpha_e/(alpha_e+alpha_i)\n",
+ "#or alpha_i/ (alpha_e+alpha_i)= 1-((n^2-1)/(n^2+2))*((Er+2)/(Er-1))= (say P)\n",
+ "#where P is fractional ionisational polarisability\n",
+ "P=1-((n**2-1)/(n**2+2))*((Er+2)/(Er-1)); #calculation of fractional ionisational polarisability\n",
+ "P1=P*100; #calculation of percentage of ionisational polarisability\n",
+ "\n",
+ "#result\n",
+ "print\"The percentage of ionisational polarisability is =\",round(P1,1),\"percent\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The percentage of ionisational polarisability is = 51.4 percent\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Applied_Physics/Chapter_08_Magnetic_Properties.ipynb b/Applied_Physics/Chapter_08_Magnetic_Properties.ipynb new file mode 100755 index 00000000..353fc099 --- /dev/null +++ b/Applied_Physics/Chapter_08_Magnetic_Properties.ipynb @@ -0,0 +1,323 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:aecad4e644a378cab0352aa29e9045a53c52646739e4f4bd6cdff80f1d0741e6"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 8:Magnetic Properties"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.1 , Page no:236"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "X=-0.5E-5; #magnetic susceptibility of silicon\n",
+ "H=0.9E4; #in A/m (magnetic field intensity)\n",
+ "mu0=4*3.14*1E-7; #in H/m (absolute permeability)\n",
+ "\n",
+ "#calculate\n",
+ "I=X*H; #calculation of intensity of magnetism\n",
+ "B=mu0*H*(1+X); #calculation of magnetic flux density\n",
+ "\n",
+ "#result\n",
+ "print\"The intensity of magnetism is I=\",I,\"A/m\";\n",
+ "print\"The magnetic flux density is B=\",round(B,3),\"Wb/m^2\";\n",
+ "print \"NOTE: The answer in the textbook is wrong\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The intensity of magnetism is I= -0.045 A/m\n",
+ "The magnetic flux density is B= 0.011 Wb/m^2\n",
+ "NOTE: The answer in the textbook is wrong\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.2 , Page no:236"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "r=0.052; #in nm (radius of orbit)\n",
+ "B=1; #in Wb/m^2 (magnetic field of induction)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "m=9.1E-31; #in Kg (mass of electron)\n",
+ "\n",
+ "#calculate\n",
+ "r=0.052*1E-9; #changing unit from nm to m\n",
+ "d_mu=(e**2*r**2*B)/(4*m); #calculation of change in magnetic moment\n",
+ "\n",
+ "#result\n",
+ "print\"The change in magnetic moment is =\",d_mu,\"Am^2\";\n",
+ "print \"NOTE: The answer in the textbook is wrong\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The change in magnetic moment is = 1.90171428571e-29 Am^2\n",
+ "NOTE: The answer in the textbook is wrong\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.3 , Page no:236"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "H=220; #in A/m (magnetic field intensity)\n",
+ "I=3300; #in A/m (intensity of magnetisation)\n",
+ "\n",
+ "#calculate\n",
+ "mu_r=1+(I/H); #calculation of relative permeability\n",
+ "\n",
+ "#result\n",
+ "print\"The relative permeability of a ferromagentic material is =\",mu_r;"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The relative permeability of a ferromagentic material is = 16.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.4 , Page no:236"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "I=3000; #in A/m (intensity of magnetisation)\n",
+ "B=0.005; #in Wb/m^2 (magnetic flus intensity)\n",
+ "pi=3.14; #value of pi used in the solution\n",
+ "mu0=4*pi*1E-7; #in H/m (absolute permeability)\n",
+ "\n",
+ "#calculate\n",
+ "H=(B/mu0)-I; #calculation of magnetic force\n",
+ "mu_r=(I/H)+1; #calculation of relative permeability\n",
+ "\n",
+ "#result\n",
+ "print\"The magnetic force is H=\",H;\n",
+ "print\"The relative permeability is =\",mu_r;"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The magnetic force is H= 980.891719745\n",
+ "The relative permeability is = 4.05844155844\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.5 , Page no:237"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "H=4E3; #in A/m (magnetic field intensity)\n",
+ "N=60; #number of turns\n",
+ "l=12; #in cm (length of solenoid)\n",
+ "\n",
+ "#calculate\n",
+ "n=N/(l*1E-2); #calculation of number of turns per unit metre\n",
+ "#Snice H=n*i;\n",
+ "i=H/n; #calculation of current through the solenoid\n",
+ "\n",
+ "#result\n",
+ "print\"The current through the solenoid is i=\",i,\"A\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The current through the solenoid is i= 8.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.6 , Page no:237"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "l=30; #in cm (length of solenoid)\n",
+ "A=1; #in cm^2 (cross-sectional area)\n",
+ "N=300; #number of turns\n",
+ "i=0.032; #in A (current through the winding)\n",
+ "phi_B=2E-6; #in Wb (magnetic flux)\n",
+ "pi=3.14; #value of pi used in the solution\n",
+ "mu0=4*pi*1E-7; #in H/m (absolute permeability)\n",
+ "\n",
+ "#calculate\n",
+ "l=l*1E-2; #changing unit from cm to m\n",
+ "A=A*1E-4; #changing unit from cm^2 to m^2\n",
+ "B=phi_B/A; #calculation of flux density\n",
+ "H=N*i/l; #calculation of magnetic intensity\n",
+ "mu=B/H; #calcluation of absolute permeability of iron\n",
+ "mu_r=mu/mu0; #calcluation of relative permeability of iron\n",
+ "\n",
+ "#result\n",
+ "print\"The flux density is B=\",B,\"Wb/m^2\";\n",
+ "print\"The magnetic intensity is H=\",H,\"A-turns/m\";\n",
+ "print\"The relative permeability of iron is =\",round(mu_r),\" (roundoff error)\";\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The flux density is B= 0.02 Wb/m^2\n",
+ "The magnetic intensity is H= 32.0 A-turns/m\n",
+ "The relative permeability of iron is = 498.0 (roundoff error)\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.7 , Page no:238"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "A=100; #in m^2 (area of Hysteresis loop)\n",
+ "B=0.01; #in Wb/m^2 (unit space along vertical axis or magnetic flux density)\n",
+ "H=40; #in A/m (unit space along horizontal axis or magnetic fild ntensity)\n",
+ "\n",
+ "#calculate\n",
+ "H_L=A*B*H; #calculation of magnetic intensity\n",
+ "\n",
+ "#result\n",
+ "print\"The Hystersis loss per cycle is =\",round(H_L),\"J/m^2\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Hystersis loss per cycle is = 40.0 J/m^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Applied_Physics/Chapter_08_Magnetic_Properties_1.ipynb b/Applied_Physics/Chapter_08_Magnetic_Properties_1.ipynb new file mode 100755 index 00000000..b2a8def3 --- /dev/null +++ b/Applied_Physics/Chapter_08_Magnetic_Properties_1.ipynb @@ -0,0 +1,323 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:708d3f2a9798f4d28f6ce90536d50509c591ba2ba31051121346b4739b2aea82"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 8:Magnetic Properties"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.1 , Page no:236"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "X=-0.5E-5; #magnetic susceptibility of silicon\n",
+ "H=0.9E4; #in A/m (magnetic field intensity)\n",
+ "mu0=4*3.14*1E-7; #in H/m (absolute permeability)\n",
+ "\n",
+ "#calculate\n",
+ "I=X*H; #calculation of intensity of magnetism\n",
+ "B=mu0*H*(1+X); #calculation of magnetic flux density\n",
+ "\n",
+ "#result\n",
+ "print\"The intensity of magnetism is I=\",I,\"A/m\";\n",
+ "print\"The magnetic flux density is B=\",round(B,3),\"Wb/m^2\";\n",
+ "print \"NOTE: The answer in the textbook is wrong\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The intensity of magnetism is I= -0.045 A/m\n",
+ "The magnetic flux density is B= 0.011 Wb/m^2\n",
+ "NOTE: The answer in the textbook is wrong\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.2 , Page no:236"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "r=0.052; #in nm (radius of orbit)\n",
+ "B=1; #in Wb/m^2 (magnetic field of induction)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "m=9.1E-31; #in Kg (mass of electron)\n",
+ "\n",
+ "#calculate\n",
+ "r=0.052*1E-9; #changing unit from nm to m\n",
+ "d_mu=(e**2*r**2*B)/(4*m); #calculation of change in magnetic moment\n",
+ "\n",
+ "#result\n",
+ "print\"The change in magnetic moment is =\",'%.3E'%d_mu,\"Am^2\";\n",
+ "print \"NOTE: The answer in the textbook is wrong\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The change in magnetic moment is = 1.902E-29 Am^2\n",
+ "NOTE: The answer in the textbook is wrong\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.3 , Page no:236"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "H=220; #in A/m (magnetic field intensity)\n",
+ "I=3300; #in A/m (intensity of magnetisation)\n",
+ "\n",
+ "#calculate\n",
+ "mu_r=1+(I/H); #calculation of relative permeability\n",
+ "\n",
+ "#result\n",
