From 41f1f72e9502f5c3de6ca16b303803dfcf1df594 Mon Sep 17 00:00:00 2001 From: Thomas Stephen Lee Date: Fri, 4 Sep 2015 22:04:10 +0530 Subject: add/remove/update books --- .../Chapter_01_Bonding_in_Solids__1.ipynb | 248 ++++ .../Chapter_02_Crystal_Structure_1.ipynb | 291 +++++ .../Chapter_03_Planes_in_Crystals_1.ipynb | 144 +++ .../Chapter_04_Crystal_Diffraction_1.ipynb | 503 ++++++++ ...pter_05_Principles_of_Quantum_Mechanics_1.ipynb | 1258 ++++++++++++++++++++ ...ectron_Theory_and_Band_Theory_of_Metals_1.ipynb | 561 +++++++++ .../Chapter_07_Dielectric_Properties_1.ipynb | 488 ++++++++ .../Chapter_08_Magnetic_Properties_1.ipynb | 323 +++++ .../Chapter_09_Semiconductors_1.ipynb | 782 ++++++++++++ .../Chapter_10_Superconductivity_1.ipynb | 316 +++++ .../Chapter_12_Fibre_Optics.ipynb | 365 ++++++ ...alculation_of__mean_free_path_of_electron_1.png | Bin 0 -> 140005 bytes .../Calculation_of__potential_energy_1.png | Bin 0 -> 149135 bytes .../Calculation_of_lattice_constant_1.png | Bin 0 -> 141934 bytes 14 files changed, 5279 insertions(+) create mode 100755 Applied_Physics_by_P_K_Mittal/Chapter_01_Bonding_in_Solids__1.ipynb create mode 100755 Applied_Physics_by_P_K_Mittal/Chapter_02_Crystal_Structure_1.ipynb create mode 100755 Applied_Physics_by_P_K_Mittal/Chapter_03_Planes_in_Crystals_1.ipynb create mode 100755 Applied_Physics_by_P_K_Mittal/Chapter_04_Crystal_Diffraction_1.ipynb create mode 100755 Applied_Physics_by_P_K_Mittal/Chapter_05_Principles_of_Quantum_Mechanics_1.ipynb create mode 100755 Applied_Physics_by_P_K_Mittal/Chapter_06_Electron_Theory_and_Band_Theory_of_Metals_1.ipynb create mode 100755 Applied_Physics_by_P_K_Mittal/Chapter_07_Dielectric_Properties_1.ipynb create mode 100755 Applied_Physics_by_P_K_Mittal/Chapter_08_Magnetic_Properties_1.ipynb create mode 100755 Applied_Physics_by_P_K_Mittal/Chapter_09_Semiconductors_1.ipynb create mode 100755 Applied_Physics_by_P_K_Mittal/Chapter_10_Superconductivity_1.ipynb create mode 100755 Applied_Physics_by_P_K_Mittal/Chapter_12_Fibre_Optics.ipynb create mode 100755 Applied_Physics_by_P_K_Mittal/screenshots/Calculation_of__mean_free_path_of_electron_1.png create mode 100755 Applied_Physics_by_P_K_Mittal/screenshots/Calculation_of__potential_energy_1.png create mode 100755 Applied_Physics_by_P_K_Mittal/screenshots/Calculation_of_lattice_constant_1.png (limited to 'Applied_Physics_by_P_K_Mittal') diff --git a/Applied_Physics_by_P_K_Mittal/Chapter_01_Bonding_in_Solids__1.ipynb b/Applied_Physics_by_P_K_Mittal/Chapter_01_Bonding_in_Solids__1.ipynb new file mode 100755 index 00000000..db06e802 --- /dev/null +++ b/Applied_Physics_by_P_K_Mittal/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_by_P_K_Mittal/Chapter_02_Crystal_Structure_1.ipynb b/Applied_Physics_by_P_K_Mittal/Chapter_02_Crystal_Structure_1.ipynb new file mode 100755 index 00000000..b6038072 --- /dev/null +++ b/Applied_Physics_by_P_K_Mittal/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_by_P_K_Mittal/Chapter_03_Planes_in_Crystals_1.ipynb b/Applied_Physics_by_P_K_Mittal/Chapter_03_Planes_in_Crystals_1.ipynb new file mode 100755 index 00000000..92dba19b --- /dev/null +++ b/Applied_Physics_by_P_K_Mittal/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_by_P_K_Mittal/Chapter_04_Crystal_Diffraction_1.ipynb