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| author | hardythe1 | 2014-08-06 17:26:32 +0530 |
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| committer | hardythe1 | 2014-08-06 17:26:32 +0530 |
| commit | 98bff1c301dd3b8b14983037a8a483e3eae1796d (patch) | |
| tree | 1f6a831fd79192d4cbc0c4738c77f8fbf29d0552 /Engineering_Physics_Vijaya | |
| parent | 90bb10608fa3697134121eabc32cfae69e7686d8 (diff) | |
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adding books
Diffstat (limited to 'Engineering_Physics_Vijaya')
25 files changed, 5455 insertions, 0 deletions
diff --git a/Engineering_Physics_Vijaya/README.txt b/Engineering_Physics_Vijaya/README.txt new file mode 100755 index 00000000..a2116ca8 --- /dev/null +++ b/Engineering_Physics_Vijaya/README.txt @@ -0,0 +1,10 @@ +Contributed By: ADITYA DHARMASAGAR KALIDAS +Course: btech +College/Institute/Organization: vaageswari college of engineering +Department/Designation: ECE +Book Title: Engineering Physics +Author: Dr. K. Vijaya Kumar +Publisher: S. Chand & Company Ltd., New Delhi +Year of publication: 2011 +Isbn: 81-219-3852-X +Edition: 1
\ No newline at end of file diff --git a/Engineering_Physics_Vijaya/chapter1_2.ipynb b/Engineering_Physics_Vijaya/chapter1_2.ipynb new file mode 100755 index 00000000..bd2e1aac --- /dev/null +++ b/Engineering_Physics_Vijaya/chapter1_2.ipynb @@ -0,0 +1,1232 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:18ac31f959977ef2080ed3a1b1a6990ce93e604dcfb0f72ab45c0c28a2428e0e" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Quantum Mechanics and Quantum Computing" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 1.1, Page number 41" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "#Variable declaration\n", + "c=3*10**8 #velocity of light in m/s\n", + "h=6.626*10**-34 #planks constant \n", + "m=1.67*10**-27 #mass of proton\n", + "\n", + "#Calculation\n", + "v=c/10 #velocity of proton\n", + "lamda=h/(m*v) #de Broglie wave length\n", + "\n", + "#Result\n", + "print(\"the de Broglie wavelength in m is \",lamda);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('the de Broglie wavelength in m is ', 1.3225548902195607e-14)\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 1.2, Page number 42" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#importing modules\n", + "import math\n", + "\n", + "#Variable declaration\n", + "V=400; #potential in Volts\n", + "\n", + "#Calculation\n", + "lamda=12.56/math.sqrt(V); #de Broglie wavelength\n", + "\n", + "#Result\n", + "print(\"The de Broglie wavelength in Armstrong is\",lamda);\n", + "\n", + "#answer given in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('The de Broglie wavelength in Armstrong is', 0.628)\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 1.3, Page number 42\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#importing modules\n", + "import math\n", + "\n", + "#Variable declaration\n", + "m=1.674*10**(-27); #mass of neutron in kg\n", + "h=6.626*10**(-34);\n", + "E=0.025; #kinetic energy in eV\n", + "\n", + "#Calculation\n", + "Ej=E*1.6*10**-19; #kinetic energy in J\n", + "lamda=h/math.sqrt(2*m*Ej); #de Broglie wavelength\n", + "lamdaA=lamda*10**10; #converting wavelength from m to Armstrong\n", + "lamdaA=math.ceil(lamdaA*10**3)/10**3; #rounding off to 3 decimals\n", + "\n", + "#Result\n", + "print(\"The de Broglie wavelength in metres is\",lamda);\n", + "print(\"The de Broglie wavelength in Armstrong is\",lamdaA);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('The de Broglie wavelength in metres is', 1.81062582829353e-10)\n", + "('The de Broglie wavelength in Armstrong is', 1.811)\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 1.4, Page number 42" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#importing modules\n", + "import math\n", + "\n", + "#Variable declaration\n", + "V=1600; #potential in Volts\n", + "\n", + "#Calculation\n", + "lamda=12.56/math.sqrt(V); #de Broglie wavelength\n", + "lamda=math.ceil(lamda*10**2)/10**2; #rounding off to 2 decimals\n", + "\n", + "#Result\n", + "print(\"The de Broglie wavelength in Armstrong is\",lamda);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('The de Broglie wavelength in Armstrong is', 0.32)\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 1.5, Page number 42" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#importing modules\n", + "import math\n", + "\n", + "#Variable declaration\n", + "deltax=0.2; #distance in armstrong\n", + "h=6.626*10**(-34);\n", + "\n", + "#Calculation\n", + "delta_xm=deltax*10**-10; #distance in m\n", + "delta_p=h/(2*math.pi*delta_xm);\n", + "\n", + "#Result\n", + "print(\"The uncertainity in momentum of electron in kg m/sec is\",delta_p);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('The uncertainity in momentum of electron in kg m/sec is', 5.2728032646344916e-24)\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 1.6, Page number 43" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#importing modules\n", + "import math\n", + "\n", + "#Variable declaration\n", + "n1=1;\n", + "n2=1;\n", + "n3=1; #values in lowest energy\n", + "h=6.62*10**(-34);\n", + "M=9.1*10**-31; #mass in kg\n", + "L=0.1; #side in nm\n", + "\n", + "#Calculation\n", + "L=L*10**-9; #side in m\n", + "n=(n1**2)+(n2**2)+(n3**2);\n", + "E1=(n*h**2)/(8*M*L**2); #energy in j\n", + "E1eV=E1/(1.6*10**-19); #energy in eV\n", + "E1eV=math.ceil(E1eV*10)/10; #rounding off to 1 decimals\n", + "\n", + "#Result\n", + "print(\"lowest energy of electron in Joule is\",E1);\n", + "print(\"lowest energy of electron is eV\",E1eV);\n", + "\n", + "#answer for lowest energy in eV given in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('lowest energy of electron in Joule is', 1.8059505494505486e-17)\n", + "('lowest energy of electron is eV', 112.9)\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 1.7, Page number 43" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#importing modules\n", + "import math\n", + "\n", + "#Variable declaration\n", + "M=9.1*10**(-31); #mass of electron in kg\n", + "h=6.66*10**(-34);\n", + "E=2000; #kinetic energy in eV\n", + "\n", + "#Calculation\n", + "Ej=E*1.6*10**-19; #kinetic energy in J\n", + "lamda=h/math.sqrt(2*M*Ej); #de Broglie wavelength\n", + "lamdaA=lamda*10**9; #converting wavelength from m to nm\n", + "lamdaA=math.ceil(lamdaA*10**3)/10**3; #rounding off to 3 decimals\n", + "\n", + "#Result\n", + "print(\"The de Broglie wavelength in nm is\",lamdaA);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('The de Broglie wavelength in nm is', 0.028)\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 1.8, Page number 43" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#importing modules\n", + "import math\n", + "\n", + "#Variable declaration\n", + "n=1; #for minimum energy\n", + "h=6.626*10**(-34);\n", + "m=9.1*10**-31; #mass in kg\n", + "L=4*10**-10; #size in m\n", + "\n", + "#Calculation\n", + "E1=(n*h**2)/(8*m*L**2); #energy in j\n", + "\n", + "#Result\n", + "print(\"lowest energy of electron in Joule is\",E1);\n", + "\n", + "#answer given in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('lowest energy of electron in Joule is', 3.7692201236263733e-19)\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 1.9, Page number 44" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#importing modules\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "h=6.626*10**(-34);\n", + "m=9.1*10**-31; #mass in kg\n", + "lamda=1.66*10**-10; #wavelength in m\n", + "\n", + "#Calculation\n", + "v=h/(m*lamda); #velocity in m/sec\n", + "v_km=v*10**-3; #velocity in km/sec\n", + "E=(1/2)*m*v**2; #kinetic energy in joule\n", + "EeV=E/(1.6*10**-19); #energy in eV\n", + "EeV=math.ceil(EeV*10**3)/10**3; #rounding off to 3 decimals\n", + "\n", + "#Result\n", + "print(\"velocity of electron in m/sec is\",round(v));\n", + "print(\"velocity of electron in km/sec is\",round(v_km));\n", + "print(\"kinetic energy of electron in Joule is\",E);\n", + "print(\"kinetic energy of electron in eV is\",EeV);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('velocity of electron in m/sec is', 4386337.0)\n", + "('velocity of electron in km/sec is', 4386.0)\n", + "('kinetic energy of electron in Joule is', 8.754176510091736e-18)\n", + "('kinetic energy of electron in eV is', 54.714)\n" + ] + } + ], + "prompt_number": 26 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 1.10, Page number 44" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#importing modules\n", + "import math\n", + "\n", + "#Variable decleration\n", + "V=15; #potential in kV\n", + "\n", + "#Calculation\n", + "v=V*10**3; #potential in V\n", + "lamda=12.26/math.sqrt(v); #de Broglie wavelength\n", + "lamda=math.ceil(lamda*10**2)/10**2 #rounding off to 2 decimals\n", + "\n", + "#result\n", + "print(\"The de Broglie wavelength in Armstrong is\",lamda);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('The de Broglie wavelength in Armstrong is', 0.11)\n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 1.11, Page number 44\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#importing modules\n", + "import math\n", + "\n", + "#Calculation\n", + "m=1.675*10**-27; #mass of neutron in kg\n", + "h=6.626*10**-34;\n", + "E=10; #kinetic energy in keV\n", + "\n", + "#Calculation\n", + "EeV=E*10**3; #Energy in eV\n", + "Ej=EeV*1.6*10**-19; #kinetic energy in J\n", + "v=math.sqrt(2*Ej/m); #velocity in m/s\n", + "lamda=h/(m*v); #de broglie wavelength in m\n", + "lamda_A=lamda*10**10; #de broglie wavelength in armstrong\n", + "lamda_A=math.ceil(lamda_A*10**4)/10**4 #rounding off to 4 decimals\n", + "\n", + "#Result\n", + "print(\"The velocity in m/sec is\",round(v));\n", + "print(\"The de Broglie wavelength in metres is\",lamda);\n", + "print(\"The de Broglie wavelength in Armstrong is\",lamda_A);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('The velocity in m/sec is', 1382189.0)\n", + "('The de Broglie wavelength in metres is', 2.861996093951046e-13)\n", + "('The de Broglie wavelength in Armstrong is', 0.0029)\n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 1.12, Page number 45" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#importing modules\n", + "import math\n", + "\n", + "#Variable decleration\n", + "m=9.1*10**-31; #mass of electron in kg\n", + "h=6.6*10**-34;\n", + "E=2; #kinetic energy in keV\n", + "\n", + "#Calculation\n", + "EeV=E*10**3; #Energy in eV\n", + "Ej=EeV*1.6*10**-19; #kinetic energy in J\n", + "p=math.sqrt(2*m*Ej); #momentum\n", + "lamda=h/p; #de broglie wavelength in m\n", + "lamda_A=lamda*10**10; #de broglie wavelength in armstrong\n", + "lamda_A=math.ceil(lamda_A*10**4)/10**4 #rounding off to 4 decimals\n", + "\n", + "#Result\n", + "print(\"The de Broglie wavelength in metres is\",lamda);\n", + "print(\"The de Broglie wavelength in Armstrong is\",lamda_A);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('The de Broglie wavelength in metres is', 2.7348483695436575e-11)\n", + "('The de Broglie wavelength in Armstrong is', 0.2735)\n" + ] + } + ], + "prompt_number": 29 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 1.13, Page number 45" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "\n", + "#Variable decleration\n", + "m=1.676*10**-27; #mass of neutron in kg\n", + "h=6.62*10**-34;\n", + "E=0.025; #kinetic energy in eV\n", + "\n", + "#Calculation\n", + "Ej=E*1.6*10**-19; #kinetic energy in J\n", + "v=math.sqrt(2*Ej/m); #velocity in m/s\n", + "lamda=h/(m*v); #wavelength in m\n", + "lamda_A=lamda*10**10; #de broglie wavelength in armstrong\n", + "lamda_A=math.ceil(lamda_A*10**5)/10**5 #rounding off to 5 decimals\n", + "\n", + "#Result\n", + "print(\"The neutrons wavelength in metres is\",lamda);\n", + "print(\"The wavelength in Armstrong is\",lamda_A);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('The neutrons wavelength in metres is', 1.8079065940980725e-10)\n", + "('The wavelength in Armstrong is', 1.80791)\n" + ] + } + ], + "prompt_number": 30 