+ "print\"The relative permeability of a ferromagentic material is =\",mu_r;"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The relative permeability of a ferromagentic material is = 16.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.4 , Page no:236"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "I=3000; #in A/m (intensity of magnetisation)\n",
+ "B=0.005; #in Wb/m^2 (magnetic flus intensity)\n",
+ "pi=3.14; #value of pi used in the solution\n",
+ "mu0=4*pi*1E-7; #in H/m (absolute permeability)\n",
+ "\n",
+ "#calculate\n",
+ "H=(B/mu0)-I; #calculation of magnetic force\n",
+ "mu_r=(I/H)+1; #calculation of relative permeability\n",
+ "\n",
+ "#result\n",
+ "print\"The magnetic force is H=\",round(H,3);\n",
+ "print\"The relative permeability is =\",round(mu_r,3);"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The magnetic force is H= 980.892\n",
+ "The relative permeability is = 4.058\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.5 , Page no:237"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "H=4E3; #in A/m (magnetic field intensity)\n",
+ "N=60; #number of turns\n",
+ "l=12; #in cm (length of solenoid)\n",
+ "\n",
+ "#calculate\n",
+ "n=N/(l*1E-2); #calculation of number of turns per unit metre\n",
+ "#Snice H=n*i;\n",
+ "i=H/n; #calculation of current through the solenoid\n",
+ "\n",
+ "#result\n",
+ "print\"The current through the solenoid is i=\",i,\"A\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The current through the solenoid is i= 8.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.6 , Page no:237"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "l=30; #in cm (length of solenoid)\n",
+ "A=1; #in cm^2 (cross-sectional area)\n",
+ "N=300; #number of turns\n",
+ "i=0.032; #in A (current through the winding)\n",
+ "phi_B=2E-6; #in Wb (magnetic flux)\n",
+ "pi=3.14; #value of pi used in the solution\n",
+ "mu0=4*pi*1E-7; #in H/m (absolute permeability)\n",
+ "\n",
+ "#calculate\n",
+ "l=l*1E-2; #changing unit from cm to m\n",
+ "A=A*1E-4; #changing unit from cm^2 to m^2\n",
+ "B=phi_B/A; #calculation of flux density\n",
+ "H=N*i/l; #calculation of magnetic intensity\n",
+ "mu=B/H; #calcluation of absolute permeability of iron\n",
+ "mu_r=mu/mu0; #calcluation of relative permeability of iron\n",
+ "\n",
+ "#result\n",
+ "print\"The flux density is B=\",B,\"Wb/m^2\";\n",
+ "print\"The magnetic intensity is H=\",H,\"A-turns/m\";\n",
+ "print\"The relative permeability of iron is =\",round(mu_r),\" (roundoff error)\";\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The flux density is B= 0.02 Wb/m^2\n",
+ "The magnetic intensity is H= 32.0 A-turns/m\n",
+ "The relative permeability of iron is = 498.0 (roundoff error)\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.7 , Page no:238"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "A=100; #in m^2 (area of Hysteresis loop)\n",
+ "B=0.01; #in Wb/m^2 (unit space along vertical axis or magnetic flux density)\n",
+ "H=40; #in A/m (unit space along horizontal axis or magnetic fild ntensity)\n",
+ "\n",
+ "#calculate\n",
+ "H_L=A*B*H; #calculation of magnetic intensity\n",
+ "\n",
+ "#result\n",
+ "print\"The Hystersis loss per cycle is =\",round(H_L),\"J/m^2\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Hystersis loss per cycle is = 40.0 J/m^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Applied_Physics/Chapter_09_Semiconductors.ipynb b/Applied_Physics/Chapter_09_Semiconductors.ipynb new file mode 100755 index 00000000..f3c26825 --- /dev/null +++ b/Applied_Physics/Chapter_09_Semiconductors.ipynb @@ -0,0 +1,782 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:eb4f284d8d12c103b674cc454947abe9c7a1a76e2d11836f2841ba55d4adeef7"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 9:Semiconductors"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.2 , Page no:272"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "Eg=0.67; #in eV (Energy band gap)\n",
+ "k=1.38E-23; #in J/K (Boltzmann\u2019s constant)\n",
+ "T1=298; #in K (room temperature)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "K=10; #ratio of number of electrons at different temperature\n",
+ "\n",
+ "#calculate\n",
+ "Eg=Eg*e; #changing unit from eV to Joule\n",
+ "#since ne=Ke*exp(-Eg/(2*k*T))\n",
+ "#and ne/ne1=exp(-Eg/(2*k*T))/exp(-Eg/(2*k*T1)) and ne/ne1=K=10\n",
+ "#therefore we have 10=exp(-Eg/(2*k*T))/exp(-Eg/(2*k*T1))\n",
+ "#re-arranging the equation for T, we get T2=1/((1/T1)-((2*k*log(10))/Eg))\n",
+ "T=1/((1/T1)-((2*k*math.log(10))/Eg)); #calculation of the temperature\n",
+ "\n",
+ "#result\n",
+ "print\"The temperature at which number of electrons in the conduction band of a semiconductor increases by a factor of 10 is T=\",round(T),\"K (roundoff error)\";\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The temperature at which number of electrons in the conduction band of a semiconductor increases by a factor of 10 is T= 362.0 K (roundoff error)\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.3 , Page no:272"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "ni=2.5E13; #in /cm^3 (intrinsic carrier density)\n",
+ "ue=3900; #in cm^2/(V-s) (electron mobilities)\n",
+ "uh=1900; #in cm^2/(V-s) (hole mobilities)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "l=1; #in cm (lenght of the box)\n",
+ "b=1;h=1; #in mm (dimensions of germanium rod )\n",
+ "\n",
+ "#calculate\n",
+ "ni=ni*1E6; #changing unit from 1/cm^3 to 1/m^3\n",
+ "ue=ue*1E-4; #changing unit from cm^2 to m^2\n",
+ "uh=uh*1E-4; #changing unit from cm^2 to m^2\n",
+ "sigma=ni*e*(ue+uh); #calculation of conductivity\n",
+ "rho=1/sigma; #calculation of resistivity\n",
+ "l=l*1E-2; #changing unit from mm to m for length\n",
+ "A=(b*1E-3)*(h*1E-3); #changing unit from mm to m for width and height and calculation of cross-sectional area\n",
+ "R=rho*l/A; #calculation of resistance\n",
+ "\n",
+ "#result\n",
+ "print\"The resistance of intrinsic germanium is R=\",R,\"ohm\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The resistance of intrinsic germanium is R= 4310.34482759 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.4 , Page no:273"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "ne=2.5E19; #in /m^3 (electron density)\n",
+ "nh=2.5E19; #in /m^3 (hole density)\n",
+ "ue=0.36; #in m^2/(V-s) (electron mobilities)\n",
+ "uh=0.17; #in m^2/(V-s) (hole mobilities)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "#since ne=nh=ni, therefore we have \n",
+ "ni=nh;\n",
+ "sigma=ni*e*(ue+uh); #calculation of conductivity\n",
+ "rho=1/sigma; #calculation of resistivity\n",
+ "\n",
+ "#result\n",
+ "print\"The conductivity of germanium is =\",round(sigma,2),\"/ohm-m\";\n",
+ "print\"The resistivity of germanium is =\",round(rho,2),\"ohm-m\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The conductivity of germanium is = 2.12 /ohm-m\n",
+ "The resistivity of germanium is = 0.47 ohm-m\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.5 , Page no:273"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "ni=1.5E16; #in /m^3 (intrinsic carrier density)\n",
+ "ue=0.135; #in m^2/(V-s) (electron mobilities)\n",
+ "uh=0.048; #in m^2/(V-s) (hole mobilities)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "ND=1E23; #in atom/m^3 (doping concentration)\n",
+ "\n",
+ "#calculate\n",
+ "sigma_i=ni*e*(ue+uh); #calculation of intrinsic conductivity\n",
+ "sigma=ND*ue*e; #calculation of conductivity after doping\n",
+ "rho=ni**2/ND; #calculation of equilibrium hole concentration\n",
+ "\n",
+ "#result\n",
+ "print\"The intrinsic conductivity for silicon is =\",sigma_i,\"S\";\n",
+ "print\"The conductivity after doping with phosphorus atoms is =\",sigma,\"S\";\n",
+ "print\"The equilibrium hole concentration is =\",rho,\"/m^3\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The intrinsic conductivity for silicon is = 0.0004392 S\n",
+ "The conductivity after doping with phosphorus atoms is = 2160.0 S\n",
+ "The equilibrium hole concentration is = 2250000000.0 /m^3\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.6 , Page no:274"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "ni=1.5E16; #in /m^3 (intrinsic carrier density)\n",
+ "ue=0.13; #in m^2/(V-s) (electron mobilities)\n",
+ "uh=0.05; #in m^2/(V-s) (hole mobilities)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "ne=5E20; #in /m^3 (concentration of donor type impurity)\n",
+ "nh=5E20; #in /m^3 (concentration of acceptor type impurity)\n",
+ "\n",
+ "#calculate\n",
+ "#part-i\n",
+ "sigma=ni*e*(ue+uh); #calculation of intrinsic conductivity\n",
+ "#part-ii\n",
+ "#since 1 donor atom is in 1E8 Si atoms, hence holes concentration can be neglected\n",
+ "sigma1=ne*e*ue; #calculation of conductivity after doping with donor type impurity\n",
+ "#part-iii\n",
+ "#since 1 acceptor atom is in 1E8 Si atoms, hence electron concentration can be neglected\n",
+ "sigma2=nh*e*uh; #calculation of conductivity after doping with acceptor type impurity\n",
+ "\n",