b/Applied_Physics_by_P_K_Mittal/Chapter_04_Crystal_Diffraction_1.ipynb new file mode 100755 index 00000000..fef74a69 --- /dev/null +++ b/Applied_Physics_by_P_K_Mittal/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_by_P_K_Mittal/Chapter_05_Principles_of_Quantum_Mechanics_1.ipynb b/Applied_Physics_by_P_K_Mittal/Chapter_05_Principles_of_Quantum_Mechanics_1.ipynb new file mode 100755 index 00000000..66041c45 --- /dev/null +++ b/Applied_Physics_by_P_K_Mittal/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_by_P_K_Mittal/Chapter_06_Electron_Theory_and_Band_Theory_of_Metals_1.ipynb b/Applied_Physics_by_P_K_Mittal/Chapter_06_Electron_Theory_and_Band_Theory_of_Metals_1.ipynb new file mode 100755 index 00000000..d52f3a45 --- /dev/null +++ b/Applied_Physics_by_P_K_Mittal/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_by_P_K_Mittal/Chapter_07_Dielectric_Properties_1.ipynb b/Applied_Physics_by_P_K_Mittal/Chapter_07_Dielectric_Properties_1.ipynb new file mode 100755 index 00000000..85a277ef --- /dev/null +++ b/Applied_Physics_by_P_K_Mittal/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_by_P_K_Mittal/Chapter_08_Magnetic_Properties_1.ipynb b/Applied_Physics_by_P_K_Mittal/Chapter_08_Magnetic_Properties_1.ipynb new file mode 100755 index 00000000..b2a8def3 --- /dev/null +++ b/Applied_Physics_by_P_K_Mittal/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_by_P_K_Mittal/Chapter_09_Semiconductors_1.ipynb b/Applied_Physics_by_P_K_Mittal/Chapter_09_Semiconductors_1.ipynb new file mode 100755 index 00000000..9ce2ea83 --- /dev/null +++ b/Applied_Physics_by_P_K_Mittal/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_by_P_K_Mittal/Chapter_10_Superconductivity_1.ipynb b/Applied_Physics_by_P_K_Mittal/Chapter_10_Superconductivity_1.ipynb new file mode 100755 index 00000000..9b11ce44 --- /dev/null +++ b/Applied_Physics_by_P_K_Mittal/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_by_P_K_Mittal/Chapter_12_Fibre_Optics.ipynb b/Applied_Physics_by_P_K_Mittal/Chapter_12_Fibre_Optics.ipynb new file mode 100755 index 00000000..f573b0a2 --- /dev/null +++ b/Applied_Physics_by_P_K_Mittal/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": {} + } + ] +} \ No newline at end of file diff --git a/Applied_Physics_by_P_K_Mittal/screenshots/Calculation_of__mean_free_path_of_electron_1.png b/Applied_Physics_by_P_K_Mittal/screenshots/Calculation_of__mean_free_path_of_electron_1.png new file mode 100755 index 00000000..37995b76 Binary files /dev/null and b/Applied_Physics_by_P_K_Mittal/screenshots/Calculation_of__mean_free_path_of_electron_1.png differ diff --git a/Applied_Physics_by_P_K_Mittal/screenshots/Calculation_of__potential_energy_1.png b/Applied_Physics_by_P_K_Mittal/screenshots/Calculation_of__potential_energy_1.png new file mode 100755 index 00000000..f6086db3 Binary files /dev/null and b/Applied_Physics_by_P_K_Mittal/screenshots/Calculation_of__potential_energy_1.png differ diff --git a/Applied_Physics_by_P_K_Mittal/screenshots/Calculation_of_lattice_constant_1.png b/Applied_Physics_by_P_K_Mittal/screenshots/Calculation_of_lattice_constant_1.png new file mode 100755 index 00000000..cf42154d Binary files /dev/null and b/Applied_Physics_by_P_K_Mittal/screenshots/Calculation_of_lattice_constant_1.png differ -- cgit