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 1.14, Page number 45" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "\n", + "#Variable decleration\n", + "V=10; #potential in kV\n", + "\n", + "#Calculation\n", + "V=V*10**3; #potential in V\n", + "lamda=12.26/math.sqrt(V); #wavelength\n", + "\n", + "#Result\n", + "print(\"The wavelength in Armstrong is\",lamda);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('The wavelength in Armstrong is', 0.1226)\n" + ] + } + ], + "prompt_number": 31 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 1.15, Page number 45" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "\n", + "#Varialble decleration\n", + "h=6.626*10**-34;\n", + "m=9.1*10**-31; #mass in kg\n", + "l=1; #width in armstrong\n", + "\n", + "#Calculation\n", + "L=l*10**-10; #width in m\n", + "#permitted electron energies En=(n**2*h**2)/(8*m*L**2)\n", + "#let X = h**2/(8*m*L**2)\n", + "X = h**2/(8*m*L**2); #energy in J\n", + "XeV=X/(1.6*10**-19); #energy in eV\n", + "#in the 1st level n1=1\n", + "n1=1;\n", + "E1=(n1**2)*XeV; #energy in eV\n", + "\n", + "#in second level n2=2\n", + "n2=2;\n", + "E2=(n2**2)*XeV; #energy in eV\n", + "#in third level n3=\n", + "n3=3;\n", + "E3=(n3**2)*XeV; #energy in eV\n", + "\n", + "#Result\n", + "print(\"minimum energy the electron can have in eV is\",round(E1));\n", + "print(\"other values of energy are in eV and in eV\",round(E2),round(E3));\n", + "\n", + "#answers given in the book are wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('minimum energy the electron can have in eV is', 38.0)\n", + "('other values of energy are in eV and in eV', 151.0, 339.0)\n" + ] + } + ], + "prompt_number": 32 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 1.16, Page number 46\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "\n", + "#Variable decleration\n", + "n=1; #lowest state\n", + "L=10; #width in armstrong\n", + "\n", + "#Calculation\n", + "L=L*10**-10; #width in m\n", + "x=L/2;\n", + "delta_x=1; #interval in armstrong\n", + "delta_x=delta_x*10**-10; #interval in m\n", + "psi1=(math.sqrt(2/L))*math.sin(math.pi*x/L);\n", + "A=psi1**2;\n", + "p=A*delta_x;\n", + "p=math.ceil(p*10)/10; #de broglie wavelength in armstrong\n", + "\n", + "#Result\n", + "print(\"probability of finding the particle is \",p);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('probability of finding the particle is ', 0.2)\n" + ] + } + ], + "prompt_number": 33 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 1.17, Page number 46" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "d=970; #density of Na in kg/m^3\n", + "n=6.02*10**26;\n", + "h=6.62*10**(-34);\n", + "m=9.1*10**-31; #mass in kg\n", + "w=23; #atomic weight\n", + "\n", + "#Calculation\n", + "N=(d*n)/w; #number of atoms per m^3\n", + "A=(h**2)/(8*m);\n", + "B=(3*N)/math.pi;\n", + "Ef=A*B**(2/3);\n", + "EfeV=Ef/(1.6*10**-19);\n", + "EfeV=math.ceil(EfeV*10**2)/10**2 #rounding of to 2 decimals\n", + "\n", + "#Result\n", + "print(\"fermi energy of Na in eV is\",EfeV);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('fermi energy of Na in eV is', 3.16)\n" + ] + } + ], + "prompt_number": 34 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 1.18, Page number 46" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "n1=1;\n", + "n2=1;\n", + "n3=1; #values in lowest energy\n", + "h=6.62*10**(-34);\n", + "m=9.1*10**-31; #mass in kg\n", + "L=0.1; #side in nm\n", + "\n", + "#Calculation\n", + "L=L*10**-9; #side in m\n", + "n=(n1**2)+(n2**2)+(n3**2);\n", + "E1=(n*h**2)/(8*m*L**2); #energy in j\n", + "E1eV=E1/(1.6*10**-19); #energy in eV\n", + "E1eV=math.ceil(E1eV*10**1)/10**1 #rounding off to 2 decimals\n", + "\n", + "#Result\n", + "print(\"lowest energy of electron in Joule is\",E1);\n", + "print(\"lowest energy of electron in eV is\",E1eV);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('lowest energy of electron in Joule is', 1.8059505494505486e-17)\n", + "('lowest energy of electron in eV is', 112.9)\n" + ] + } + ], + "prompt_number": 35 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 1.19, Page number 47" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "mn=1.676*10**-27; #mass of neutron in kg\n", + "me=9.1*10**-31; #mass of electron in kg\n", + "h=6.62*10**-34;\n", + "c=3*10**8; #velocity of light in m/sec\n", + "\n", + "#Calculation\n", + "En=2*me*c**2;\n", + "lamda=h/math.sqrt(2*mn*En); #wavelength in m\n", + "lamda_A=lamda*10**10; #converting lamda from m to A\n", + "lamda_A=math.ceil(lamda_A*10**6)/10**6 #rounding off to 6 decimals\n", + "\n", + "#Result\n", + "print(\"The de broglie wavelength in Angstrom is\",lamda_A);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('The de broglie wavelength in Angstrom is', 0.000283)\n" + ] + } + ], + "prompt_number": 36 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 1.20, Page number 47 ***************************************************************************" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#import module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "n2=2; #second quantum state\n", + "n4=4; #fourth quantum state\n", + "h=6.626*10**-34;\n", + "m=9.1*10**-31; #mass in kg\n", + "a=2; #potential box length in armstrong\n", + "\n", + "#Calculation\n", + "a=a*10**-10; #length in m\n", + "A=n2**2*h**2;\n", + "B=8*m*a**2;\n", + "E2=A/B; #energy in j\n", + "E2eV=E2/(1.6*10**-19); #energy in eV\n", + "C=n4**2*h**2;\n", + "E4=C/B; #energy in j\n", + "E4eV=E4/(1.6*10**-19); #energy in eV\n", + "\n", + "#Result\n", + "print(\"energy corresponding to second quantum state in Joule is\",E2);\n", + "print(\"energy corresponding to second quantum state in eV is\",E2eV);\n", + "print(\"energy corresponding to fourth quantum state in Joule is\",E4);\n", + "print(\"energy corresponding to fourth quantum state in eV is\",E4eV);\n", + "\n", + "\n", + "#answers given in the book are wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('energy corresponding to second quantum state in Joule is', 6.030752197802197e-18)\n", + "('energy corresponding to second quantum state in eV is', 37.69220123626373)\n", + "('energy corresponding to fourth quantum state in Joule is', 2.412300879120879e-17)\n", + "('energy corresponding to fourth quantum state in eV is', 150.7688049450549)\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 1.21, Page number 48 ***********" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "V=344; #accelerated voltage in V\n", + "n=1; #first reflection\n", + "theta=60; #glancing angle in degrees\n", + "\n", + "#Calculation\n", + "lamda=12.27/math.sqrt(V);\n", + "d=(n*lamda)/(2*math.sin(theta));\n", + "\n", + "#Result\n", + "print(\"The spacing of the crystal in Angstrom is\",lamda);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('The spacing of the crystal in Angstrom is', 0.6615540636030947)\n" + ] + } + ], + "prompt_number": 38 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 1.22, Page number 49 *************" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "n2=2; #second quantum state\n", + "n3=3; #fourth quantum state\n", + "h=6.626*10**-34;\n", + "m=9.1*10**-31; #mass in kg\n", + "a=1*10**-10; #width of potential well in m\n", + "\n", + "#Calculation\n", + "B=8*m*a**2;\n", + "E1=h**2/B; #ground state energy\n", + "E1eV=E1/(1.6*10**-19); #energy in eV\n", + "A=n2**2*h**2;\n", + "E2=A/B; #energy in j\n", + "E2eV=E2/(1.6*10**-19); #energy in eV\n", + "C=n3**2*h**2;\n", + "E3=C/B; #energy in j\n", + "E3eV=E3/(1.6*10**-19); #energy in eV\n", + "E1=math.ceil(E1*10**3)/10**3 #rounding off to 3 decimals\n", + "E1eV=math.ceil(E1eV*10**3)/10**3 #rounding off to 3 decimals\n", + "E2eV=math.ceil(E2eV*10**3)/10**3 #rounding off to 3 decimals\n", + "E3eV=math.ceil(E3eV*10**3)/10**3 #rounding off to 3 decimals\n", + "\n", + "#Result\n", + "print(\"ground state energy in Joule is\",E1);\n", + "print(\"ground state energy in eV is\",E1eV);\n", + "print(\"first energy state in eV is\",E2eV);\n", + "print(\"second energy state in eV is\",E3eV);\n", + "\n", + "#answers given in the book are wrong by one decimal" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('ground state energy in Joule is', 0.001)\n", + "('ground state energy in eV is', 37.693)\n", + "('first energy state in eV is', 150.769)\n", + "('second energy state in eV is', 339.23)\n" + ] + } + ], + "prompt_number": 39 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 1.23, Page number 49" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "\n", + "#Variable decleration\n", + "n3=3; #fourth quantum state\n", + "h=6.626*10**-34;\n", + "m=9.1*10**-31; #mass in kg\n", + "\n", + "\n", + "#ground state energy E1 = h**2/(8*m*a**2)\n", + "#second excited state E3 = (9*h**2)/(8*m*a**2)\n", + "#required energy E = E3-E1\n", + "#E = (9*h**2)/(8*m*a**2) - h**2/(8*m*a**2)\n", + "#E = (h**2/(8*m*a**2))*(9-1)\n", + "#therefore E = (8*h**2)/(8*m*a**2)\n", + "#hence E = (h**2)/(m*a**2)\n", + "\n", + "#Result \n", + "# the required energy is E = (h**2)/(m*a**2)" + ], + "language": "python", + "metadata": {}, + "outputs": [], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 1.24, Page number 50" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "delta_x=10**-8; #length of box in m\n", + "h=6.626*10**-34;\n", + "m=9.1*10**-31; #mass in kg\n", + "\n", + "#Calculation\n", + "delta_v=h/(m*delta_x); #uncertainity in m/sec\n", + "delta_vk=delta_v*10**-3; #uncertainity in km/sec\n", + "delta_vk=math.ceil(delta_vk*10**2)/10**2 #rounding off to 2 decimals\n", + "\n", + "#Result\n", + "print(\"minimum uncertainity in velocity in m/sec is\",round(delta_v));\n", + "print(\"minimum uncertainity in velocity in km/sec is\",delta_vk);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('minimum uncertainity in velocity in m/sec is', 72813.0)\n", + "('minimum uncertainity in velocity in km/sec is', 72.82)\n" + ] + } + ], + "prompt_number": 40 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 1.25, Page number 50" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "mp=1.6*10**-27; #mass of proton in kg\n", + "me=9.1*10**-31; #mass of electron in kg\n", + "h=6.626*10**(-34);\n", + "c=3*10**10; #velocity of light in m/sec\n", + "\n", + "#Calculation\n", + "Ep=me*c**2;\n", + "lamda=h/math.sqrt(2*mp*Ep); #wavelength in m\n", + "lamda_A=lamda*10**10; #converting lamda from m to A\n", + "\n", + "#Result\n", + "print(\"The de broglie wavelength in Angstrom is\",lamda_A);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('The de broglie wavelength in Angstrom is', 4.092931643497047e-06)\n" + ] + } + ], + "prompt_number": 41 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 1.26, Page number 51 *************************************************" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "m=1.675*10**(-27); #mass of neutron in kg\n", + "h=6.626*10**(-34);\n", + "n=1; #diffractive order\n", + "d=0.314; #spacing in nm\n", + "E=0.04; #kinetic energy in eV\n", + "\n", + "#Calculation\n", + "d=d*10**-9; #spacing in m\n", + "Ej=E*1.6*10**-19; #kinetic energy in J\n", + "lamda=h/math.sqrt(2*m*Ej); #de Broglie wavelength\n", + "lamdaA=lamda*10**9; #converting wavelength from m to nm\n", + "theta=math.asin((n*lamda)/(2*d));\n", + "print(\"The de Broglie wavelength in metres is\",lamda);\n", + "print(\"The de Broglie wavelength in nm is\",lamdaA);\n", + "print(\"glancing angle in degrees is\",theta);\n", + "\n", + "#answer given in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('The de Broglie wavelength in metres is', 1.4309980469755228e-10)\n", + "('The de Broglie wavelength in nm is', 0.1430998046975523)\n", + "('glancing angle in degrees is', 0.2298853909391574)\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Engineering_Physics_Vijaya/chapter2_2.ipynb b/Engineering_Physics_Vijaya/chapter2_2.ipynb new file mode 100755 index 00000000..a118db3c --- /dev/null +++ b/Engineering_Physics_Vijaya/chapter2_2.ipynb @@ -0,0 +1,813 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:04561aafd347865fa8c83acfb9b60eb84db275f85862655b442f546023cadd1e" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Electron