+ "#result\n",
+ "print\"The intrinsic conductivity for silicon is =\",sigma,\"(ohm-m)^-1\";\n",
+ "print\"The conductivity after doping with donor type impurity is =\",sigma1,\"(ohm-m)^-1\";\n",
+ "print\"The conductivity after doping with acceptor type impurity is =\",sigma2,\"(ohm-m)^-1\";\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The intrinsic conductivity for silicon is = 0.000432 (ohm-m)^-1\n",
+ "The conductivity after doping with donor type impurity is = 10.4 (ohm-m)^-1\n",
+ "The conductivity after doping with acceptor type impurity is = 4.0 (ohm-m)^-1\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.7 , Page no:274"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "ni=1E20; #in /m^3 (intrinsic carrier density)\n",
+ "ND=1E21; #in /m^3 (donor impurity concentration)\n",
+ "\n",
+ "#calculate\n",
+ "nh=ni**2/ND; #calculation of density of hole carriers at room temperature\n",
+ "\n",
+ "#result\n",
+ "print\"The density of hole carriers at room temperature is nh=\",nh,\"/m^3\";\n",
+ "#Note: answer in the book is wrong due to printing mistake"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The density of hole carriers at room temperature is nh= 1e+19 /m^3\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.8 , Page no:275"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "M=72.6; #atomic mass of germanium\n",
+ "P=5400; #in Kg/m^3 (density)\n",
+ "ue=0.4; #in m^2/V-s (mobility of electrons)\n",
+ "uh=0.2; #in m^2/V-s (mobility of holes)\n",
+ "Eg=0.7; #in eV (Band gap)\n",
+ "m=9.1E-31; #in Kg (mass of electron)\n",
+ "k=1.38E-23; #in J/K (Boltzmann\u2019s constant)\n",
+ "T=300; #in K (temperature)\n",
+ "h=6.63E-34; #in J/s (Planck\u2019s constant)\n",
+ "pi=3.14; #value of pi used in the solution\n",
+ "e=1.6E-19; #in C(charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "Eg=Eg*e; #changing unit from eV to J\n",
+ "ni=2*(2*pi*m*k*T/h**2)**(3/2)*math.exp(-Eg/(2*k*T));\n",
+ "sigma=ni*e*(ue+uh);\n",
+ "\n",
+ "#result\n",
+ "print\"The intrinsic carrier density for germanium at 300K is ni=\",ni,\"/m^3\";\n",
+ "print\"The conductivity of germanium is=\",round(sigma,3),\"(ohm-m)^-1\";\n",
+ "print \"NOTE: The answer in the textbook is wrong\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The intrinsic carrier density for germanium at 300K is ni= 3.33408559508e+19 /m^3\n",
+ "The conductivity of germanium is= 3.201 (ohm-m)^-1\n",
+ "NOTE: The answer in the textbook is wrong\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9 , Page no:275"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "rho1=4.5; #in ohm-m (resistivity at 20 degree Celcius)\n",
+ "rho2=2.0; #in ohm-m (resistivity at 32 degree Celcius)\n",
+ "k=1.38E-23; #in J/K (Boltzmann\u2019s constant)\n",
+ "T1=20; T2=32; #in degree Celcius (two temperatures)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "T1=T1+273; #changing unit from degree Celius to K\n",
+ "T2=T2+273; #changing unit from degree Celius to K\n",
+ "#since sigma=e*u*C*T^(3/2)*exp(-Eg/(2*k*T))\n",
+ "#therefore sigma1/sigma2=(T1/T2)^3/2*exp((-Eg/(2*k)*((1/T1)-(1/T2))\n",
+ "#and sigma=1/rho \n",
+ "#therefore we have rho2/rho1=(T1/T2)^3/2*exp((-Eg/(2*k)*((1/T1)-(1/T2))\n",
+ "#re-arranging above equation for Eg, we get Eg=(2*k/((1/T1)-(1/T2)))*((3/2)*log(T1/T2)-log(rho2/rho1))\n",
+ "Eg=(2*k/((1/T1)-(1/T2)))*((3/2)*math.log(T1/T2)-math.log(rho2/rho1));\n",
+ "Eg1=Eg/e;#changing unit from J to eV\n",
+ "\n",
+ "#result\n",
+ "print\"The energy band gap is Eg=\",Eg,\"J\";\n",
+ "print\"\\t\\t\\t =\",round(Eg1,3),\"eV\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The energy band gap is Eg= 1.54302914521e-19 J\n",
+ "\t\t\t = 0.964 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.10 , Page no:276"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "rho=2300; #in ohm-m (resistivity of pure silicon)\n",
+ "ue=0.135; #in m^2/V-s (mobility of electron)\n",
+ "uh=0.048; #in m^2/V-s (mobility of electron)\n",
+ "Nd=1E19; #in /m^3 (doping concentration)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "#since sigma=ni*e*(ue+uh) and sigma=1/rho\n",
+ "#therefore ni=1/(rho*e*(ue+uh))\n",
+ "ni=1/(rho*e*(ue+uh)); #calculation of intrinsic concentration\n",
+ "ne=Nd; #calculation of electron concentration\n",
+ "nh=ni**2/Nd; #calculation of hole concentration\n",
+ "sigma=ne*ue*e+nh*uh*e; #calculation of conductivity\n",
+ "rho=1/sigma; #calculation of resistivity\n",
+ "\n",
+ "#result\n",
+ "print\"The electron concentration is ne=\",ne,\"/m^3\";\n",
+ "print\"The hole concentration is nh=\",nh,\"/m^3\";\n",
+ "print\"The resistivity of the specimen is =\",round(rho,3),\"ohm-m\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The electron concentration is ne= 1e+19 /m^3\n",
+ "The hole concentration is nh= 2.20496745228e+13 /m^3\n",
+ "The resistivity of the specimen is = 4.63 ohm-m\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.11 , Page no:276"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "uh=1900; #in cm^2/V-s (mobility of electron)\n",
+ "Na=2E17; #in /m^3 (acceptor doping concentration)\n",
+ "e=1.6E-19; #in C(charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "uh=uh*1E-4; #changing unit from cm^2/V-s to m^2/V-s\n",
+ "Na=Na*1E6; #changing unit from 1/cm^3 to 1/m^3\n",
+ "nh=Na; #hole concentration \n",
+ "#since sigma=ne*ue*e+nh*uh*e and nh>>ne\n",
+ "#therefore sigma=nh*uh*e\n",
+ "sigma=nh*uh*e; #calculation of conductivity\n",
+ "\n",
+ "#result\n",
+ "print\"The conductivity of p-type Ge crystal is =\",sigma,\"/ohm-m (roundoff error)\";\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The conductivity of p-type Ge crystal is = 6080.0 /ohm-m (roundoff error)\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.12 , Page no:277"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "ue=0.19; #in m^2/V-s (mobility of electron)\n",
+ "T=300; #in K (temperature)\n",
+ "k=1.38E-23; #in J/K (Boltzmann\u2019s constant)\n",
+ "e=1.6E-19; #in C(charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "Dn=ue*k*T/e; #calculation of diffusion co-efficient\n",
+ "\n",
+ "#result\n",
+ "print\"The diffusion co-efficient of electron in silicon is Dn=\",Dn,\"m^2/s\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The diffusion co-efficient of electron in silicon is Dn= 0.00491625 m^2/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.13 , Page no:277"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "Eg=0.4; #in eV (Band gap of semiconductor)\n",
+ "k=1.38E-23; #in J/K (Boltzmann\u2019s constant)\n",
+ "T1=0; #in degree Celcius (first temperature)\n",
+ "T2=50; #in degree Celcius (second temperature)\n",
+ "T3=100; #in degree Celcius (third temperature)\n",
+ "e=1.602E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "T1=T1+273; #changing temperature form Celcius to Kelvin\n",
+ "T2=T2+273; #changing temperature form Celcius to Kelvin\n",
+ "T3=T3+273; #changing temperature form Celcius to Kelvin\n",
+ "Eg=Eg*e; #changing unit from eV to Joule\n",
+ "#Using F_E=1/(1+exp(Eg/2*k*T))\n",
+ "F_E1=1/(1+math.exp(Eg/(2*k*T1))); #calculation of probability of occupation of lowest level at 0 degree Celcius\n",
+ "F_E2=1/(1+math.exp(Eg/(2*k*T2))); #calculation of probability of occupation of lowest level at 50 degree Celcius\n",
+ "F_E3=1/(1+math.exp(Eg/(2*k*T3))); #calculation of probability of occupation of lowest level at 100 degree Celcius\n",
+ "\n",
+ "#result\n",
+ "print\"The probability of occupation of lowest level in conduction band is\";\n",
+ "print\"\\t at 0 degree Celcius, F_E=\",F_E1,\"eV\";\n",
+ "print\"\\t at 50 degree Celcius, F_E=\",F_E2,\"eV\";\n",
+ "print\"\\t at 100 degree Celcius, F_E=\",F_E3,\"eV\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The probability of occupation of lowest level in conduction band is\n",
+ "\t at 0 degree Celcius, F_E= 0.000202505914236 eV\n",
+ "\t at 50 degree Celcius, F_E= 0.000754992968827 eV\n",
+ "\t at 100 degree Celcius, F_E= 0.00197639649915 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.14 , Page no:278"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "Eg=1.2; #in eV (Energy band gap)\n",
+ "k=1.38E-23; #in J/K (Boltzmann\u2019s constant)\n",
+ "T1=600; T2=300; #in K (two temperatures)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "Eg=Eg*e; #changing unit from eV to Joule\n",
+ "#since sigma is proportional to exp(-Eg/(2*k*T))\n",
+ "#therefore ratio=sigma1/sigma2=exp(-Eg/(2*k*((1/T1)-(1/T2))));\n",
+ "ratio= math.exp((-Eg/(2*k))*((1/T1)-(1/T2))); #calculation of ratio of conductivity at 600K and at 300K\n",
+ "\n",
+ "#result\n",
+ "print\"The ratio of conductivity at 600K and at 300K is =\",ratio;"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The ratio of conductivity at 600K and at 300K is = 108467.17792\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.15 , Page no:278"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "ue=0.39; #in m^2/V-s (mobility of electron)\n",
+ "n=5E13; #number of donor atoms\n",
+ "ni=2.4E19; #in atoms/m^3 (intrinsic carrier density)\n",
+ "l=10; #in mm (length of rod)\n",
+ "a=1; #in mm (side of square cross-section)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "l=l*1E-3; #changing unit from mm to m\n",
+ "a=a*1E-3; #changing unit from mm to m\n",
+ "A=a**2; #calculation of cross-section area\n",
+ "Nd=n/(l*A); #calculation of donor concentration\n",
+ "ne=Nd; #calculation of electron density\n",
+ "nh=ni**2/Nd; #calculation of hole density\n",
+ "#since sigma=ne*e*ue+nh*e*ue and since ne>>nh\n",
+ "#therefore sigma=ne*e*ue\n",
+ "sigma=ne*e*ue; #calculation of conductivity\n",
+ "rho=1/sigma; #calculation of resistivity\n",
+ "R=rho*l/A; #calculation of resistance \n",
+ "\n",
+ "#result\n",
+ "print\"The electron density is ne=\",ne,\"/m^3\";\n",
+ "print\"The hole density is nh=\",nh,\"/m^3\";\n",
+ "print\"The conductivity is =\",sigma,\"/ohm-m\";\n",
+ "print\"The resistance is R=\",round(R),\"ohm\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The electron density is ne= 5e+21 /m^3\n",
+ "The hole density is nh= 1.152e+17 /m^3\n",
+ "The conductivity is = 312.0 /ohm-m\n",
+ "The resistance is R= 32.0 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.16 , Page no:279"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "RH=3.66E-4; #in m^3/C (Hall coefficient)\n",
+ "rho=8.93E-3; #in ohm-m (resistivity)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "u=RH/rho; #calculation of mobility\n",
+ "n=1/(RH*e); #calculation of density\n",
+ "\n",
+ "#result\n",
+ "print\"The mobility is u=\",u,\"m^2/(V-s)\";\n",
+ "print\"The density is n=\",n,\"/m^3\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The mobility is u= 0.0409854423292 m^2/(V-s)\n",
+ "The density is n= 1.70765027322e+22 /m^3\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.17 , Page no:279"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "RH=3.66E-4; #in m^3/C (Hall coefficient)\n",
+ "rho=8.93E-3; #in ohm-m (resistivity)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "nh=1/(RH*e); #calculation of density of charge carrier\n",
+ "uh=1/(rho*nh*e); #calculation of mobility of charge carrier\n",
+ "\n",
+ "#result\n",
+ "print\"The density of charge carrier is nh=\",nh,\"/m^3\";\n",
+ "print\"The mobility of charge carrier is uh=\",uh,\"m^2/(V-s)\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The density of charge carrier is nh= 1.70765027322e+22 /m^3\n",
+ "The mobility of charge carrier is uh= 0.0409854423292 m^2/(V-s)\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Applied_Physics/Chapter_09_Semiconductors_1.ipynb b/Applied_Physics/Chapter_09_Semiconductors_1.ipynb new file mode 100755 index 00000000..9ce2ea83 --- /dev/null +++ b/Applied_Physics/Chapter_09_Semiconductors_1.ipynb @@ -0,0 +1,782 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:c155cecc785c23bdccd0bf715c04d53d7788e1b590b4310d6af27c13ce7b97ac"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 9:Semiconductors"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.2 , Page no:272"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "Eg=0.67; #in eV (Energy band gap)\n",
+ "k=1.38E-23; #in J/K (Boltzmann\u2019s constant)\n",
+ "T1=298; #in K (room temperature)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "K=10; #ratio of number of electrons at different temperature\n",
+ "\n",
+ "#calculate\n",
+ "Eg=Eg*e; #changing unit from eV to Joule\n",
+ "#since ne=Ke*exp(-Eg/(2*k*T))\n",
+ "#and ne/ne1=exp(-Eg/(2*k*T))/exp(-Eg/(2*k*T1)) and ne/ne1=K=10\n",
+ "#therefore we have 10=exp(-Eg/(2*k*T))/exp(-Eg/(2*k*T1))\n",
+ "#re-arranging the equation for T, we get T2=1/((1/T1)-((2*k*log(10))/Eg))\n",
+ "T=1/((1/T1)-((2*k*math.log(10))/Eg)); #calculation of the temperature\n",
+ "\n",
+ "#result\n",
+ "print\"The temperature at which number of electrons in the conduction band of a semiconductor increases by a factor of 10 is T=\",round(T),\"K (roundoff error)\";\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The temperature at which number of electrons in the conduction band of a semiconductor increases by a factor of 10 is T= 362.0 K (roundoff error)\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.3 , Page no:272"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "ni=2.5E13; #in /cm^3 (intrinsic carrier density)\n",
+ "ue=3900; #in cm^2/(V-s) (electron mobilities)\n",
+ "uh=1900; #in cm^2/(V-s) (hole mobilities)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "l=1; #in cm (lenght of the box)\n",
+ "b=1;h=1; #in mm (dimensions of germanium rod )\n",
+ "\n",
+ "#calculate\n",
+ "ni=ni*1E6; #changing unit from 1/cm^3 to 1/m^3\n",
+ "ue=ue*1E-4; #changing unit from cm^2 to m^2\n",
+ "uh=uh*1E-4; #changing unit from cm^2 to m^2\n",
+ "sigma=ni*e*(ue+uh); #calculation of conductivity\n",
+ "rho=1/sigma; #calculation of resistivity\n",
+ "l=l*1E-2; #changing unit from mm to m for length\n",
+ "A=(b*1E-3)*(h*1E-3); #changing unit from mm to m for width and height and calculation of cross-sectional area\n",
+ "R=rho*l/A; #calculation of resistance\n",
+ "\n",
+ "#result\n",
+ "print\"The resistance of intrinsic germanium is R=\",'%.3E'%R,\"ohm\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The resistance of intrinsic germanium is R= 4.310E+03 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.4 , Page no:273"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "ne=2.5E19; #in /m^3 (electron density)\n",
+ "nh=2.5E19; #in /m^3 (hole density)\n",
+ "ue=0.36; #in m^2/(V-s) (electron mobilities)\n",
+ "uh=0.17; #in m^2/(V-s) (hole mobilities)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "#since ne=nh=ni, therefore we have \n",
+ "ni=nh;\n",
+ "sigma=ni*e*(ue+uh); #calculation of conductivity\n",
+ "rho=1/sigma; #calculation of resistivity\n",
+ "\n",
+ "#result\n",
+ "print\"The conductivity of germanium is =\",round(sigma,2),\"/ohm-m\";\n",
+ "print\"The resistivity of germanium is =\",round(rho,2),\"ohm-m\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The conductivity of germanium is = 2.12 /ohm-m\n",
+ "The resistivity of germanium is = 0.47 ohm-m\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.5 , Page no:273"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "ni=1.5E16; #in /m^3 (intrinsic carrier density)\n",
+ "ue=0.135; #in m^2/(V-s) (electron mobilities)\n",
+ "uh=0.048; #in m^2/(V-s) (hole mobilities)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "ND=1E23; #in atom/m^3 (doping concentration)\n",
+ "\n",
+ "#calculate\n",
+ "sigma_i=ni*e*(ue+uh); #calculation of intrinsic conductivity\n",
+ "sigma=ND*ue*e; #calculation of conductivity after doping\n",
+ "rho=ni**2/ND; #calculation of equilibrium hole concentration\n",
+ "\n",
+ "#result\n",
+ "print\"The intrinsic conductivity for silicon is =\",'%.3E'%sigma_i,\"S\";\n",
+ "print\"The conductivity after doping with phosphorus atoms is =\",'%.3E'%sigma,\"S\";\n",
+ "print\"The equilibrium hole concentration is =\",'%.3E'%rho,\"/m^3\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The intrinsic conductivity for silicon is = 4.392E-04 S\n",
+ "The conductivity after doping with phosphorus atoms is = 2.160E+03 S\n",
+ "The equilibrium hole concentration is = 2.250E+09 /m^3\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.6 , Page no:274"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "ni=1.5E16; #in /m^3 (intrinsic carrier density)\n",
+ "ue=0.13; #in m^2/(V-s) (electron mobilities)\n",
+ "uh=0.05; #in m^2/(V-s) (hole mobilities)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "ne=5E20; #in /m^3 (concentration of donor type impurity)\n",
+ "nh=5E20; #in /m^3 (concentration of acceptor type impurity)\n",
+ "\n",
+ "#calculate\n",
+ "#part-i\n",
+ "sigma=ni*e*(ue+uh); #calculation of intrinsic conductivity\n",
+ "#part-ii\n",
+ "#since 1 donor atom is in 1E8 Si atoms, hence holes concentration can be neglected\n",
+ "sigma1=ne*e*ue; #calculation of conductivity after doping with donor type impurity\n",
+ "#part-iii\n",
+ "#since 1 acceptor atom is in 1E8 Si atoms, hence electron concentration can be neglected\n",
+ "sigma2=nh*e*uh; #calculation of conductivity after doping with acceptor type impurity\n",
+ "\n",
+ "#result\n",
+ "print\"The intrinsic conductivity for silicon is =\",'%.3E'%sigma,\"(ohm-m)^-1\";\n",
+ "print\"The conductivity after doping with donor type impurity is =\",sigma1,\"(ohm-m)^-1\";\n",
+ "print\"The conductivity after doping with acceptor type impurity is =\",sigma2,\"(ohm-m)^-1\";\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The intrinsic conductivity for silicon is = 4.320E-04 (ohm-m)^-1\n",
+ "The conductivity after doping with donor type impurity is = 10.4 (ohm-m)^-1\n",