Theory of Metals" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 2.1, Page number 69" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "#import module\n", + "import math\n", + "\n", + "#Calculation\n", + "# given that E-Ef = kT\n", + "# fermi function FE = 1/(1+exp((E-Ef)/kT)\n", + "# therefore FE = 1/(1+exp(kT/kT));\n", + "# FE = 1/(1+exp(1))\n", + "FE=1/(1+math.exp(1));\n", + "FE=math.ceil(FE*10**2)/10**2; #rounding off to 2 decimals\n", + "\n", + "#Result\n", + "print(\"fermi function is\",FE);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('fermi function is', 0.27)\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 2.2, Page number 69" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "\n", + "#Calculation\n", + "# given that E-Ef = kT\n", + "# fermi function FE = 1/(1+exp((E-Ef)/kT)\n", + "# therefore FE = 1/(1+exp(kT/kT));\n", + "# FE = 1/(1+exp(1))\n", + "FE=1/(1+math.exp(1));\n", + "FE=math.ceil(FE*10**3)/10**3; #rounding off to 3 decimals\n", + "\n", + "#Result\n", + "print(\"fermi function is\",FE);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('fermi function is', 0.269)\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 2.3, Page number 69" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "FE=10/100; #fermi function is 10%\n", + "Ef=5.5; #fermi energy of silver in eV\n", + "k=1.38*10**-23;\n", + "\n", + "#Calculation\n", + "E=Ef+(Ef/100);\n", + "#FE=1/(1+math.exp((E-Ef)/(k*T)))\n", + "#therefore 1/FE = 1+math.exp((E-Ef)/(k*T))\n", + "#therefore (1/FE)-1 = math.exp((E-Ef)/(k*T))\n", + "#therefore log((1/FE)-1) = (E-Ef)/(k*T)\n", + "#therefore T = (E-Ef)/(k*math.log((1/FE)-1))\n", + "#let X=E-Ef; \n", + "X=E-Ef; #energy in eV\n", + "X=X*1.6*10**-19; #energy in J\n", + "T = (X/(k*math.log((1/FE)-1)));\n", + "T=math.ceil(T*10**2)/10**2; #rounding off to 2 decimals\n", + "\n", + "#Result\n", + "print(\"temperature in K is\",T);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('temperature in K is', 290.23)\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 2.4, Page number 70 **************************************" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "#let X=E-Ef\n", + "X=0.5; #E-Ef=0.5 in eV\n", + "\n", + "#Calculation\n", + "X=X*1.6*10**-19; #X in J\n", + "FE=1/100; #fermi function is 1% \n", + "k=1.38*10**-23;\n", + "#FE=1/(1+exp(X/(k*T)))\n", + "#therefore 1/FE = 1+math.exp(X/(k*T))\n", + "#therefore (1/FE)-1 = math.exp(X/(k*T))\n", + "#therefore log((1/FE)-1) = X/(k*T)\n", + "#but log(x) = 2.303*math.log10(x)\n", + "#therefore T = X/(k*math.log((1/FE)-1))\n", + "#but log(x)=2.303*math.log10(x)\n", + "#therefore T = X/(k*2.303*math.log10((1/FE)-1))\n", + "T = X/(k*2.303*math.log10((1/FE)-1));\n", + "\n", + "#Result\n", + "print(\"temperature in K is\",T);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('temperature in K is', 1261.3505710887953)\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 2.5, Page number 71 *******" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "rho_s=10.5*10**3; #density in kg/m^3\n", + "NA=6.02*10**26; #avagadro number per kmol\n", + "MA=107.9; \n", + "\n", + "#Calculation\n", + "n=(rho_s*NA)/MA;\n", + "sigma=6.8*10**7;\n", + "e=1.6*10**-19; #charge in coulomb\n", + "mew=sigma/(n*e);\n", + "mew=math.ceil(mew*10**6)/10**6; #rounding off to 6 decimals\n", + "\n", + "#Result\n", + "print(\"density of electrons is\",n);\n", + "print(\"mobility of electrons in silver in m^2/Vs is\",mew);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('density of electrons is', 5.85820203892493e+28)\n", + "('mobility of electrons in silver in m^2/Vs is', 0.007255)\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 2.6, Page number 71 ***" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "d=8.92*10**3; #density in kg/m^3\n", + "rho=1.73*10**-8; #resistivity in ohm-m\n", + "m=9.1*10**-31; #mass in kg\n", + "w=63.5; #atomic weight\n", + "e=1.6*10**-19; #charge in coulomb\n", + "A=6.02*10**26; #avagadro number\n", + "\n", + "#Calculation\n", + "n=(d*A)/w;\n", + "mew=1/(rho*n*e);\n", + "tow=m/(n*(e**2)*rho);\n", + "mew=math.ceil(mew*10**6)/10**6; #rounding off to 6 decimals\n", + "\n", + "#Result\n", + "print(\"mobility of electrons in Copper in m/Vs is\",mew);\n", + "print(\"average time of collision of electrons in copper in sec is\",tow);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('mobility of electrons in Copper in m/Vs is', 0.004273)\n", + "('average time of collision of electrons in copper in sec is', 2.4297841992299697e-14)\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 2.7, Page number 72" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "rho=1.54*10**-8; #resistivity in ohm-m\n", + "n=5.8*10**28; #electron/m^3\n", + "m=9.108*10**-31; #mass in kg\n", + "e=1.602*10**-19; #charge in coulomb\n", + "\n", + "#Calculation\n", + "tow=m/(n*(e**2)*rho);\n", + "\n", + "#Result\n", + "print(\"relaxation time of conduction electrons in sec is\",tow);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('relaxation time of conduction electrons in sec is', 3.973281032516849e-14)\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 2.8, Page number 73" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "FE=10/100; #fermi function is 10%\n", + "Ef=5.5; #fermi energy of silver in eV\n", + "k=1.38*10**-23;\n", + "\n", + "#Calculation\n", + "E=Ef+(Ef/100);\n", + "#FE=1/(1+math.exp((E-Ef)/(k*T)))\n", + "#therefore 1/FE = 1+math.exp((E-Ef)/(k*T))\n", + "#therefore (1/FE)-1 = math.exp((E-Ef)/(k*T))\n", + "#therefore log((1/FE)-1) = (E-Ef)/(k*T)\n", + "#therefore T = (E-Ef)/(k*math.log((1/FE)-1))\n", + "#let X=E-Ef; \n", + "X=E-Ef; #energy in eV\n", + "X=X*1.6*10**-19; #energy in J\n", + "T = (X/(k*math.log((1/FE)-1)));\n", + "T=math.ceil(T*10**2)/10**2; #rounding off to 2 decimals\n", + "\n", + "#Result\n", + "print(\"temperature in K is\",T);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('temperature in K is', 290.23)\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 2.9, Page number 73" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "\n", + "#Calculation\n", + "# given that E-Ef = kT\n", + "# fermi function FpE = 1/(1+exp((E-Ef)/kT)\n", + "# therefore FpE = 1/(1+exp(kT/kT));\n", + "# FpE = 1/(1+exp(1))\n", + "FpE=1/(1+math.exp(1));\n", + "FpE=math.ceil(FpE*10**2)/10**2; #rounding off to 2 decimals\n", + "\n", + "#Result\n", + "print(\"fermi function is\",FpE);\n", + "#the presence of electron at that energy level is not certain" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('fermi function is', 0.27)\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 2.10, Page number 74 ****************************" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "m=9.1*10**-31; #mass in kg\n", + "h=6.626*10**-34;\n", + "A=(8*m)**(3/2);\n", + "\n", + "#Calculation\n", + "B=math.pi/(2*h**3);\n", + "EfeV=3.10; #fermi energy in eV\n", + "Ef=EfeV*1.6*10**-19; #fermi energy in J\n", + "EFeV=EfeV+0.02; #energy after interval in eV\n", + "EF=EFeV*1.6*10**-19; #energy after interval in J\n", + "function Q=f(E),Q=A*B*math.sqrt(E),endfunction\n", + "I=intg(Ef,EF,f)\n", + "\n", + "#Result\n", + "print(\"number of energy states per unit volume is\",I);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "ename": "SyntaxError", + "evalue": "invalid syntax (<ipython-input-25-15d658985351>, line 18)", + "output_type": "pyerr", + "traceback": [ + "\u001b[1;36m File \u001b[1;32m\"<ipython-input-25-15d658985351>\"\u001b[1;36m, line \u001b[1;32m18\u001b[0m\n\u001b[1;33m function Q=f(E),Q=A*B*math.sqrt(E),endfunction\u001b[0m\n\u001b[1;37m ^\u001b[0m\n\u001b[1;31mSyntaxError\u001b[0m\u001b[1;31m:\u001b[0m invalid syntax\n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 2.11, Page number 74" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "T=300; #temperature in K\n", + "n=8.5*10**28; #density per m^3\n", + "rho=1.69*10**-8; #resistivity in ohm/m^3\n", + "me=9.11*10**-31; #mass of electron in kg\n", + "e=1.6*10**-19; #charge in coulomb\n", + "KB=1.38*10**-23; #boltzmann constant in J/k\n", + "\n", + "#Calculation\n", + "lamda=math.sqrt(3*KB*me*T)/(n*(e**2)*rho);\n", + "\n", + "#Result\n", + "print(\"mean free path of electron in m is\",lamda);\n", + "\n", + "#answer given in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('mean free path of electron in m is', 2.892506814374228e-09)\n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 2.12, Page number 75" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "rho=1.43*10**-8; #resistivity in ohm-m\n", + "n=6.5*10**28; #electron/m^3\n", + "m=9.11*10**-34; #mass in kg\n", + "e=1.6*10**-19; #charge in coulomb\n", + "\n", + "#Calculation\n", + "tow=m/(n*(e**2)*rho);\n", + "\n", + "#Result\n", + "print(\"relaxation time of conduction electrons in sec is\",tow);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('relaxation time of conduction electrons in sec is', 3.8285032275416887e-17)\n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 2.13, Page number 75 ******" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "d=8.92*10**3; #density in kg/m^3\n", + "rho=1.73*10**-8; #resistivity in ohm-m\n", + "m=9.1*10**-31; #mass in kg\n", + "M=63.5; #atomic weight\n", + "e=1.6*10**-19; #charge in coulomb\n", + "A=6.02*10**26; #avagadro number\n", + "\n", + "#Calculation\n", + "n=(d*A)/M;\n", + "mew=1/(rho*n*e);\n", + "tow=m/(n*(e**2)*rho);\n", + "mew=math.ceil(mew*10**6)/10**6; #rounding off to 6 decimals\n", + "\n", + "#Result\n", + "print(\"mobility of electrons in Copper in m/Vs is\",mew);\n", + "print(\"average time of collision of electrons in copper in sec is\",tow);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('mobility of electrons in Copper in m/Vs is', 0.004273)\n", + "('average time of collision of electrons in copper in sec is', 2.4297841992299697e-14)\n" + ] + } + ], + "prompt_number": 31 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 2.14, Page number 76" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "MH=1.008*2*1.67*10**-27; #mass in kg\n", + "T=30; #temperature in C\n", + "\n", + "#Calculation\n", + "T=T+273; #temperature in K\n", + "KB=1.38*10**-23; #boltzmann constant in J/k\n", + "KE=(3/2)*KB*T; #kinetic energy in J\n", + "KEeV=KE*6.24*10**18; #kinetic energy in eV\n", + "cbar=math.sqrt((3*KB*T)/MH);\n", + "\n", + "#Result\n", + "print(\"average kinetic energy in J is\",KE);\n", + "print(\"average kinetic energy in eV is\",KEeV);\n", + "print(\"velocity of molecules in m/s is\",cbar);\n", + "\n", + "#answers for average kinetic energy in eV and velocity of electrons given in the book are wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('average kinetic energy in J is', 6.2720999999999986e-21)\n", + "('average kinetic energy in eV is', 0.039137903999999994)\n", + "('velocity of molecules in m/s is', 1930.269663853336)\n" + ] + } + ], + "prompt_number": 33 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 2.15, Page number 77 ****" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "Ee=10; #electron kinetic energy in eV\n", + "Ep=10; #proton kinetic energy in eV\n", + "me=9.1*10**-31; #mass of electron in kg\n", + "mp=1.67*10**-27; #mass of proton in kg\n", + "\n", + "#Calculation\n", + "EeeV=Ee*1.6*10**-19; #electron kinetic energy in J\n", + "EpeV=Ep*1.6*10**-19; #proton kinetic energy in J\n", + "cebar=math.sqrt((2*EeeV)/me);\n", + "cpbar=math.sqrt((2*EpeV)/mp);\n", + "\n", + "#Result\n", + "print(\"velocity of electron in m/s is\",cebar);\n", + "print(\"velocity of proton in m/s is\",cpbar);\n", + "\n", + "#answers given in the book are wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('velocity of electron in m/s is', 1875228.9237539817)\n", + "('velocity of proton in m/s is', 43774.05241316662)\n" + ] + } + ], + "prompt_number": 35 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 2.16, Page number 77" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "A=10; #area of cross section in mm^2\n", + "A=A*10**-6; #area of cross section in m^2\n", + "i=100; #current in amp\n", + "n=8.5*10**28; #number of electrons per mm^3\n", + "e=1.6*10**-19; #electron charge in coulumb\n", + "\n", + "#Calculation\n", + "vd=1/(n*A*e);\n", + "\n", + "#Result\n", + "print(\"drift velocity in m/s is\",vd);\n", + "\n", + "#answer given in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('drift velocity in m/s is', 7.3529411764705884e-06)\n" + ] + } + ], + "prompt_number": 36 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 2.17, Page number 78" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "tow=3*10**-14; #relaxation time in sec\n", + "n=8*10**28; #density of electrons per m^3\n", + "KB=1.38*10**-23; #boltzmann constant in J/k\n", + "T=0; #temperature in C\n", + "\n", + "#Calculation\n", + "T=T+273; #temperature in K\n", + "m=9.1*10**-31; #mass of electron in kg\n", + "sigma_T=((3*n*tow*(KB**2)*T)/(2*m));\n", + "sigma_T=math.ceil(sigma_T*10**2)/10**2; #rounding off to 2 decimals\n", + "\n", + "#Result\n", + "print(\"thermal conductivity of copper in ohm-1 is\",sigma_T);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('thermal conductivity of copper in ohm-1 is', 205.68)\n" + ] + } + ], + "prompt_number": 38 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Engineering_Physics_Vijaya/chapter4_2.ipynb b/Engineering_Physics_Vijaya/chapter4_2.ipynb new file mode 100755 index 00000000..80203b2d --- /dev/null +++ b/Engineering_Physics_Vijaya/chapter4_2.ipynb @@ -0,0 +1,756 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:544912fca601384def1f6da3b02bc7431b47e0d8f9efa5f2e7d2a367448daaa6" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Magnetic Properties" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 4.1, Page number 119" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "H=6.5*10**-4; #magnetic field in T\n", + "M=1.4; #field with iron\n", + "\n", + "#Calculation\n", + "chi=M/H;\n", + "mew_r=1+chi;\n", + "mew_r=math.ceil(mew_r*10**2)/10**2; #rounding off to 2 decimals\n", + "\n", + "#Result\n", + "print(\"relative permeability of iron is\",mew_r);\n", + "\n", + "#answer given in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('relative permeability of iron is', 2154.85)\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 4.2, Page number 119" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "H=220; #field in amp/m\n", + "M=3300; #magnetisation in amp/m\n", + "\n", + "#Calculation\n", + "chi=M/H;\n", + "mew_r=1+chi;\n", + "\n", + "#Result\n", + "print(\"relative permeability is\",mew_r);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('relative permeability is', 16.0)\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 4.3, Page number 120 *****" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "r=5.29*10**-11; #radius of orbit in m\n", + "B=2; #applied field in Tesla\n", + "e=1.602*10**-19; #charge of electron in coulomb\n", + "m=9.108*10**-31; #mass of electron in kg\n", + "\n", + "#Calculation\n", + "mew=(e**2)*(r**2)*B/(4*m);\n", + "\n", + "#Result\n", + "print(\"magnetic moment in Am^2 is\",mew);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('magnetic moment in Am^2 is', 3.94260574090909e-29)\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 4.4, Page number 120" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "\n", + "#Variable decleration\n", + "chi=0.5*10**-5; #susceptibility \n", + "H=10**6; #field strength in amp/m\n", + "\n", + "#Calculation\n", + "mew_0=4*math.pi*10**-7;\n", + "I=chi*H;\n", + "B=mew_0*(I+H);\n", + "B=math.ceil(B*10**3)/10**3; #rounding off to 3 decimals\n", + "\n", + "#Result\n", + "print(\"intensity of magnetisation in Amp/m is\",I);\n", + "print(\"flux density in Weber/m^2 is\",B);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('intensity of magnetisation in Amp/m is', 5.0)\n", + "('flux density in Weber/m^2 is', 1.257)\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 4.5, Page number 120" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + " \n", + "#import module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "e=2.86; #edge in armstrong\n", + "e=e*10**-10; #edge in m\n", + "Is=1.76*10**6; #magnetisation in amp/m\n", + "mewB=9.27*10**-24; #1 bohr magneton in amp m^2\n", + "\n", + "#Calculation\n", + "N=2/(e**3); #density per m^3\n", + "mewbar=Is/N;\n", + "mew_bar=mewbar/mewB;\n", + "mew_bar=math.ceil(mew_bar*10**3)/10**3; #rounding off to 3 decimals\n", + "\n", + "#Result\n", + "print(\"average dipole moment in mewB is\",mew_bar);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('average dipole moment in mewB is', 2.221)\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 4.6, Page number 121 ***" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "H=10**6; #magnetic field in amp/m\n", + "chi=1.5*10**-3; #susceptibility\n", + "\n", + "#Calculation\n", + "mew_0=4*math.pi*10**-7;\n", + "M=chi*H;\n", + "B=mew_0*(M+H);\n", + "\n", + "#Result\n", + "print(\"magnetisation in Amp/m is\",M);\n", + "print(\"flux density in Tesla is\",B);\n", + "\n", + "#answer for flux density given in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('magnetisation in Amp/m is', 1500.0)\n", + "('flux density in Tesla is', 1.258522017028071)\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 4.7, Page number 121" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "chi=3.7*10**-3; #susceptibility \n", + "H=10**4; #field strength in amp/m\n", + "\n", + "#Calculation\n", + "mew_0=4*math.pi*10**-7;\n", + "M=chi*H;\n", + "B=mew_0*(M+H);\n", + "B=math.ceil(B*10**5)/10**5; #rounding off to 5 decimals\n", + "\n", + "#Result\n", + "print(\"magnetisation in Amp/m is\",M);\n", + "print(\"flux density in Weber/m^2 is\",B);\n", + "\n", + "#answer for flux density given in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('magnetisation in Amp/m is', 37.0)\n", + "('flux density in Weber/m^2 is', 0.01262)\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 4.8, Page number 121" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "r=0.052*10**-9; #radius of orbit in m\n", + "B=1; #magnetic field in Wb/m^2\n", + "e=1.6*10**-19; #charge of electron in coulomb\n", + "m=9.1*10**-31; #mass of electron in kg\n", + "\n", + "#Calculation\n", + "dmew=(e**2)*(r**2)*B/(4*m);\n", + "\n", + "#Result\n", + "print(\"magnetic moment in Am^2 is\",dmew);\n", + "\n", + "#answer given in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('magnetic moment in Am^2 is', 1.901714285714286e-29)\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 4.9, Page number 122" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "\n", + "#Variable decleration\n", + "chi=-0.5*10**-5; #susceptibility \n", + "H=9.9*10**4; #field strength in amp/m\n", + "\n", + "#Calculation\n", + "mew_0=4*math.pi*10**-7;\n", + "I=chi*H;\n", + "B=mew_0*H*(1+chi);\n", + "I=math.ceil(I*10**4)/10**4; #rounding off to 4 decimals\n", + "B=math.ceil(B*10**4)/10**4; #rounding off to 4 decimals\n", + "\n", + "#Result\n", + "print(\"intensity of magnetisation in Amp/m is\",I);\n", + "print(\"flux density in Weber/m^2 is\",B);\n", + "\n", + "#answer for flux density given in the book is wrong " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('intensity of magnetisation in Amp/m is', -0.495)\n", + "('flux density in Weber/m^2 is', 0.1245)\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 4.10, Page number 122" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "r=6.1*10**-11; #radius of H atom in m\n", + "new=8.8*10**15; #frequency in rev/sec\n", + "e=1.6*10**-19;\n", + "\n", + "#Calculation\n", + "mew0=4*math.pi*10**-7;\n", + "i=e*new;\n", + "B=(mew0*i)/(2*r);\n", + "mew=i*math.pi*(r**2);\n", + "i=math.ceil(i*10**7)/10**7; #rounding off to 7 decimals\n", + "B=math.ceil(B*10**3)/10**3; #rounding off to 3 decimals\n", + "\n", + "#Result\n", + "print(\"current in amp is\",i);\n", + "print(\"magnetic induction in weber/m^2 is\",B);\n", + "print(\"dipole moment in amp m^2 is\",mew);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('current in amp is', 0.0014081)\n", + "('magnetic induction in weber/m^2 is', 14.503)\n", + "('dipole moment in amp m^2 is', 1.645933169972273e-23)\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 4.11, Page number 123" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "Is=1.96*10**6; #saturation magnetisation in amp/m\n", + "a=3; #cube edge of iron in armstrong\n", + "a=a*10**-10; #cube edge of iron in m\n", + "mew_b=9.27*10**-24; #bohr magneton in amp/m^2\n", + "n=2; #number of atoms per unit cell\n", + "\n", + "#Calculation\n", + "N=n/(a**3);\n", + "mewbar=Is/N;\n", + "mew_ab=mewbar/mew_b;\n", + "mew_ab=math.ceil(mew_ab*10**4)/10**4; #rounding off to 4 decimals\n", + "\n", + "#Result\n", + "print(\"average number of Bohr magnetons in bohr magneton per atom is\",mew_ab);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('average number of Bohr magnetons in bohr magneton per atom is', 2.8544)\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 4.12, Page number 123" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "I=3000; #magnetisation in amp/m\n", + "B=0.005; #flux density in weber/m^2\n", + "\n", + "#Calculation\n", + "mew0=4*math.pi*10**-7;\n", + "H=(B/mew0)-I;\n", + "mew_r=(I/H)+1;\n", + "H=math.ceil(H*10**3)/10**3; #rounding off to 3 decimals\n", + "mew_r=math.ceil(mew_r*10**3)/10**3; #rounding off to 3 decimals\n", + "\n", + "#Result\n", + "print(\"magnetic force in amp/m is\",H);\n", + "print(\"relative permeability is\",mew_r);\n", + "\n", + "#answer given in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('magnetic force in amp/m is', 978.874)\n", + "('relative permeability is', 4.065)\n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 4.13, Page number 124" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "H=1800; #magnetising field in amp/m\n", + "phi=3*10**-5; #magnetic flux in weber\n", + "A=0.2; #cross sectional area in cm^2\n", + "\n", + "#Calculation\n", + "A=A*10**-4; #cross sectional area in m^2\n", + "B=phi/A;\n", + "mew=B/H;\n", + "mew=math.ceil(mew*10**8)/10**8 #rounding off to 8 decimals\n", + "\n", + "#Result\n", + "print(\"the permeability in Henry/m is\",mew);\n", + "\n", + "#answer given in the book is wron" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('the permeability in Henry/m is', 0.00083334)\n" + ] + } + ], + "prompt_number": 31 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 4.14, Page number 124 ********************" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "r=0.04; #radius of circular loop in m\n", + "i=1000; #current in mA\n", + "i=i*10**-3; #current in amp\n", + "B=10**-3; #magnetic flux density in Wb/m^2\n", + "theta=45; #angle in degrees\n", + "\n", + "#Calculation\n", + "A=math.pi*(r**2);\n", + "mew=i*A;\n", + "tow=i*B*math.cos(theta);\n", + "mew=math.ceil(mew*10**6)/10**6 #rounding off to 6 decimals\n", + "\n", + "#Result\n", + "print(\"the magnetic dipole moment in amp m^2 is\",mew);\n", + "print(\"the torque in Nm is\",tow);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('the magnetic dipole moment in amp m^2 is', 0.005027)\n", + "('the torque in Nm is', 0.0005253219888177298)\n" + ] + } + ], + "prompt_number": 36 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 4.15, Page number 125" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "\n", + "#Variable decleration\n", + "A=100; #area of hysteris loop in m^2\n", + "B=0.01; #flux density in wb/m^2\n", + "H=40; #magnetic field in amp/m\n", + "M=7650; #atomic weight in kg/m^3\n", + "\n", + "#Calculation\n", + "hl=A*B*H;\n", + "\n", + "#Result\n", + "print(\"the hysterisis loss per cycle in J/m^3 is\",hl);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('the hysterisis loss per cycle in J/m^3 is', 40.0)\n" + ] + } + ], + "prompt_number": 38 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 4.17, Page number 125" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "hl=200; #hysterisis loss per cycle in J/m^3\n", + "M=7650; #atomic weight