+ "The conductivity after doping with acceptor type impurity is = 4.0 (ohm-m)^-1\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.7 , Page no:274"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "ni=1E20; #in /m^3 (intrinsic carrier density)\n",
+ "ND=1E21; #in /m^3 (donor impurity concentration)\n",
+ "\n",
+ "#calculate\n",
+ "nh=ni**2/ND; #calculation of density of hole carriers at room temperature\n",
+ "\n",
+ "#result\n",
+ "print\"The density of hole carriers at room temperature is nh=\",nh,\"/m^3\";\n",
+ "#Note: answer in the book is wrong due to printing mistake"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The density of hole carriers at room temperature is nh= 1e+19 /m^3\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.8 , Page no:275"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "M=72.6; #atomic mass of germanium\n",
+ "P=5400; #in Kg/m^3 (density)\n",
+ "ue=0.4; #in m^2/V-s (mobility of electrons)\n",
+ "uh=0.2; #in m^2/V-s (mobility of holes)\n",
+ "Eg=0.7; #in eV (Band gap)\n",
+ "m=9.1E-31; #in Kg (mass of electron)\n",
+ "k=1.38E-23; #in J/K (Boltzmann\u2019s constant)\n",
+ "T=300; #in K (temperature)\n",
+ "h=6.63E-34; #in J/s (Planck\u2019s constant)\n",
+ "pi=3.14; #value of pi used in the solution\n",
+ "e=1.6E-19; #in C(charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "Eg=Eg*e; #changing unit from eV to J\n",
+ "ni=2*(2*pi*m*k*T/h**2)**(3/2)*math.exp(-Eg/(2*k*T));\n",
+ "sigma=ni*e*(ue+uh);\n",
+ "\n",
+ "#result\n",
+ "print\"The intrinsic carrier density for germanium at 300K is ni=\",'%.3E'%ni,\"/m^3\";\n",
+ "print\"The conductivity of germanium is=\",round(sigma,3),\"(ohm-m)^-1\";\n",
+ "print \"NOTE: The answer in the textbook is wrong\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The intrinsic carrier density for germanium at 300K is ni= 3.334E+19 /m^3\n",
+ "The conductivity of germanium is= 3.201 (ohm-m)^-1\n",
+ "NOTE: The answer in the textbook is wrong\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9 , Page no:275"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "rho1=4.5; #in ohm-m (resistivity at 20 degree Celcius)\n",
+ "rho2=2.0; #in ohm-m (resistivity at 32 degree Celcius)\n",
+ "k=1.38E-23; #in J/K (Boltzmann\u2019s constant)\n",
+ "T1=20; T2=32; #in degree Celcius (two temperatures)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "T1=T1+273; #changing unit from degree Celius to K\n",
+ "T2=T2+273; #changing unit from degree Celius to K\n",
+ "#since sigma=e*u*C*T^(3/2)*exp(-Eg/(2*k*T))\n",
+ "#therefore sigma1/sigma2=(T1/T2)^3/2*exp((-Eg/(2*k)*((1/T1)-(1/T2))\n",
+ "#and sigma=1/rho \n",
+ "#therefore we have rho2/rho1=(T1/T2)^3/2*exp((-Eg/(2*k)*((1/T1)-(1/T2))\n",
+ "#re-arranging above equation for Eg, we get Eg=(2*k/((1/T1)-(1/T2)))*((3/2)*log(T1/T2)-log(rho2/rho1))\n",
+ "Eg=(2*k/((1/T1)-(1/T2)))*((3/2)*math.log(T1/T2)-math.log(rho2/rho1));\n",
+ "Eg1=Eg/e;#changing unit from J to eV\n",
+ "\n",
+ "#result\n",
+ "print\"The energy band gap is Eg=\",'%.3E'%Eg,\"J\";\n",
+ "print\"\\t\\t\\t =\",round(Eg1,3),\"eV\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The energy band gap is Eg= 1.543E-19 J\n",
+ "\t\t\t = 0.964 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.10 , Page no:276"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "rho=2300; #in ohm-m (resistivity of pure silicon)\n",
+ "ue=0.135; #in m^2/V-s (mobility of electron)\n",
+ "uh=0.048; #in m^2/V-s (mobility of electron)\n",
+ "Nd=1E19; #in /m^3 (doping concentration)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "#since sigma=ni*e*(ue+uh) and sigma=1/rho\n",
+ "#therefore ni=1/(rho*e*(ue+uh))\n",
+ "ni=1/(rho*e*(ue+uh)); #calculation of intrinsic concentration\n",
+ "ne=Nd; #calculation of electron concentration\n",
+ "nh=ni**2/Nd; #calculation of hole concentration\n",
+ "sigma=ne*ue*e+nh*uh*e; #calculation of conductivity\n",
+ "rho=1/sigma; #calculation of resistivity\n",
+ "\n",
+ "#result\n",
+ "print\"The electron concentration is ne=\",ne,\"/m^3\";\n",
+ "print\"The hole concentration is nh=\",'%.3E'%nh,\"/m^3\";\n",
+ "print\"The resistivity of the specimen is =\",round(rho,3),\"ohm-m\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The electron concentration is ne= 1e+19 /m^3\n",
+ "The hole concentration is nh= 2.205E+13 /m^3\n",
+ "The resistivity of the specimen is = 4.63 ohm-m\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.11 , Page no:276"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "uh=1900; #in cm^2/V-s (mobility of electron)\n",
+ "Na=2E17; #in /m^3 (acceptor doping concentration)\n",
+ "e=1.6E-19; #in C(charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "uh=uh*1E-4; #changing unit from cm^2/V-s to m^2/V-s\n",
+ "Na=Na*1E6; #changing unit from 1/cm^3 to 1/m^3\n",
+ "nh=Na; #hole concentration \n",
+ "#since sigma=ne*ue*e+nh*uh*e and nh>>ne\n",
+ "#therefore sigma=nh*uh*e\n",
+ "sigma=nh*uh*e; #calculation of conductivity\n",
+ "\n",
+ "#result\n",
+ "print\"The conductivity of p-type Ge crystal is =\",sigma,\"/ohm-m (roundoff error)\";\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The conductivity of p-type Ge crystal is = 6080.0 /ohm-m (roundoff error)\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.12 , Page no:277"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "ue=0.19; #in m^2/V-s (mobility of electron)\n",
+ "T=300; #in K (temperature)\n",
+ "k=1.38E-23; #in J/K (Boltzmann\u2019s constant)\n",
+ "e=1.6E-19; #in C(charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "Dn=ue*k*T/e; #calculation of diffusion co-efficient\n",
+ "\n",
+ "#result\n",
+ "print\"The diffusion co-efficient of electron in silicon is Dn=\",'%.3E'%Dn,\"m^2/s\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The diffusion co-efficient of electron in silicon is Dn= 4.916E-03 m^2/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.13 , Page no:277"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "Eg=0.4; #in eV (Band gap of semiconductor)\n",
+ "k=1.38E-23; #in J/K (Boltzmann\u2019s constant)\n",
+ "T1=0; #in degree Celcius (first temperature)\n",
+ "T2=50; #in degree Celcius (second temperature)\n",
+ "T3=100; #in degree Celcius (third temperature)\n",
+ "e=1.602E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "T1=T1+273; #changing temperature form Celcius to Kelvin\n",
+ "T2=T2+273; #changing temperature form Celcius to Kelvin\n",
+ "T3=T3+273; #changing temperature form Celcius to Kelvin\n",
+ "Eg=Eg*e; #changing unit from eV to Joule\n",
+ "#Using F_E=1/(1+exp(Eg/2*k*T))\n",
+ "F_E1=1/(1+math.exp(Eg/(2*k*T1))); #calculation of probability of occupation of lowest level at 0 degree Celcius\n",
+ "F_E2=1/(1+math.exp(Eg/(2*k*T2))); #calculation of probability of occupation of lowest level at 50 degree Celcius\n",
+ "F_E3=1/(1+math.exp(Eg/(2*k*T3))); #calculation of probability of occupation of lowest level at 100 degree Celcius\n",
+ "\n",
+ "#result\n",
+ "print\"The probability of occupation of lowest level in conduction band is\";\n",
+ "print\"\\t at 0 degree Celcius, F_E=\",'%.3E'%F_E1,\"eV\";\n",
+ "print\"\\t at 50 degree Celcius, F_E=\",'%.3E'%F_E2,\"eV\";\n",
+ "print\"\\t at 100 degree Celcius, F_E=\",'%.3E'%F_E3,\"eV\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The probability of occupation of lowest level in conduction band is\n",
+ "\t at 0 degree Celcius, F_E= 2.025E-04 eV\n",
+ "\t at 50 degree Celcius, F_E= 7.550E-04 eV\n",
+ "\t at 100 degree Celcius, F_E= 1.976E-03 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.14 , Page no:278"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "Eg=1.2; #in eV (Energy band gap)\n",
+ "k=1.38E-23; #in J/K (Boltzmann\u2019s constant)\n",
+ "T1=600; T2=300; #in K (two temperatures)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "Eg=Eg*e; #changing unit from eV to Joule\n",
+ "#since sigma is proportional to exp(-Eg/(2*k*T))\n",
+ "#therefore ratio=sigma1/sigma2=exp(-Eg/(2*k*((1/T1)-(1/T2))));\n",
+ "ratio= math.exp((-Eg/(2*k))*((1/T1)-(1/T2))); #calculation of ratio of conductivity at 600K and at 300K\n",
+ "\n",
+ "#result\n",
+ "print\"The ratio of conductivity at 600K and at 300K is =\",'%.3E'%ratio;"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The ratio of conductivity at 600K and at 300K is = 1.085E+05\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.15 , Page no:278"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "ue=0.39; #in m^2/V-s (mobility of electron)\n",
+ "n=5E13; #number of donor atoms\n",
+ "ni=2.4E19; #in atoms/m^3 (intrinsic carrier density)\n",
+ "l=10; #in mm (length of rod)\n",
+ "a=1; #in mm (side of square cross-section)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "l=l*1E-3; #changing unit from mm to m\n",
+ "a=a*1E-3; #changing unit from mm to m\n",
+ "A=a**2; #calculation of cross-section area\n",