in kg/m^3\n", + "m=100; #magnetisation cycles per second\n", + "\n", + "#Calculation\n", + "hpl=hl*m;\n", + "pl=hpl/M;\n", + "pl=math.ceil(pl*10**4)/10**4 #rounding off to 4 decimals\n", + "\n", + "#Result\n", + "print(\"hysterisis power loss per second in watt/m^3 is\",hpl);\n", + "print(\"the power loss in watt/kg is\",pl); \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('hysterisis power loss per second in watt/m^3 is', 20000)\n", + "('the power loss in watt/kg is', 2.6144)\n" + ] + } + ], + "prompt_number": 40 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Engineering_Physics_Vijaya/chapter5_2.ipynb b/Engineering_Physics_Vijaya/chapter5_2.ipynb new file mode 100755 index 00000000..14018aea --- /dev/null +++ b/Engineering_Physics_Vijaya/chapter5_2.ipynb @@ -0,0 +1,639 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:19dabe1afe46093105a84b4746899bd5b483ca26e3b557510765740ff72179af" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Superconductivity" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 5.1, Page number 148" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#importing modules\n", + "import math\n", + "\n", + "#Variable declaration\n", + "Tc=3.7; #in kelvin\n", + "Hc_0=0.0306; \n", + "T=2\n", + "\n", + "#Calculation\n", + "Hc_2k=Hc_0*(1-((T/Tc)**2));\n", + "Hc_2k=math.ceil(Hc_2k*10**5)/10**5; #rounding off to 5 decimals\n", + "\n", + "#Result\n", + "print(\"the critical feild at 2K in tesla is\",Hc_2k);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('the critical feild at 2K in tesla is', 0.02166)\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 5.2, Page number 149\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#importing modules\n", + "import math\n", + "\n", + "#Variable declaration\n", + "T=4.2; #in kelvin\n", + "Tc=7.18; #in kelvin\n", + "Hc_0=6.5*10**4; #in amp per meter\n", + "D=10**-3\n", + "\n", + "#Calculation\n", + "R=D/2; #radius is equal to half of diameter\n", + "Hc_T=Hc_0*(1-((T/Tc)**2));\n", + "Hc_T=math.ceil(Hc_T*10)/10; #rounding off to 1 decimals\n", + "Ic=2*math.pi*R*Hc_T #critical current is calculated by 2*pi*r*Hc(T)\n", + "Ic=math.ceil(Ic*10**2)/10**2; #rounding off to 2 decimals\n", + "\n", + "#Result\n", + "print(\"the critical feild in Tesla is\",round(Hc_T));\n", + "print(\"the critical current in Amp is\",Ic);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('the critical feild in Tesla is', 42759.0)\n", + "('the critical current in Amp is', 134.34)\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 5.3, Page number 149\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#importing modules\n", + "import math\n", + "\n", + "#Variable declaration\n", + "lamda_T=75 #in nm\n", + "T=3.5 \n", + "HgTc=4.12 #in K\n", + "\n", + "#Calculation\n", + "lamda_o=lamda_T*math.sqrt(1-((T/HgTc)**4));\n", + "lamda_o=math.ceil(lamda_o*10**2)/10**2; #rounding off to 2 decimals\n", + "\n", + "#Result\n", + "print(\"the pentration depth at 0k is\",lamda_o);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('the pentration depth at 0k is', 51.92)\n" + ] + } + ], + "prompt_number": 29 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 5.4, Page number 150" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#importing modules\n", + "import math\n", + "\n", + "#Variable declaration\n", + "lamda_T1=396 #pentration depth in armstrong\n", + "lamda_T2=1730 #pentration depth in armstrong\n", + "T1=3 #temperature in K\n", + "T2=7.1 #temperature in K\n", + "\n", + "#Calculation\n", + "#lamda_T2**2=lamda_0**2*(((Tc**4-T2**4)/Tc**4)**-1)\n", + "#lamda_T1**2=lamda_0**2*(((Tc**4-T1**4)/Tc**4)**-1)\n", + "#dividing lamda_T2**2 by lamda_T1**2 = (Tc**4-T1**4)/(Tc**4-T2**4)\n", + "#let A=lamda_T2**2 and B=lamda_T1**2\n", + "A=lamda_T2**2\n", + "B=lamda_T1**2\n", + "C=A/B\n", + "C=math.ceil(C*10**4)/10**4; #rounding off to 4 decimals\n", + "X=T1**4\n", + "Y=T2**4\n", + "Y=math.ceil(Y*10**2)/10**2; #rounding off to 2 decimals\n", + "#C*((TC**4)-Y)=(Tc**4)-X\n", + "#C*(Tc**4)-(Tc**4)=C*Y-X\n", + "#(Tc**4)*(C-1)=(C*Y)-X\n", + "#let Tc**4 be D\n", + "#D*(C-1)=(C*Y)-X\n", + "D=((C*Y)-X)/(C-1)\n", + "D=math.ceil(D*10)/10; #rounding off to 1 decimals\n", + "Tc=D**(1/4)\n", + "Tc=math.ceil(Tc*10**4)/10**4; #rounding off to 4 decimals\n", + "\n", + "#Result\n", + "print(\"the pentration depth at 0k is\",Tc);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('the pentration depth at 0k is', 7.1932)\n" + ] + } + ], + "prompt_number": 44 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 5.5, Page number 150" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#importing modules\n", + "import math\n", + "\n", + "#Variable declaration\n", + "Tc=7.2 #in K\n", + "Ho=6.5*10**3 #in amp per m\n", + "T=5 #in K\n", + "\n", + "#Calculation\n", + "Hc=Ho*(1-((T/Tc)**2))\n", + "Hc=math.ceil(Hc*10**2)/10**2; #rounding off to 2 decimals\n", + "\n", + "#Result\n", + "print(\"the critical magnetic feild at 5K in amp per m is\",Hc)\n", + "\n", + "# answer given in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('the critical magnetic feild at 5K in amp per m is', 3365.36)\n" + ] + } + ], + "prompt_number": 33 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 5.6, Page number 151" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#importing modules\n", + "import math\n", + "\n", + "#Variable declaration\n", + "Tc=3.5 #in K\n", + "Ho=3.2*10**3 #in amp per m\n", + "T=2.5 #in K\n", + "\n", + "#Calculation\n", + "Hc=Ho*(1-((T/Tc)**2))\n", + "Hc=math.ceil(Hc*10**2)/10**2; #rounding off to 2 decimals\n", + "\n", + "#Result\n", + "print(\"the critical magnetic feild at 5K in amp per m is\",Hc)\n", + "\n", + "#answer in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('the critical magnetic feild at 5K in amp per m is', 1567.35)\n" + ] + } + ], + "prompt_number": 45 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 5.7, Page number 151" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#importing modules\n", + "import math\n", + "\n", + "#Variable declaration\n", + "Hc=5*10**3 #in amp per m\n", + "Ho=2*10**4 #in amp per m\n", + "T=6 #in K\n", + "\n", + "#Calculation\n", + "Tc=T/math.sqrt(1-(Hc/Ho))\n", + "Tc=math.ceil(Tc*10**2)/10**2; #rounding off to 2 decimals\n", + "\n", + "#Result\n", + "print(\"the critical magnetic feild at 5K in amp per m is\",Tc)\n", + "\n", + "#answer in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('the critical magnetic feild at 5K in amp per m is', 6.93)\n" + ] + } + ], + "prompt_number": 66 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 5.8, Page number 152" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#importing modules\n", + "import math\n", + "\n", + "#Variable declaration\n", + "Hc=2*10**3 #in amp per m\n", + "R=0.02 #in m\n", + "\n", + "#Calculation\n", + "Ic=2*math.pi*R*Hc\n", + "Ic=math.ceil(Ic*10**2)/10**2; #rounding off to 2 decimals\n", + "\n", + "#Result\n", + "print(\"the critical current is\",Ic)\n", + "\n", + "#answer in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('the critical magnetic feild at 5K in amp per m is', 251.33)\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 5.9, Page number 152" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#importing modules\n", + "import math\n", + "\n", + "#Variable declaration\n", + "M1=199.5 #in a.m.u\n", + "T1=5 #in K\n", + "T2=5.1 #in K\n", + "\n", + "#Calculation\n", + "M2=((T1/T2)**2)*M1\n", + "M2=math.ceil(M2*10**3)/10**3; #rounding off to 3 decimals\n", + "\n", + "#Result\n", + "print(\"the isotopic mass of M2 is\",M2)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('the isotopic mass of M2 is', 191.754)\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 5.10, Page number 152" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "D=3*10**-3 #in meters\n", + "Tc=8 #in K \n", + "T=5 #in K \n", + "Ho=5*10**4\n", + "\n", + "#Calculation\n", + "R=D/2\n", + "Hc=Ho*(1-((T/Tc)**2))\n", + "Ic=2*math.pi*R*Hc\n", + "Ic=math.ceil(Ic*10**3)/10**3; #rounding off to 3 decimals\n", + "\n", + "#Result\n", + "print(\"critical magnetic feild in amp per m is\",round(Hc));\n", + "print(\"critical current in amp is\",Ic);\n", + "\n", + "#answer in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('critical magnetic feild in amp per m is', 30469.0)\n", + "('critical current in amp is', 287.162)\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 5.11, Page number 153" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#importing modules\n", + "import math\n", + "\n", + "#Variable declaration\n", + "M1=199.5 \n", + "M2=203.4 \n", + "Tc1=4.185 #in K\n", + "\n", + "#Calculation\n", + "Tc2=Tc1*math.sqrt(M1/M2)\n", + "Tc2=math.ceil(Tc2*10**3)/10**3; #rounding off to 3 decimals\n", + "\n", + "#Result\n", + "print(\"the critical temperature is\",Tc2)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('the critical temperature is', 4.145)\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 5.12, Page number 154" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#importing modules\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "V=8.5*10**-6 #in volts\n", + "e=1.6*10**-19 #in C\n", + "h=6.626*10**-24\n", + "\n", + "#Calculation\n", + "new=2*e*V/h\n", + "new=math.ceil(new*10**5)/10**5; #rounding off to 5 decimals\n", + "\n", + "#Result\n", + "print(\"EM wave generated frequency in Hz is\",new)\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('EM wave generated frequency in Hz is', 0.41051)\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 5.13, Page number 154" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#Variable declaration\n", + "p1=1 #in mm\n", + "p2=6 #in mm\n", + "Tc1=5 #in K\n", + "\n", + "#Calculation\n", + "Tc2=Tc1*(p2/p1);\n", + "\n", + "#Result\n", + "print(\"the critical temperature in K is\",round(Tc2))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('the critical temperature in K is', 30.0)\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 5.14, Page number 154\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#Variable declaration\n", + "Tc=8.7 #in K\n", + "Hc=6*10**5 #in A per m\n", + "Ho=3*10**6 #in A per m\n", + "\n", + "#Calculation\n", + "T=Tc*(math.sqrt(1-(Hc/Ho)))\n", + "\n", + "#Result\n", + "print(\" maximum critical temperature in K is\",T)\n", + "\n", + "#answer given in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(' maximum critical temperature in K is', 7.781516561699267)\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Engineering_Physics_Vijaya/chapter6_2.ipynb b/Engineering_Physics_Vijaya/chapter6_2.ipynb new file mode 100755 index 00000000..4c7f2be8 --- /dev/null +++ b/Engineering_Physics_Vijaya/chapter6_2.ipynb @@ -0,0 +1,238 @@ +{ + "metadata": { + "name": "chapter6 (1)" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Dielectric Properties" + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example number 6.1, Page number 187" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# To calculate the energy stored in the condenser and polarizing the dielectric\n\n#import module\nimport math\nfrom __future__ import division\n\n#Variable decleration\nC=2; #capacitance in micro farad\nC=C*10**-6; #capacitance in farad\nV=1000; #voltage in Volts\nepsilon_r=100; \n\n#Calculation\nW=(C*(V**2))/2;\nC0=C/epsilon_r;\nW0=(C0*(V**2))/2;\nW_0=1-W0;\n\n#Result\nprint(\"energy stored in the condenser in Joule is\",W);\nprint(\"energy stored in the dielectric in Joule is\",W_0);", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "('energy stored in the condenser in Joule is', 1.0)\n('energy stored in the dielectric in Joule is', 0.99)\n" + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example number 6.2, Page number 188" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# To calculate the ratio between electronic and ionic polarizability\n\n#import module\nimport math\nfrom __future__ import division\n\n#Variable decleration\nepsilon_r=4.94;\nN=2.69; #let n**2 be N\n\n#Calculaion\n#(epsilon_r-1)/(epsilon_r+2) = (N*alpha)/(3*epsilon_0)\n#alpha = alpha_e+alpha_i\n#therefore (epsilon_r-1)/(epsilon_r+2) = (N*(alpha_e+alpha_i))/(3*epsilon_0)\n#let (N*(alpha_e+alpha_i))/(3*epsilon_0) be X\nX=(epsilon_r-1)/(epsilon_r+2);\n#Ez=n^2\n#therefore (N-1)/(N+2) = (N*alpha_e)/(3*epsilon_0)\n#let (N*alpha_e)/(3*epsilon_0) be Y\nY=(N-1)/(N+2);\n#dividing X/Y = (N*(alpha_e+alpha_i))/(N*alpha_e)\n#therefore X/Y = 1+(alpha_i/alpha_e)\n#let alpha_i/alpha_e be A\nR=(X/Y)-1;\nR=math.ceil(R*10**4)/10**4; #rounding off to 4 decimals\n\n#Result\nprint(\"ratio between electronic and ionic polarizability is\",R);\n\n#answer given in the book is wrong in the second part", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "('ratio between electronic and ionic polarizability is', 0.5756)\n" + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example number 6.3, Page number 188" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# To calculate the dielectric constant of the material\n\n#import module\nimport math\nfrom __future__ import division\n\n#Variable decleration\nN=3*10**28; #atoms per m^3\nalpha_e=10**-40; #farad m^2\nepsilon_0=8.854*10**-12; #f/m\n\n#Calculation\nepsilon_r=1+(N*alpha_e/epsilon_0);\nepsilon_r=math.ceil(epsilon_r*10**3)/10**3; #rounding off to 3 decimals\n\n#Result\nprint(\"dielectric constant of the material is\",epsilon_r);", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "('dielectric constant of the material is', 1.339)\n" + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example number 6.4, Page number 189" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# To calculate the electronic polarizability of He atoms\n\n#import module\nimport math\nfrom __future__ import division\n\n#Variable decleration\nepsilon_0=8.854*10**-12; #f/m\nepsilon_r=1.0000684;\n\n#Calculation\nN=2.7*10**25; #atoms per m^3\nalpha_e=(epsilon_0*(epsilon_r-1))/N;\n\n#Result\nprint(\"electronic polarizability of He atoms in Fm^2 is\",alpha_e);", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "('electronic polarizability of He atoms in Fm^2 is', 2.2430133333322991e-41)\n" + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example number 6.5, Page number 189" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# To calculate the capacitance and charge\n\n#import module\nimport math\nfrom __future__ import division\n\n#Variable decleration\nepsilon_0=8.854*10**-12; #f/m\nA=100; #area in cm^2\nA=A*10**-4; #area in m^2\nV=100; #potential in V\nd=1; #plate seperation in cm\n\n#Calculation\nd=d*10**-2; #plate seperation in m\nC=(epsilon_0*A)/d;\nQ=C*V;\n\n#Result\nprint(\"charge on the plates in F is\",C);\nprint(\"charge on the capacitor in coulomb is\",Q);", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "('charge on the plates in F is', 8.854e-12)\n('charge on the capacitor in coulomb is', 8.854e-10)\n" + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example number 6.6, Page number 190" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# To calculate the resultant voltage across the capacitors\n\n#import module\nimport math\nfrom __future__ import division\n\n\n#Variable decleration\nQ=2*10**-10; #charge in coulomb\nd=4; #plate seperation in mm\nd=d*10**-3; #plate seperation in m\nepsilon_r=3.5;\nepsilon_0=8.85*10**-12; #f/m\nA=650; #area in mm^2\n\n#Calculation\nA=A*10**-6; #area in m^2\nV=(Q*d)/(epsilon_0*epsilon_r*A);\nV=math.ceil(V*10**3)/10**3; #rounding off to 3 decimals\n\n#Result\nprint(\"voltage across the capacitor in Volts is\",V);", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "('voltage across the capacitor in Volts is', 39.735)\n" + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example number 6.7, Page number 190" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# To calculate the dielectric displacement\n\n#import module\nimport math\nfrom __future__ import division\n\n\n#Variable decleration\nV=10; #potential in volts\nd=2*10**-3; #plate seperation in m\nepsilon_r=6; #dielectric constant\nepsilon_0=8.85*10**-12; #f/m\n\n#Calculation\nE=V/d;\nD=epsilon_0*epsilon_r*E;\n\n#Result\nprint(\"dielectric displacement in cm^-2 is\",D);\n\n#answer given in the book is wrong in the 7th decimal point", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "('dielectric displacement in cm^-2 is', 2.6549999999999994e-07)\n" + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example number 6.8, Page number 191" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# To calculate the polarizability and relative permittivity of He\n\n#import module\nimport math\nfrom __future__ import division\n\n\n#Variable decleration\nR=0.55; #radius of He atom in angstrom\nR=R*10**-10; #radius of He atom in m\nepsilon_0=8.84*10**-12; #f/m\nN=2.7*10**25;\n\n#Calculation\nalpha_e=4*math.pi*epsilon_0*R**3;\nepsilon_r=(N*alpha_e/epsilon_0)+1;\nepsilon_r=math.ceil(epsilon_r*10**6)/10**6; #rounding off to 6 decimals\n\n#Result\nprint(\"polarizability in farad m^2 is\",alpha_e);\nprint(\"relative permitivity is\",epsilon_r);", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "('polarizability in farad m^2 is', 1.848205241292183e-41)\n('relative permitivity is', 1.000057)\n" + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example number 6.9, Page number 191" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# To calculate the field strength and total dipole moment\n\n#import module\nimport math\nfrom __future__ import division\n\n\n#Variable decleration\nV=15; #potential difference in volts\nC=6; #capacity in micro farad\nC=C*10**-6; #capacity in farad\nepsilon_0=8.84*10**-12; #f/m\nepsilon_r=8;\nA=360; #surface area in cm^2\n\n#Calculation\nA=A*10**-4; #surface area in m^2\nE=(V*C)/(epsilon_0*epsilon_r*A);\nd=epsilon_0*(epsilon_r-1)*V*A;\n\n#Result\nprint(\"field strength in V/m is\",E);\nprint(\"total dipole moment in cm is\",d);\n\n#answer for field strength E given in the book is wrong ", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "('field strength in V/m is', 35350678.73303167)\n('total dipole moment in cm is', 3.34152e-11)\n" + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example number 6.10, Page number 191" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "# To calculate the complex polarisability of material\n\n#import module\nimport math\nfrom __future__ import division\n\n#Variable decleration\nepsilonr=4.36; #dielectric constant\nt=2.8*10**-2; #loss tangent(t)\nN=4*10**28; #number of electrons\nepsilon0=8.84*10**-12; \n\n#Calculation\nepsilon_r = epsilonr*t;\nepsilonstar = (complex(epsilonr,-epsilon_r));\nalphastar = (epsilonstar-1)/(epsilonstar+2);\nalpha_star = 3*epsilon0*alphastar/N; #complex polarizability(Fm**2)\n\n#Result\nprint(\"the complex polarizability in F-m^2 is\"'alphastar',alpha_star);\n#disp('j',I,R);\n#by taking 10^-40 common we get alphastar = (3.5-j0.06)*10^-40 F-m^2", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "('the complex polarizability in F-m^2 isalphastar', (3.5037933503257222e-40-6.000743833211258e-42j))\n" + } + ], + "prompt_number": 11 + }, + { + "cell_type": "code", + "collapsed": false, + "input": "", + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Engineering_Physics_Vijaya/chapter7_2.ipynb b/Engineering_Physics_Vijaya/chapter7_2.ipynb new file mode 100755 index 00000000..d4161b18 --- /dev/null +++ b/Engineering_Physics_Vijaya/chapter7_2.ipynb @@ -0,0 +1,1514 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:b26f0e8151a54ecdc596868a34547e181ac6dce2c5aea4a02c15b80e1401fd4f" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Semiconductors" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 7.1, Page number 251" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "T1=300; #temp in K\n", + "T2=310; #temp in K\n", + "ni1=2.5*10**19; #per cubic metre\n", + "EgeV1=0.72; #value of Eg in eV\n", + "EgeV2=1.12; #value of Eg in eV\n", + "\n", + "#Calculation\n", + "Eg1=EgeV1*1.6*10**-19; #Eg in J\n", + "Eg2=EgeV2*1.6*10**-19; #Eg in J\n", + "KB=1.38*10**-23; #boltzmann constant in J/k\n", + "#density of electron hole pair is ni = A*(T**(3/2))*exp(-Eg/(2*KB*T))\n", + "#let (T**(3/2))*exp(-Eg/(2*KB*T)) be X\n", + "X1=(T1**(3/2))*math.exp(-Eg1/(2*KB*T1));\n", + "X2=(T2**(3/2))*math.exp(-Eg2/(2*KB*T2));\n", + "#therefore ni1=A*X1 and ni2=A*X2. dividing ni2/ni1 we get X2/X1\n", + "ni2=ni1*(X2/X1);\n", + "\n", + "#Result\n", + "print(\"the number of electron hole pairs per cubic metre is\",ni2);\n", + "\n", + "#answer given in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('the number of electron hole pairs per cubic metre is', 2.3207901206362184e+16)\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 7.2, Page number 251" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "RH=3.66*10**-4; #hall coefficient in m^3/coulomb\n", + "sigma=112; #conductivity in ohm-1 m-1\n", + "e=1.6*10**-19;\n", + "\n", + "#Calculation\n", + "ne=1/(RH*e);\n", + "#sigma = e*ne*(mew_e+mew_h)\n", + "#assuming mew_h = 0\n", + "mew_e=sigma/(e*ne);\n", + "\n", + "#Result\n", + "print(\"the charge carrier density per m^3 is\",ne);\n", + "print(\"electron mobility in m^2/Vs is\",mew_e);\n", + "\n", + "#answer given in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('the charge carrier density per m^3 is', 1.7076502732240434e+22)\n", + "('electron mobility in m^2/Vs is', 0.040992)\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 7.3, Page number 252" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "ni=1.5*10**16; #intrinsic concentration per m^3\n", + "e=1.6*10**-19;\n", + "mew_e=0.13; #mobility of electrons in m^2/Vs\n", + "mew_h=0.05; #mobility of holes in m^2/Vs\n", + "ND=5*10**20; #conductivity in atoms/m^3\n", + "\n", + "#Calculation\n", + "sigma1=ni*e*(mew_e+mew_h);\n", + "nd=(ni**2)/ND;\n", + "sigma2=ND*e*mew_e;\n", + "NA=5*10**20;\n", + "na=(ni**2)/NA;\n", + "sigma3=NA*e*mew_h;\n", + "sigma1=math.ceil(sigma1*10**7)/10**7; #rounding off to 7 decimals\n", + "sigma2=math.ceil(sigma2*10**2)/10**2; #rounding off to 2 decimals\n", + "\n", + "#Result\n", + "print(\"intrinsic conductivity of Si in ohm-1 m-1 is\",sigma1);\n", + "print(\"conductivity of Si during donor impurity in ohm-1 m-1 is\",sigma2);\n", + "print(\"conductivity of Si during acceptor impurity in ohm-1 m-1 is\",round(sigma3));" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('intrinsic conductivity of Si in ohm-1 m-1 is', 0.000432)\n", + "('conductivity of Si during donor impurity in ohm-1 m-1 is', 10.41)\n", + "('conductivity of Si during acceptor impurity in ohm-1 m-1 is', 4.0)\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 7.4, Page number 253" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "sigma1=2; #conductivity in ohm-1 m-1\n", + "EgeV=0.72; #band gap in eV\n", + "KB=1.38*10**-23; #boltzmann constant\n", + "T1=20; #temp in C\n", + "T2=40; #temp in C\n", + "\n", + "#Calculation\n", + "Eg=EgeV*1.6*10**-19; #in J\n", + "T1=T1+273; #temp in K\n", + "T2=T2+273; #temp in K\n", + "#sigma2/sigma1 = exp((-Eg/(2*KB))*((1/T2)-(1/T1)))\n", + "#by taking log on both sides we get 2.303*log10(sigma2/sigma1) = (Eg/(2*KB))*((1/T1)-(1/T2))\n", + "#let (Eg/(2*KB))*((1/T1)-(1/T2)) be X\n", + "X=(Eg/(2*KB))*((1/T1)-(1/T2));\n", + "#let log10(sigma2/sigma1) be Y\n", + "Y=X/2.303;\n", + "#log10(sigma2/sigma1) = log10(sigma2)-log10(sigma1)\n", + "#let log10(sigma2) be A\n", + "A=Y+math.log10(sigma1);\n", + "sigma2=10**A;\n", + "sigma2=math.ceil(sigma2*10**2)/10**2; #rounding off to 2 decimals\n", + "\n", + "#Result\n", + "print(\"the conductivity in mho m-1 is\",sigma2);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('the conductivity in mho m-1 is', 4.97)\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 7.5, Page number 253" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "mew_n=1300*10**-4; #in m^2/Vs\n", + "mew_p=500*10**-4; #in m^2/Vs\n", + "sigma=3*10**4; #conductivity in ohm-1 m-1\n", + "e=1.6*10**-19;\n", + "\n", + "#Calculation\n", + "N=sigma/(e*mew_n);\n", + "ni=1.5*10**16; #per m^3\n", + "p=(ni**2)/N;\n", + "P=sigma/(e*mew_p);\n", + "n=(ni**2)/P;\n", + "N=math.ceil(N*10**4)/10**4; #rounding