+ "Nd=n/(l*A); #calculation of donor concentration\n",
+ "ne=Nd; #calculation of electron density\n",
+ "nh=ni**2/Nd; #calculation of hole density\n",
+ "#since sigma=ne*e*ue+nh*e*ue and since ne>>nh\n",
+ "#therefore sigma=ne*e*ue\n",
+ "sigma=ne*e*ue; #calculation of conductivity\n",
+ "rho=1/sigma; #calculation of resistivity\n",
+ "R=rho*l/A; #calculation of resistance \n",
+ "\n",
+ "#result\n",
+ "print\"The electron density is ne=\",ne,\"/m^3\";\n",
+ "print\"The hole density is nh=\",nh,\"/m^3\";\n",
+ "print\"The conductivity is =\",sigma,\"/ohm-m\";\n",
+ "print\"The resistance is R=\",round(R),\"ohm\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The electron density is ne= 5e+21 /m^3\n",
+ "The hole density is nh= 1.152e+17 /m^3\n",
+ "The conductivity is = 312.0 /ohm-m\n",
+ "The resistance is R= 32.0 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.16 , Page no:279"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "RH=3.66E-4; #in m^3/C (Hall coefficient)\n",
+ "rho=8.93E-3; #in ohm-m (resistivity)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "u=RH/rho; #calculation of mobility\n",
+ "n=1/(RH*e); #calculation of density\n",
+ "\n",
+ "#result\n",
+ "print\"The mobility is u=\",round(u,4),\"m^2/(V-s)\";\n",
+ "print\"The density is n=\",'%.3E'%n,\"/m^3\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The mobility is u= 0.041 m^2/(V-s)\n",
+ "The density is n= 1.708E+22 /m^3\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.17 , Page no:279"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "RH=3.66E-4; #in m^3/C (Hall coefficient)\n",
+ "rho=8.93E-3; #in ohm-m (resistivity)\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "\n",
+ "#calculate\n",
+ "nh=1/(RH*e); #calculation of density of charge carrier\n",
+ "uh=1/(rho*nh*e); #calculation of mobility of charge carrier\n",
+ "\n",
+ "#result\n",
+ "print\"The density of charge carrier is nh=\",'%.3E'%nh,\"/m^3\";\n",
+ "print\"The mobility of charge carrier is uh=\",round(uh,4),\"m^2/(V-s)\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The density of charge carrier is nh= 1.708E+22 /m^3\n",
+ "The mobility of charge carrier is uh= 0.041 m^2/(V-s)\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Applied_Physics/Chapter_10_Superconductivity.ipynb b/Applied_Physics/Chapter_10_Superconductivity.ipynb new file mode 100755 index 00000000..7b7470a8 --- /dev/null +++ b/Applied_Physics/Chapter_10_Superconductivity.ipynb @@ -0,0 +1,316 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:03ee5374207ea96e9a4f31e16f771d187a11f07daf287d94fc54e352b5db8bff"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 10:Superconductivity"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.1 , Page no:313"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "Tc=7.2; #in K (critical temperature)\n",
+ "T=5; #in K (given temperature)\n",
+ "H0=6.5E3; #in A/m (critical magnetic field at 0K)\n",
+ "\n",
+ "#calculate\n",
+ "Hc=H0*(1-(T/Tc)**2); #calculation of magnitude of critical magnetic field\n",
+ "\n",
+ "#result\n",
+ "print\"The magnitude of critical magnetic field is Hc=\",round(Hc,2),\"A/m\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The magnitude of critical magnetic field is Hc= 3365.35 A/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.2 , Page no:313"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "r=0.02; #in m (radius of ring)\n",
+ "Hc=2E3; #in A/m (critical magnetic field at 5K)\n",
+ "pi=3.14; #value of pi used in the solutiion\n",
+ "\n",
+ "#calculate\n",
+ "Ic=2*pi*r*Hc; #calculation of critical current value\n",
+ "\n",
+ "#result\n",
+ "print\"The critical current value is Ic=\",Ic,\"A\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The critical current value is Ic= 251.2 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.3 , Page no:313"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "M1=199.5; #in amu (isotropic mass at 5K)\n",
+ "T1=5; #in K (first critical temperature)\n",
+ "T2=5.1; #in K (second critical temperature)\n",
+ "#calculate\n",
+ "#since Tc=C*(1/sqrt(M)\n",
+ "#therefore T1*sqrt(M1)=T2*sqrt(M2)\n",
+ "#therefore we have M2=(T1/T2)^2*M1\n",
+ "M2=(T1/T2)**2*M1; #calculation of isotropic mass at 5.1K\n",
+ "\n",
+ "#result\n",
+ "print\"The isotropic mass at 5.1K is M2=\",round(M2,3),\"a.m.u.\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The isotropic mass at 5.1K is M2= 191.753 a.m.u.\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.4 , Page no:314"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "T=6; #in K (given temperature)\n",
+ "Hc=5E3; #in A/m (critical magnetic field at 5K)\n",
+ "H0=2E4; #in A/m (critical magnetic field at 0K)\n",
+ "\n",
+ "#calculate\n",
+ "#since Hc=H0*(1-(T/Tc)^2)\n",
+ "#therefor we have Tc=T/sqrt(1-(Hc/H)^2)\n",
+ "Tc=T/math.sqrt(1-(Hc/H0)); #calculation of transition temperature\n",
+ "\n",
+ "#result\n",
+ "print\"The transition temperature is Tc=\",round(Tc,3),\"K\";\n",
+ "print \"NOTE: The answer in the textbook is wrong\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The transition temperature is Tc= 6.928 K\n",
+ "NOTE: The answer in the textbook is wrong\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.5 , Page no:314"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "T=5; #in K (given temperature)\n",
+ "d=3; #in mm (diameter of the wire)\n",
+ "Tc=8; #in K (critical temperature for Pb)\n",
+ "H0=5E4; #in A/m (critical magnetic field at 0K)\n",
+ "pi=3.14; #value of pi used in the solution\n",
+ "\n",
+ "#calculate\n",
+ "Hc=H0*(1-(T/Tc)**2); #calculation of critical magnetic field at 5K\n",
+ "r=(d*1E-3)/2; #calculation of radius in m\n",
+ "Ic=2*pi*r*Hc; #calculation of critical current at 5K\n",
+ "\n",
+ "#result\n",
+ "print\"The critical magnetic field at 5K is Hc=\",Hc,\"A/m\";\n",
+ "print\"The critical current at 5K is Ic=\",Ic,\"A\";\n",
+ "print \" (roundoff error)\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The critical magnetic field at 5K is Hc= 30468.75 A/m\n",
+ "The critical current at 5K is Ic= 287.015625 A\n",
+ " (roundoff error)\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.6 , Page no:314"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "V=8.50; #in micro V (voltage across Josephson junction )\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "h=6.626E-34; #in J/s (Planck\u2019s constant)\n",
+ "\n",
+ "#calculate\n",
+ "V=V*1E-6; #changing unit from V to microVolt\n",
+ "v1=2*e*V/h; #calculation of frequency of EM waves\n",
+ "\n",
+ "#result\n",
+ "print\"The frequency of EM waves is v=\",v1,\"Hz\";\n",
+ "print \"NOTE: The answer in the textbook is wrong\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The frequency of EM waves is v= 4105040748.57 Hz\n",
+ "NOTE: The answer in the textbook is wrong\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.7 , Page no:315"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "M1=200.59; #in amu (average atomic mass at 4.153K)\n",
+ "Tc1=4.153; #in K (first critical temperature)\n",
+ "M2=204; #in amu (average atomic mass of isotopes)\n",
+ "\n",
+ "#calculate\n",
+ "#since Tc=C*(1/sqrt(M)\n",
+ "#therefore T1*sqrt(M1)=T2*sqrt(M2)\n",
+ "#therefore we have Tc2=Tc1*sqrt(M1/M2)\n",
+ "Tc2=Tc1*math.sqrt(M1/M2); #calculation of transition temperature of the isotopes\n",
+ "\n",
+ "#result\n",
+ "print\"The transition temperature of the isotopes is Tc2=\",round(Tc2,3),\"K\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The transition temperature of the isotopes is Tc2= 4.118 K\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Applied_Physics/Chapter_10_Superconductivity_1.ipynb b/Applied_Physics/Chapter_10_Superconductivity_1.ipynb new file mode 100755 index 00000000..9b11ce44 --- /dev/null +++ b/Applied_Physics/Chapter_10_Superconductivity_1.ipynb @@ -0,0 +1,316 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:2419f2161da8f36d7fa10fabb5f5aa4b318a9912237faaa5f03ac922aa4b8ac9"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 10:Superconductivity"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.1 , Page no:313"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "Tc=7.2; #in K (critical temperature)\n",
+ "T=5; #in K (given temperature)\n",
+ "H0=6.5E3; #in A/m (critical magnetic field at 0K)\n",
+ "\n",
+ "#calculate\n",
+ "Hc=H0*(1-(T/Tc)**2); #calculation of magnitude of critical magnetic field\n",
+ "\n",
+ "#result\n",
+ "print\"The magnitude of critical magnetic field is Hc=\",round(Hc,2),\"A/m\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The magnitude of critical magnetic field is Hc= 3365.35 A/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.2 , Page no:313"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "r=0.02; #in m (radius of ring)\n",
+ "Hc=2E3; #in A/m (critical magnetic field at 5K)\n",
+ "pi=3.14; #value of pi used in the solutiion\n",