off to 4 decimals\n", + "\n", + "#Result\n", + "print(\"concentration of electrons in n-type per cubic metre are\",N);\n", + "print(\"concentration of holes in n-type per cubic metre are\",round(p));\n", + "print(\"concentration of electrons in p-type per cubic metre are\",round(n));\n", + "print(\"concentration of holes in p-type per cubic metre are\",P);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('concentration of electrons in n-type per cubic metre are', 1.4423076923076921e+24)\n", + "('concentration of holes in n-type per cubic metre are', 156000000.0)\n", + "('concentration of electrons in p-type per cubic metre are', 60000000.0)\n", + "('concentration of holes in p-type per cubic metre are', 3.7499999999999995e+24)\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 7.6, Page number 254" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "ni=2.37*10**19; #intrinsic carrier density per m^3\n", + "mew_e=0.38; #in m**2/Vs\n", + "mew_n=0.18; #in m**2/Vs\n", + "\n", + "#Calculation\n", + "e=1.6*10**-19;\n", + "sigmai=ni*e*(mew_e+mew_n);\n", + "rho=1/sigmai;\n", + "rho=math.ceil(rho*10**3)/10**3; #rounding off to 3 decimals\n", + "\n", + "#Result\n", + "print(\"resistivity in ohm m is\",rho);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('resistivity in ohm m is', 0.471)\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 7.7, Page number 254" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "Eg=1.12; #band gap in eV\n", + "K=1.38*10**-23;\n", + "T=300; #temp in K\n", + "\n", + "#Calculation\n", + "#EF = (Eg/2)+(3*K*T/4)*log(mh/me)\n", + "#given me=0.12m0 and mh=0.28m0. therefore mh/me = 0.28/0.12 \n", + "#let mh/me be X. therefore X=0.28/0.12 \n", + "X=0.28/0.12;\n", + "EF=(Eg/2)+((3*K*T/4)*math.log(X));\n", + "\n", + "#Result\n", + "print(\"the position of fermi level in eV is\",EF);\n", + "\n", + "#answer given in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('the position of fermi level in eV is', 0.56)\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 7.8, Page number 254" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "KB=1.38*10**-23;\n", + "T=300; #temp in K\n", + "h=6.626*10**-34;\n", + "m0=9.11*10**-31;\n", + "mh=m0;\n", + "me=m0;\n", + "EgeV=0.7; #energy gap in eV\n", + "\n", + "#Calculation\n", + "Eg=EgeV*1.6*10**-19; #in J\n", + "A=((2*math.pi*KB/(h**2))**(3/2))*(me*mh)**(3/4);\n", + "B=T**(3/2);\n", + "C=math.exp(-Eg/(2*KB*T));\n", + "ni=2*A*B*C;\n", + "\n", + "#Result\n", + "print(\"concentration of intrinsic charge carriers per cubic metre is\",ni);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('concentration of intrinsic charge carriers per cubic metre is', 3.3481803992458756e+19)\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 7.9, Page number 255" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "ni=2.4*10**19;\n", + "mew_e=0.39;\n", + "mew_h=0.19;\n", + "e=1.6*10**-19;\n", + "\n", + "#Result\n", + "sigmai=ni*e*(mew_e+mew_h);\n", + "rhoi=1/sigmai;\n", + "rhoi=math.ceil(rhoi*10**2)/10**2; #rounding off to 2 decimals\n", + "\n", + "#Result\n", + "print(\"resistivity in ohm m is\",rhoi);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('resistivity in ohm m is', 0.45)\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 7.10, Page number 255" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "l=1; #length in cm\n", + "l=l*10**-2; #length in m\n", + "e=1.6*10**-19;\n", + "w=1; #width in mm\n", + "t=1; #thickness in mm\n", + "\n", + "#Calculation\n", + "w=w*10**-3; #width in m\n", + "t=t*10**-3; #thickness in m\n", + "A=w*t;\n", + "ni=2.5*10**19;\n", + "mew_e=0.39;\n", + "mew_p=0.19;\n", + "sigma=ni*e*(mew_p+mew_e);\n", + "R=l/(sigma*A);\n", + "\n", + "#Result\n", + "print(\"resistance of intrinsic Ge rod in ohm is\",R);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('resistance of intrinsic Ge rod in ohm is', 4310.3448275862065)\n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 7.11, Page number 255" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "Eg=1.1; #energy gap in eV\n", + "m=9.109*10**-31;\n", + "k=1.38*10**-23;\n", + "T=300;\n", + "e=1.6*10**-19;\n", + "h=6.626*10**-34;\n", + "mew_e=0.48; #electron mobility\n", + "mew_h=0.013; #hole mobility\n", + "\n", + "#Calculation\n", + "C=2*(2*math.pi*m*k/(h**2))**(3/2);\n", + "X=2*k*T/e;\n", + "Y=-Eg/X;\n", + "A=math.exp(Y);\n", + "ni=C*(T**(3/2))*A;\n", + "sigma=ni*e*(mew_e+mew_h);\n", + "sigma=math.ceil(sigma*10**6)/10**6 #rounding off to 6 decimals\n", + "\n", + "#Result\n", + "print(\"conductivity in ohm-1 m-1 is\",sigma);\n", + "\n", + "# answer given in the book is wrong, Page number 255" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('conductivity in ohm-1 m-1 is', 0.001162)\n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 7.12, Page number 256" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "m=9.109*10**-31;\n", + "k=1.38*10**-23;\n", + "T=300;\n", + "e=1.6*10**-19;\n", + "h=6.626*10**-34;\n", + "Eg=0.7;\n", + "mew_e=0.4; #electron mobility\n", + "mew_h=0.2; #hole mobility\n", + "\n", + "#Calculation\n", + "C=2*(2*math.pi*m*k/((h**2)))**(3/2);\n", + "X=2*k*T/e;\n", + "ni=C*(T**(3/2))*math.exp(-Eg/X);\n", + "sigma=ni*e*(mew_e+mew_h);\n", + "sigma=math.ceil(sigma*10**3)/10**3 #rounding off to 3 decimals\n", + "\n", + "#Result\n", + "print(\"conductivity in ohm-1 m-1\",sigma);\n", + "\n", + "#answer given in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('conductivity in ohm-1 m-1', 3.214)\n" + ] + } + ], + "prompt_number": 30 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 7.13, Page number 256" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "k=8.616*10**-5;\n", + "T1=20; #temp in C\n", + "T1=T1+273; #temp in K\n", + "T2=32; #temp in C\n", + "rho2=4.5; #resistivity in ohm m\n", + "rho1=2; #resistivity in ohm m\n", + "\n", + "#Calculation\n", + "T2=T2+273; #temp in K\n", + "dy=math.log10(rho2)-math.log10(rho1);\n", + "dx=(1/T1)-(1/T2);\n", + "Eg=2*k*dy/dx;\n", + "Eg=math.ceil(Eg*10**3)/10**3 #rounding off to 3 decimals\n", + "\n", + "#Result\n", + "print(\"energy band gap in eV is\",Eg);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('energy band gap in eV is', 0.452)\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 7.13, Page number 256" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "k=8.616*10**-5;\n", + "T1=20; #temp in C\n", + "T2=32; ##temp in C\n", + "rho2=4.5; #resistivity in ohm m\n", + "rho1=2; #resistivity in ohm m\n", + "\n", + "#Calculation\n", + "T1=T1+273; #temp in K\n", + "T2=T2+273; #temp in K\n", + "dy=math.log10(rho2)-math.log10(rho1);\n", + "dx=(1/T1)-(1/T2);\n", + "Eg=2*k*dy/dx;\n", + "Eg=math.ceil(Eg*10**3)/10**3 #rounding off to 3 decimals\n", + "\n", + "#Result\n", + "print(\"energy band gap in eV is\",Eg);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('energy band gap in eV is', 0.452)\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 7.14, Page number 257" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "EgeV=1; #energy in eV\n", + "k=1.38*10**-23;\n", + "Eg=EgeV*1.602*10**-19; #in J\n", + "#EF can be taken as (Ev+0.5)eV\n", + "#therefore (Ev+0.5)eV = (Ec+Ev)/2--------(1)\n", + "#let fermi level shift by 10% then (Ev+0.6)eV = ((Ec+Ev)/2)+((3*k*T/4)*log(4))-----(2)\n", + "#subtracting (1) from (2)\n", + "#0.1 eV = (3*k*T/4)*math.log(4)\n", + "E=0.1; #energy in eV\n", + "E=E*1.602*10**-19; #energy in J\n", + "T=(4*E)/(3*k*math.log(4));\n", + "\n", + "#Result\n", + "print(\"temperature in K is\",T);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('temperature in K is', 1116.520509905372)\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 7.15, Page number 257" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "ni=1.5*10**16;\n", + "e=1.6*10**-19;\n", + "mew_e=0.13;\n", + "mew_h=0.05;\n", + "\n", + "#Calculation\n", + "sigma=ni*e*(mew_e+mew_h);\n", + "M=28.1; #atomic weight of Si\n", + "d=2.33*10**3; #density in kg/m^3\n", + "v=M/d;\n", + "N=6.02*10**26;\n", + "N1=N/v;\n", + "#1 donor type impurity is added to 1 impurity atom\n", + "ND=N1/(10**8);\n", + "p=(ni**2)/ND;\n", + "sigma_exd=ND*e*mew_e;\n", + "#1 acceptor type impurity is added to 1 impurity atom\n", + "Na=N1/(10**8);\n", + "n=(ni**2)/Na;\n", + "sigma_exa=Na*e*mew_h;\n", + "sigma=math.ceil(sigma*10**7)/10**7 #rounding off to 7 decimals\n", + "sigma_exd=math.ceil(sigma_exd*10**3)/10**3 #rounding off to 3 decimals\n", + "sigma_exa=math.ceil(sigma_exa*10**3)/10**3 #rounding off to 3 decimals\n", + "\n", + "#Result\n", + "print(\"conductivity in ohm-1 m-1 is\",sigma);\n", + "print(\"number of Si atoms per m^3 is\",N1);\n", + "print(\"conductivity for donor type impurity in ohm-1 m-1 is\",sigma_exd);\n", + "print(\"conductivity for acceptor type impurity in ohm-1 m-1 is\",sigma_exa);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('conductivity in ohm-1 m-1 is', 0.000432)\n", + "('number of Si atoms per m^3 is', 4.991672597864769e+28)\n", + "('conductivity for donor type impurity in ohm-1 m-1 is', 10.383)\n", + "('conductivity for acceptor type impurity in ohm-1 m-1 is', 3.994)\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 7.16, Page number 258" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "T=300; #temperature in K\n", + "KB=1.38*10**-23;\n", + "e=1.6*10**-19;\n", + "mew_e=0.19; #mobility of electrons in m^2/Vs\n", + "\n", + "#Calculation\n", + "Dn=mew_e*KB*T/e;\n", + "Dn=math.ceil(Dn*10**6)/10**6 #rounding off to 6 decimals\n", + "\n", + "#Result\n", + "print(\"diffusion coefficient of electrons in m^2/s is\",Dn);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('diffusion coefficient of electrons in m^2/s is', 0.004917)\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 7.17, Page number 259" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "\n", + "#import module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "RH=3.66*10**-4; #hall coefficient in m^3/coulomb\n", + "I=10**-2; #current in amp\n", + "B=0.5; #magnetic field in wb/m^2\n", + "t=1; #thickness in mm\n", + "\n", + "#Calculation\n", + "t=t*10**-3; #thickness in m\n", + "VH=(RH*I*B)/t;\n", + "VH=VH*10**3; #converting from Volts to mV\n", + "\n", + "#Result\n", + "print(\"Hall voltage in mV is\",VH);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('Hall voltage in mV is', 1.83)\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 7.18, Page number 259" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "RH=-7.35*10**-5; #hall coefficient\n", + "e=1.6*10**-19;\n", + "sigma=200;\n", + "\n", + "#Calculation\n", + "n=(-1/(RH*e));\n", + "mew=sigma/(n*e);\n", + "\n", + "#Result\n", + "print(\"density of charge carriers in m^3 is\",n);\n", + "print(\"mobility of charge carriers in m^2/Vs is\",mew);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('density of charge carriers in m^3 is', 8.503401360544217e+22)\n", + "('mobility of charge carriers in m^2/Vs is', 0.0147)\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 7.19, Page number 259" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "I=50; #current in amp\n", + "B=1.5; #magnetic field in T\n", + "n=8.4*10**28; #free electron concentration in electron/m^3\n", + "t=0.5; #thickness in cm\n", + "e=1.6*10**-19;\n", + "\n", + "#Calculation\n", + "t=t*10**-2; #thickness in m\n", + "VH=(I*B)/(n*e*t);\n", + "VH=VH*10**6; #converting VH from V to micro V\n", + "VH=math.ceil(VH*10**4)/10**4 #rounding off to 4 decimals\n", + "\n", + "#Result\n", + "print(\"magnitude of Hall voltage in microVolt is\",VH);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('magnitude of Hall voltage in microVolt is', 1.1161)\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 7.20, Page number 260" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "\n", + "#import module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "RH=3.66*10**-4;\n", + "e=1.6*10**-19;\n", + "rho_n=8.93*10**-3;\n", + "\n", + "#Calculation\n", + "n=1/(RH*e);\n", + "mew_e=RH/rho_n;\n", + "mew_e=math.ceil(mew_e*10**5)/10**5 #rounding off to 5 decimals\n", + "\n", + "#Result\n", + "print(\"n per m^3 is\",n);\n", + "print(\"mew_e in m^2/V is\",mew_e);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('n per m^3 is', 1.7076502732240434e+22)\n", + "('mew_e in m^2/V is', 0.04099)\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 7.21, Page number 260" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "mew_e=0.13; #electron mobility in m^2/Vs\n", + "mew_h=0.048; #hole mobility in m^2/Vs\n", + "ni=1.5*10**16;\n", + "e=1.6*10**-19;\n", + "T=300; #temp in K\n", + "ND=10**23; #density per m^3\n", + "\n", + "#Calculation\n", + "sigmai=ni*e*(mew_e+mew_h);\n", + "sigma=ND*mew_e*e;\n", + "p=(ni**2)/ND;\n", + "sigmai=math.ceil(sigmai*10**5)/10**5 #rounding off to 5 decimals\n", + "\n", + "#Result\n", + "print(\"conductivity of intrinsic Si in s is\",sigmai);\n", + "print(\"conductivity in s is\",sigma);\n", + "print(\"equilibrium hole concentration per m^3 is\",round(p));\n", + "\n", + "#answers for sigmai and sigma given in the book are wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('conductivity of intrinsic Si in s is', 0.00043)\n", + "('conductivity in s is', 2080.0)\n", + "('equilibrium hole concentration per m^3 is', 2250000000.0)\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 7.22, Page number 261" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "T=300; #temp in K\n", + "kB=1.38*10**-23;\n", + "mew_e=0.36; #mobility of electrons in m^2/Vs\n", + "e=1.6*10**-19;\n", + "mew_h=0.7; #mobility of electrons in m^2/Vs\n", + "sigma=2.12; #conductivity in ohm-1 m-1\n", + "C=4.83*10**21; #proportional constant\n", + "\n", + "#Calculation\n", + "ni=sigma/(e*(mew_e+mew_h));\n", + "#exp(-Eg/(2*kB*T)) = (C*(T^(3/2)))/ni\n", + "#let X be (C*(T^(3/2)))/ni\n", + "X=(C*(T**(3/2)))/ni;\n", + "#exp(-Eg/(2*kB*T)) = X \n", + "#applyinf log on both sides\n", + "#Eg/(2*kB*T) = log(X)\n", + "Eg=2*kB*T*math.log(X);\n", + "\n", + "#Result\n", + "print(\"forbidden energy gap in eV is\",Eg);\n", + "\n", + "#answer given in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('forbidden energy gap in eV is', 1.2016388762259164e-19)\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 7.23, Page number 261" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "Eg=0.4; #energy gap in eV\n", + "Eg=Eg*1.6*10**-19; #Eg in J\n", + "KB=1.38*10**-23;\n", + "T1=0; #temp 1 in C\n", + "T2=50; #temp 2 in C\n", + "T3=100; #temp 3 in C\n", + "\n", + "#Calculation\n", + "T1k=T1+273; #temp 1 in K\n", + "T2k=T2+273; #temp 2 in K\n", + "T3k=T3+273; #temp 3 in K\n", + "#F(E) = 1/(1+(exp((E-Ep)/(KB*T))))\n", + "#but E-Ep = (1/2)*Eg\n", + "#therefore F(E) = 1/(1+(exp(Eg/(2*KB*T))))\n", + "FE1=1/(1+(math.exp(Eg/(2*KB*T1k))));\n", + "FE2=1/(1+(math.exp(Eg/(2*KB*T2k))));\n", + "FE3=1/(1+(math.exp(Eg/(2*KB*T3k))));\n", + "FE1=math.ceil(FE1*10**6)/10**6 #rounding off to 6 decimals\n", + "FE2=math.ceil(FE2*10**6)/10**6 #rounding off to 6 decimals\n", + "FE3=math.ceil(FE3*10**6)/10**6 #rounding off to 6 decimals\n", + "\n", + "#Result\n", + "print(\"probability of occupation at 0 C in eV is\",FE1);\n", + "print(\"probability of occupation at 50 C in eV is\",FE2);\n", + "print(\"probability of occupation at 100 C in eV is\",FE3);\n", + "\n", + "#answers given in the book are wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('probability of occupation at 0 C in eV is', 0.000205)\n", + "('probability of occupation at 50 C in eV is', 0.000762)\n", + "('probability of occupation at 100 C in eV is', 0.001992)\n" + ] + } + ], + "prompt_number": 26 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 7.24, Page number 262" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "Eg=1.2; #energy in eV\n", + "Eg=Eg*1.6*10**-19; #in J\n", + "KB=1.38*10**-23;\n", + "T1=600; #temp in K\n", + "T2=300; #temp in K\n", + "\n", + "#Calculation\n", + "#sigma is proportional to exp(-Eg/(2*KB*T))\n", + "#let sigma1/sigma2 be R\n", + "R=math.exp((Eg/(2*KB))*((1/T2)-(1/T1)));\n", + "\n", + "#Result\n", + "print(\"the ratio between conductivity is\",round(R));\n", + "\n", + "#answer given in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('the ratio between conductivity is', 108467.0)\n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 7.25, Page number 263" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "ni=2.5*10**19; #density of charge carriers in m^3\n", + "r=1/(10**6); #ratio\n", + "e=1.6*10**-19;\n", + "mew_e=0.36; #mobility of electrons in m^2/Vs\n", + "mew_h=0.18; #mobility of holes in m^2/Vs\n", + "N=4.2*10**28; #number of Si atoms per m^3\n", + "\n", + "#Calculation\n", + "Ne=r*N;\n", + "Nh=(ni**2)/Ne;\n", + "sigma=(Ne*e*mew_e)+(Nh*e*mew_h);\n", + "rho=1/sigma;\n", + "rho=math.ceil(rho*10**8)/10**8 #rounding off to 8 decimals\n", + "\n", + "#Result\n", + "print(\"number of impurity atoms per m^3 is\",Ne);\n", + "print(\"the resistivity of doped Ge in ohm m is\",rho);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('number of impurity atoms per m^3 is', 4.2e+22)\n", + "('the resistivity of doped Ge in ohm m is', 0.00041336)\n" + ] + } + ], + "prompt_number": 30 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 7.26, Page number 264" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "n=5*10**17; #concentration in m^3\n", + "vd=350; #drift velocity in m/s\n", + "E=1000; #electric field in V/m\n", + "e=1.6*10**-19;\n", + "\n", + "#Calculation\n", + "mew=vd/E;\n", + "sigma=n*e*mew;\n", + "sigma=math.ceil(sigma*10**4)/10**4 #rounding off to 4 decimals\n", + "\n", + "#Result\n", + "print(\"the conductivity of material in ohm m is\",sigma);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('the conductivity of material in ohm m is', 0.028)\n" + ] + } + ], + "prompt_number": 32 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 7.27, Page number 264" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "sigma_e=2.2*10**-4; #conductivity\n", + "mew_e=125*10**-3; #mobility of electrons in m^2/Vs\n", + "e=1.602*10**-19;\n", + "\n", + "#Calculation\n", + "ne=sigma_e/(e*mew_e);\n", + "\n", + "#Result\n", + "print(\"concentration in m^3 is\",ne);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('concentration in m^3 is', 1.0986267166042448e+16)\n" + ] + } + ], + "prompt_number": 33 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 7.28, Page number 265" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "RH=3.66*10**-4; #hall coefficient in m^3/c\n", + "rho_i=8.93*10**-3; #resistivity in ohm m\n", + "e=1.6*10**-19;\n", + "\n", + "#Calculation\n", + "nh=1/(RH*e);\n", + "mew_h=1/(rho_i*nh*e);\n", + "mew_h=math.ceil(mew_h*10**4)/10**4 #rounding off to 4 decimals\n", + "\n", + "#Result\n", + "print(\"density of charge carriers in m^3 is\",nh);\n", + "print(\"mobility of charge carriers is %f m^2/Vs\",mew_h);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('density of charge carriers in m^3 is', 1.7076502732240434e+22)\n", + "('mobility of charge carriers is %f m^2/Vs', 0.041)\n" + ] + } + ], + "prompt_number": 35 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 7.29, Page number 265" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "I=3; #current in mA\n", + "I=I*10**-3; #current in amp\n", + "e=1.6*10**-19;\n", + "RH=3.66*10**-4; #hall coefficient in m^3/C\n", + "B=1; #flux density in w/m^2\n", + "d=2; #dimension along Y in cm\n", + "z=1; #dimension along z in mm\n", + "\n", + "#Calculation\n", + "d=d*10**-2; #dimension along Y in m\n", + "z=z*10**-3; #dimension along z in m\n", + "A=d*z; #area in m^2\n", + "EH=RH*I*B/A;\n", + "VH=EH*d;\n", + "VH=VH*10**3; #converting from V to mV\n", + "n=1/(RH*e);\n", + "VH=math.ceil(VH*10**2)/10**2 #rounding off to 2 decimals\n", + "\n", + "#Result\n", + "print(\"Hall voltage in mV is\",VH);\n", + "print(\"charge carrier concentration in m^3 is\",n);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('Hall voltage in mV is', 1.1)\n", + "('charge carrier concentration in m^3 is', 1.7076502732240434e+22)\n" + ] + } + ], + "prompt_number": 37 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Engineering_Physics_Vijaya/chapter8_2.ipynb b/Engineering_Physics_Vijaya/chapter8_2.ipynb new file mode 100755 index 00000000..2dc13b1f --- /dev/null +++ b/Engineering_Physics_Vijaya/chapter8_2.ipynb @@ -0,0 +1,253 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:064d55405a5d05f007b28f32cf39a9f99d10f303fc4084e2d14d99aaeb87858c" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Physics of Nano Materials" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 8.1, Page number 320" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import modules\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "r=5; #radius in m\n", + "pi=3.14;\n", + "\n", + "#Calculation \n", + "SA=4*pi*r**2; #surface area of sphere in m^2\n", + "V=(4/3)*pi*r**3; #volume of sphere in m^3\n", + "R=SA/V; #ratio\n", + "#surface area to volume ratio can also be given by 3/radius\n", + "\n", + "#Result\n", + "print(\"surface area to volume ratio of sphere in m-1 is\",R);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('surface area to volume ratio of sphere in m-1 is', 0.6)\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 8.2, Page number 321" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import modules\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "d=26; #distance in m\n", + "r=d/2; #radius in m\n", + "pi=3.14;\n", + "\n", + "#Calculation\n", + "SA=4*pi*r**2; #surface area of sphere in m^2\n", + "V=(4/3)*pi*r**3; #volume of sphere in m^3\n", + "R=SA/V; #ratio\n", + "R=math.ceil(R*10**3)/10**3; #rounding off to 3 decimals\n", + "#surface area to volume ratio can also be given by 3/radius\n", + "\n", + "#Result\n", + "print(\"surface area to volume ratio of sphere in m-1 is\",R);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('surface area to volume ratio of sphere in m-1 is', 0.231)\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 8.3, Page number 321" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import modules\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "r=1; #radius in m\n", + "h=1; #height in m\n", + "pi=3.14\n", + "\n", + "#Calculation\n", + "V=(1/3)*pi*(r**2)*h;\n", + "V=math.ceil(V*10**2)/10**2; #rounding off to 2 decimals\n", + "\n", + "#Result\n", + "print(\"volume of cone in m^3 is\",V); " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('volume of cone in m^3 is', 1.05)\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 8.4, Page number 321" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import modules\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "r=3; # radius in m\n", + "h=4; # height in m\n", + "pi=3.14\n", + "\n", + "#Calculation\n", + "SA=pi*r*math.sqrt((r**2)+(h**2));\n", + "TSA=SA+(pi*r**2);\n", + "\n", + "#Result\n", + "print(\"total surface area of cone in m^2 is\",TSA);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('total surface area of cone in m^2 is', 75.36)\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 8.5, Page number 322" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#import modules\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable decleration\n", + "V=100; #volume of cone in cubic inches\n", + "r=5; #radius of cone in inches\n", + "pi=3.14;\n", + "\n", + "#Calculation\n", + "r_m=r*0.0254; #radius of cone in m\n", + "#volume V=(1/3)*pi*(r**2)*h\n", + "#therefore h = (3*V)/(pi*r**2)\n", + "h=(3*V)/(pi*r**2); #height in inches\n", + "R=3/r_m;\n", + "h=math.ceil(h*10**3)/10**3; #rounding off to 3 decimals\n", + "\n", + "#Result\n", + "print(\"height of the cone in inches is\",h);\n", + "print(\"surface area to volume ratio in m-1 is\",R);\n", + "\n", + "#answer for the surface area to volume ratio given in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "('height of the cone in inches is', 3.822)\n", + "('surface area to volume ratio in m-1 is', 23.62204724409449)\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
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