+ "\n",
+ "#calculate\n",
+ "Ic=2*pi*r*Hc; #calculation of critical current value\n",
+ "\n",
+ "#result\n",
+ "print\"The critical current value is Ic=\",Ic,\"A\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The critical current value is Ic= 251.2 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.3 , Page no:313"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "M1=199.5; #in amu (isotropic mass at 5K)\n",
+ "T1=5; #in K (first critical temperature)\n",
+ "T2=5.1; #in K (second critical temperature)\n",
+ "#calculate\n",
+ "#since Tc=C*(1/sqrt(M)\n",
+ "#therefore T1*sqrt(M1)=T2*sqrt(M2)\n",
+ "#therefore we have M2=(T1/T2)^2*M1\n",
+ "M2=(T1/T2)**2*M1; #calculation of isotropic mass at 5.1K\n",
+ "\n",
+ "#result\n",
+ "print\"The isotropic mass at 5.1K is M2=\",round(M2,3),\"a.m.u.\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The isotropic mass at 5.1K is M2= 191.753 a.m.u.\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.4 , Page no:314"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "T=6; #in K (given temperature)\n",
+ "Hc=5E3; #in A/m (critical magnetic field at 5K)\n",
+ "H0=2E4; #in A/m (critical magnetic field at 0K)\n",
+ "\n",
+ "#calculate\n",
+ "#since Hc=H0*(1-(T/Tc)^2)\n",
+ "#therefor we have Tc=T/sqrt(1-(Hc/H)^2)\n",
+ "Tc=T/math.sqrt(1-(Hc/H0)); #calculation of transition temperature\n",
+ "\n",
+ "#result\n",
+ "print\"The transition temperature is Tc=\",round(Tc,3),\"K\";\n",
+ "print \"NOTE: The answer in the textbook is wrong\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The transition temperature is Tc= 6.928 K\n",
+ "NOTE: The answer in the textbook is wrong\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.5 , Page no:314"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "T=5; #in K (given temperature)\n",
+ "d=3; #in mm (diameter of the wire)\n",
+ "Tc=8; #in K (critical temperature for Pb)\n",
+ "H0=5E4; #in A/m (critical magnetic field at 0K)\n",
+ "pi=3.14; #value of pi used in the solution\n",
+ "\n",
+ "#calculate\n",
+ "Hc=H0*(1-(T/Tc)**2); #calculation of critical magnetic field at 5K\n",
+ "r=(d*1E-3)/2; #calculation of radius in m\n",
+ "Ic=2*pi*r*Hc; #calculation of critical current at 5K\n",
+ "\n",
+ "#result\n",
+ "print\"The critical magnetic field at 5K is Hc=\",'%.3E'%Hc,\"A/m\";\n",
+ "print\"The critical current at 5K is Ic=\",round(Ic,4),\"A\";\n",
+ "print \" (roundoff error)\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The critical magnetic field at 5K is Hc= 3.047E+04 A/m\n",
+ "The critical current at 5K is Ic= 287.0156 A\n",
+ " (roundoff error)\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.6 , Page no:314"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "V=8.50; #in micro V (voltage across Josephson junction )\n",
+ "e=1.6E-19; #in C (charge of electron)\n",
+ "h=6.626E-34; #in J/s (Planck\u2019s constant)\n",
+ "\n",
+ "#calculate\n",
+ "V=V*1E-6; #changing unit from V to microVolt\n",
+ "v1=2*e*V/h; #calculation of frequency of EM waves\n",
+ "\n",
+ "#result\n",
+ "print\"The frequency of EM waves is v=\",'%.3E'%v1,\"Hz\";\n",
+ "print \"NOTE: The answer in the textbook is wrong\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The frequency of EM waves is v= 4.105E+09 Hz\n",
+ "NOTE: The answer in the textbook is wrong\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 10.7 , Page no:315"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "M1=200.59; #in amu (average atomic mass at 4.153K)\n",
+ "Tc1=4.153; #in K (first critical temperature)\n",
+ "M2=204; #in amu (average atomic mass of isotopes)\n",
+ "\n",
+ "#calculate\n",
+ "#since Tc=C*(1/sqrt(M)\n",
+ "#therefore T1*sqrt(M1)=T2*sqrt(M2)\n",
+ "#therefore we have Tc2=Tc1*sqrt(M1/M2)\n",
+ "Tc2=Tc1*math.sqrt(M1/M2); #calculation of transition temperature of the isotopes\n",
+ "\n",
+ "#result\n",
+ "print\"The transition temperature of the isotopes is Tc2=\",round(Tc2,3),\"K\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The transition temperature of the isotopes is Tc2= 4.118 K\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Applied_Physics/Chapter_12Fibre_Optics.ipynb b/Applied_Physics/Chapter_12Fibre_Optics.ipynb new file mode 100755 index 00000000..95067254 --- /dev/null +++ b/Applied_Physics/Chapter_12Fibre_Optics.ipynb @@ -0,0 +1,365 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:13927a6aa5f4b8929243c9ca2c470f4da4ee5b45564e616337e68523dd18d625"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 12:Fibre Optics"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.1 , Page no:360"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "u1=1.563; #refractive index of core\n",
+ "u2=1.498; #refractive index of cladding\n",
+ "\n",
+ "#calculate\n",
+ "d=(u1-u2)/u1; #calculation of fractional index change\n",
+ "\n",
+ "#result\n",
+ "print\"The fractional index change for a given optical fibre is =\",round(d,4);"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The fractional index change for a given optical fibre is = 0.0416\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.2 , Page no:360"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "u1=1.55; #refractive index of core\n",
+ "u2=1.50; #refractive index of cladding\n",
+ "\n",
+ "#calculate\n",
+ "d=(u1-u2)/u1; #calculation of fractional index change\n",
+ "NA=u1*math.sqrt(2*d); #calculation of numerical aperture\n",
+ "theta=math.asin(NA); #calculation of acceptance angle\n",
+ "theta1=theta*180/3.14;\n",
+ "\n",
+ "#result\n",
+ "print\"The numerical aperture of the fibre is NA=\",round(NA,3);\n",
+ "print\"The acceptance angle of the optical fibre is =\",round(theta1,2),\"degree\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The numerical aperture of the fibre is NA= 0.394\n",
+ "The acceptance angle of the optical fibre is = 23.2 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.3 , Page no:360"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "u1=1.563; #refractive index of core\n",
+ "u2=1.498; #refractive index of cladding\n",
+ "\n",
+ "#calculate\n",
+ "NA=math.sqrt(u1**2-u2**2); #calculation of numerical aperture\n",
+ "theta=math.asin(NA); #calculation of acceptance angle\n",
+ "theta1=theta*180/3.14;\n",
+ "#result\n",
+ "print\"The numerical aperture of the fibre is NA=\",round(NA,4);\n",
+ "print\"The acceptance angle of the optical fibre is =\",round(theta1,2),\"degree\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The numerical aperture of the fibre is NA= 0.4461\n",
+ "The acceptance angle of the optical fibre is = 26.5 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.4 , Page no:360"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "NA=0.39; #numerical aperture of the optical fibre\n",
+ "d=0.05; #difference in the refractive index of the material of the core and cladding\n",
+ "\n",
+ "#calculate\n",
+ "#since NA=u1*sqrt(2*d)\n",
+ "#we have u1=NA/sqrt(2*d)\n",
+ "u1= NA/math.sqrt(2*d); #calculation of refractive index of material of the core\n",
+ "\n",
+ "#result\n",
+ "print\"The refractive index of material of the core is u1=\",round(u1,3);"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The refractive index of material of the core is u1= 1.233\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.5 , Page no:361"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "u1=1.50; #refractive index of core\n",
+ "u2=1.45; #refractive index of cladding\n",
+ "\n",
+ "#calculate\n",
+ "d=(u1-u2)/u1; #calculation of fractional index change\n",
+ "NA=u1*math.sqrt(2*d); #calculation of numerical aperture\n",
+ "theta_0=math.asin(NA); #calculation of acceptance angle\n",
+ "theta_01=theta_0*180/3.14;\n",
+ "theta_c=math.asin(u2/u1); #calculation of critical angle\n",
+ "theta_c1=theta_c*180/3.14;\n",
+ "\n",
+ "#result\n",
+ "print\"The numerical aperture of the fibre is NA=\",round(NA,3);\n",
+ "print\"The acceptance angle of the optical fibre is =\",round(theta_01,2),\"degree\";\n",
+ "print\"The critical angle of the optical fibre is =\",round(theta_c1,2),\"degree\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The numerical aperture of the fibre is NA= 0.387\n",
+ "The acceptance angle of the optical fibre is = 22.8 degree\n",
+ "The critical angle of the optical fibre is = 75.2 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.6 , Page no:361"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "NA=0.33; #numerical aperture\n",
+ "d=0.02; #difference in the refractive index of the core and cladding of the material\n",
+ "\n",
+ "#calculate\n",
+ "#since NA=u1*sqrt(2*d)\n",
+ "#therefore we have\n",
+ "u1=NA/math.sqrt(2*d); #calculation of refractive index of the core\n",
+ "#since d=(u1-u2)/u2\n",
+ "#therefore we have\n",
+ "u2=(1-d)*u1; #calculation of refractive index of the cladding\n",
+ "\n",
+ "#result\n",
+ "print\"The refractive index of the core is u1=\",round(u1,2);\n",
+ "print\"The refractive index of the cladding is u2=\",round(u2,3);\n",
+ "print \"NOTE: The answer in the textbook is wrong\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The refractive index of the core is u1= 1.65\n",
+ "The refractive index of the cladding is u2= 1.617\n",
+ "NOTE: The answer in the textbook is wrong\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.7 , Page no:361"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "u1=3.5; #refractive index of core\n",
+ "u2=3.45; #refractive index of cladding\n",
+ "u0=1; #refractive index of the air\n",
+ "\n",
+ "#calculate\n",
+ "NA=math.sqrt(u1**2-u2**2); #calculation of numerical aperture\n",
+ "NA1=NA/u0;\n",
+ "alpha=math.asin(NA); #calculation of acceptance angle\n",
+ "alpha1=alpha*180/3.14;\n",
+ "#result\n",
+ "print\"The numerical aperture of the fibre is NA=\",round(NA1,2);\n",
+ "print\"The acceptance angle of the optical fibre is =\",round(alpha1,2),\"degree\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The numerical aperture of the fibre is NA= 0.59\n",
+ "The acceptance angle of the optical fibre is = 36.14 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.8 , Page no:361"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "u1=1.48; #refractive index of core\n",
+ "u2=1.45; #refractive index of cladding\n",
+ "\n",
+ "#calculate\n",
+ "NA=math.sqrt(u1**2-u2**2); #calculation of numerical aperture\n",
+ "theta=math.asin(NA); #calculation of acceptance angle\n",
+ "theta1=theta*180/3.14;\n",
+ "#result\n",
+ "print\"The numerical aperture of the fibre is NA=\",round(NA,3);\n",
+ "print\"The acceptance angle of the optical fibre is =\",round(theta1,2),\"degree\";\n",
+ "print \" (roundoff error)\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The numerical aperture of the fibre is NA= 0.296\n",
+ "The acceptance angle of the optical fibre is = 17.26 degree\n",
+ " (roundoff error)\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Applied_Physics/Chapter_12_Fibre_Optics.ipynb b/Applied_Physics/Chapter_12_Fibre_Optics.ipynb new file mode 100755 index 00000000..f573b0a2 --- /dev/null +++ b/Applied_Physics/Chapter_12_Fibre_Optics.ipynb @@ -0,0 +1,365 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:dac7a0b06f622e783ea6c469bf1c100c0c8157cc5e77bffea65e378f258808b2"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 12:Fibre Optics"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.1 , Page no:360"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "u1=1.563; #refractive index of core\n",
+ "u2=1.498; #refractive index of cladding\n",
+ "\n",
+ "#calculate\n",
+ "d=(u1-u2)/u1; #calculation of fractional index change\n",
+ "\n",
+ "#result\n",
+ "print\"The fractional index change for a given optical fibre is =\",round(d,4);"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The fractional index change for a given optical fibre is = 0.0416\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.2 , Page no:360"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "u1=1.55; #refractive index of core\n",
+ "u2=1.50; #refractive index of cladding\n",
+ "\n",
+ "#calculate\n",
+ "d=(u1-u2)/u1; #calculation of fractional index change\n",
+ "NA=u1*math.sqrt(2*d); #calculation of numerical aperture\n",
+ "theta=math.asin(NA); #calculation of acceptance angle\n",
+ "theta1=theta*180/3.14;\n",
+ "\n",
+ "#result\n",
+ "print\"The numerical aperture of the fibre is NA=\",round(NA,3);\n",
+ "print\"The acceptance angle of the optical fibre is =\",round(theta1,2),\"degree\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The numerical aperture of the fibre is NA= 0.394\n",
+ "The acceptance angle of the optical fibre is = 23.2 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.3 , Page no:360"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "u1=1.563; #refractive index of core\n",
+ "u2=1.498; #refractive index of cladding\n",
+ "\n",
+ "#calculate\n",
+ "NA=math.sqrt(u1**2-u2**2); #calculation of numerical aperture\n",
+ "theta=math.asin(NA); #calculation of acceptance angle\n",
+ "theta1=theta*180/3.14;\n",
+ "#result\n",
+ "print\"The numerical aperture of the fibre is NA=\",round(NA,4);\n",
+ "print\"The acceptance angle of the optical fibre is =\",round(theta1,2),\"degree\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The numerical aperture of the fibre is NA= 0.4461\n",
+ "The acceptance angle of the optical fibre is = 26.5 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.4 , Page no:360"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "NA=0.39; #numerical aperture of the optical fibre\n",
+ "d=0.05; #difference in the refractive index of the material of the core and cladding\n",
+ "\n",
+ "#calculate\n",
+ "#since NA=u1*sqrt(2*d)\n",
+ "#we have u1=NA/sqrt(2*d)\n",
+ "u1= NA/math.sqrt(2*d); #calculation of refractive index of material of the core\n",
+ "\n",
+ "#result\n",
+ "print\"The refractive index of material of the core is u1=\",round(u1,3);"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The refractive index of material of the core is u1= 1.233\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.5 , Page no:361"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "u1=1.50; #refractive index of core\n",
+ "u2=1.45; #refractive index of cladding\n",
+ "\n",
+ "#calculate\n",
+ "d=(u1-u2)/u1; #calculation of fractional index change\n",
+ "NA=u1*math.sqrt(2*d); #calculation of numerical aperture\n",
+ "theta_0=math.asin(NA); #calculation of acceptance angle\n",
+ "theta_01=theta_0*180/3.14;\n",
+ "theta_c=math.asin(u2/u1); #calculation of critical angle\n",
+ "theta_c1=theta_c*180/3.14;\n",
+ "\n",
+ "#result\n",
+ "print\"The numerical aperture of the fibre is NA=\",round(NA,3);\n",
+ "print\"The acceptance angle of the optical fibre is =\",round(theta_01,2),\"degree\";\n",
+ "print\"The critical angle of the optical fibre is =\",round(theta_c1,2),\"degree\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The numerical aperture of the fibre is NA= 0.387\n",
+ "The acceptance angle of the optical fibre is = 22.8 degree\n",
+ "The critical angle of the optical fibre is = 75.2 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.6 , Page no:361"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "NA=0.33; #numerical aperture\n",
+ "d=0.02; #difference in the refractive index of the core and cladding of the material\n",
+ "\n",
+ "#calculate\n",
+ "#since NA=u1*sqrt(2*d)\n",
+ "#therefore we have\n",
+ "u1=NA/math.sqrt(2*d); #calculation of refractive index of the core\n",
+ "#since d=(u1-u2)/u2\n",
+ "#therefore we have\n",
+ "u2=(1-d)*u1; #calculation of refractive index of the cladding\n",
+ "\n",
+ "#result\n",
+ "print\"The refractive index of the core is u1=\",round(u1,2);\n",
+ "print\"The refractive index of the cladding is u2=\",round(u2,3);\n",
+ "print \"NOTE: The answer in the textbook is wrong\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The refractive index of the core is u1= 1.65\n",
+ "The refractive index of the cladding is u2= 1.617\n",
+ "NOTE: The answer in the textbook is wrong\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.7 , Page no:361"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "u1=3.5; #refractive index of core\n",
+ "u2=3.45; #refractive index of cladding\n",
+ "u0=1; #refractive index of the air\n",
+ "\n",
+ "#calculate\n",
+ "NA=math.sqrt(u1**2-u2**2); #calculation of numerical aperture\n",
+ "NA1=NA/u0;\n",
+ "alpha=math.asin(NA); #calculation of acceptance angle\n",
+ "alpha1=alpha*180/3.14;\n",
+ "#result\n",
+ "print\"The numerical aperture of the fibre is NA=\",round(NA1,2);\n",
+ "print\"The acceptance angle of the optical fibre is =\",round(alpha1,2),\"degree\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The numerical aperture of the fibre is NA= 0.59\n",
+ "The acceptance angle of the optical fibre is = 36.14 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12.8 , Page no:361"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#given\n",
+ "u1=1.48; #refractive index of core\n",
+ "u2=1.45; #refractive index of cladding\n",
+ "\n",
+ "#calculate\n",
+ "NA=math.sqrt(u1**2-u2**2); #calculation of numerical aperture\n",
+ "theta=math.asin(NA); #calculation of acceptance angle\n",
+ "theta1=theta*180/3.14;\n",
+ "#result\n",
+ "print\"The numerical aperture of the fibre is NA=\",round(NA,3);\n",
+ "print\"The acceptance angle of the optical fibre is =\",round(theta1,2),\"degree\";\n",
+ "print \" (roundoff error)\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The numerical aperture of the fibre is NA= 0.296\n",
+ "The acceptance angle of the optical fibre is = 17.26 degree\n",
+ " (roundoff error)\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
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