From 3ee42f5fa3bf4b934d8a2eeb7ae533518bb15e62 Mon Sep 17 00:00:00 2001 From: Trupti Kini Date: Tue, 5 Jan 2016 23:30:04 +0600 Subject: Added(A)/Deleted(D) following books A Engineering_Physics_(Volume-2)_by_S.K._Gupta/chapter1_1.ipynb A Engineering_Physics_(Volume-2)_by_S.K._Gupta/chapter2_1.ipynb A Engineering_Physics_(Volume-2)_by_S.K._Gupta/chapter3_1.ipynb A Engineering_Physics_(Volume-2)_by_S.K._Gupta/chapter4_1.ipynb A Engineering_Physics_(Volume-2)_by_S.K._Gupta/chapter5_1.ipynb A Engineering_Physics_(Volume-2)_by_S.K._Gupta/chapter6_1.ipynb A Engineering_Physics_(Volume-2)_by_S.K._Gupta/chapter7_1.ipynb A Engineering_Physics_(Volume-2)_by_S.K._Gupta/screenshots/X-ray_diffraction_1.png A Engineering_Physics_(Volume-2)_by_S.K._Gupta/screenshots/ultrasonics_1.png A Engineering_Physics_(Volume-2)_by_S.K._Gupta/screenshots/wave_mechanics_1.png A sample_notebooks/MohdAsif/Ch2.ipynb --- .../chapter1_1.ipynb | 2156 ++++++++++++++++++++ .../chapter2_1.ipynb | 1050 ++++++++++ .../chapter3_1.ipynb | 685 +++++++ .../chapter4_1.ipynb | 441 ++++ .../chapter5_1.ipynb | 129 ++ .../chapter6_1.ipynb | 865 ++++++++ .../chapter7_1.ipynb | 503 +++++ .../screenshots/X-ray_diffraction_1.png | Bin 0 -> 152419 bytes .../screenshots/ultrasonics_1.png | Bin 0 -> 157652 bytes .../screenshots/wave_mechanics_1.png | Bin 0 -> 155664 bytes sample_notebooks/MohdAsif/Ch2.ipynb | 1269 ++++++++++++ 11 files changed, 7098 insertions(+) create mode 100644 Engineering_Physics_(Volume-2)_by_S.K._Gupta/chapter1_1.ipynb create mode 100644 Engineering_Physics_(Volume-2)_by_S.K._Gupta/chapter2_1.ipynb create mode 100644 Engineering_Physics_(Volume-2)_by_S.K._Gupta/chapter3_1.ipynb create mode 100644 Engineering_Physics_(Volume-2)_by_S.K._Gupta/chapter4_1.ipynb create mode 100644 Engineering_Physics_(Volume-2)_by_S.K._Gupta/chapter5_1.ipynb create mode 100644 Engineering_Physics_(Volume-2)_by_S.K._Gupta/chapter6_1.ipynb create mode 100644 Engineering_Physics_(Volume-2)_by_S.K._Gupta/chapter7_1.ipynb create mode 100644 Engineering_Physics_(Volume-2)_by_S.K._Gupta/screenshots/X-ray_diffraction_1.png create mode 100644 Engineering_Physics_(Volume-2)_by_S.K._Gupta/screenshots/ultrasonics_1.png create mode 100644 Engineering_Physics_(Volume-2)_by_S.K._Gupta/screenshots/wave_mechanics_1.png create mode 100644 sample_notebooks/MohdAsif/Ch2.ipynb diff --git a/Engineering_Physics_(Volume-2)_by_S.K._Gupta/chapter1_1.ipynb b/Engineering_Physics_(Volume-2)_by_S.K._Gupta/chapter1_1.ipynb new file mode 100644 index 00000000..7c196e63 --- /dev/null +++ b/Engineering_Physics_(Volume-2)_by_S.K._Gupta/chapter1_1.ipynb @@ -0,0 +1,2156 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:ee520d8248bc6c6ff0e419554e336661165329b56b79f9a6fd4a16790c44f027" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter1:WAVE MECHANICS" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg1:pg-12" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from sympy import *\n", + "h=6.62*10**-34 #planck constant in joule-sec\n", + "m=9.1*10**-31 #mass of electron in kg\n", + "q=1.6*10**-19 #charge of electron in coulombs\n", + "V=Symbol('V') #energy of electron in eV\n", + "lamda=round(((h/sqrt(2*m*q))*10**10),2)/sqrt(V) #lamda=h/sqrt(2mE)=h/sqrt(2mqV)\n", + "print\"de-Broglie wavelength for an electron of energy V ev is=\",lamda,\"Angstrom\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "de-Broglie wavelength for an electron of energy V ev is= 12.27/sqrt(V) Angstrom\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg2:pg-12" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "h=6.63*10**-34 #planck constant in joule-sec\n", + "m=9.1*10**-31 #mass of electron in kg\n", + "q=1.6*10**-19 #charge of electron in coulombs\n", + "V=50. #potential difference in volts(given)\n", + "lamda=int(((h/sqrt(2*m*q))*10**10)*1e2)*1e-2/sqrt(V) #lamda=h/sqrt(2mE)=h/sqrt(2mqV)\n", + "print\"de-Broglie wavelength = \", round(lamda,4),\"Angstrom\" " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "de-Broglie wavelength = 1.7367 Angstrom\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg3:pg-12" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "h=6.62*10**-34 #planck constant in joule-sec\n", + "m=9.1*10**-31 #mass of electron in kg\n", + "q=1.6*10**-19 #charge of electron in coulombs\n", + "V=100. #potential difference in volts(given)\n", + "E=q*V\n", + "lamda=round(((h/sqrt(2*m*q))*10**10),2)/sqrt(V) #lamda=h/sqrt(2mE)=h/sqrt(2mqV)\n", + "print\"de-Broglie wavelength =\",round(lamda,3),\"Angstrom\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "de-Broglie wavelength = 1.227 Angstrom\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg4:pg-12" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "h=6.6*10**-34 #planck constant in joule-sec\n", + "m=9.0*10**-31 #mass of electron in kg\n", + "KE=(15*10**3)*(1.6*10**-19) #Kinetic Energy of electron in joule\n", + "v=sqrt((2*KE)/m) #KE=1/2(mv**2) joule\n", + "p=m*v #momentum of electron in Kg-m/sec\n", + "lamda=h/p #de-broglie wavelength\n", + "print\"de-Broglie wavelength is=\",round(lamda*10**10,1),\"Angstrom\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "de-Broglie wavelength is= 0.1 Angstrom\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg5:pg-13" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "h=6.62*10**-34 #planck constant in joule-sec\n", + "m=1.67*10**-27 #mass of neutron in Kg\n", + "lamda=10**-10 #de-broglie wavelength in meter(given)\n", + "v=h/(m*lamda) #since lamda=h/mv \n", + "KE=(1./2)*m*v**2 #in joule\n", + "KE=KE/(1.6*10**-19) #in eV\n", + "print\"Velocity of neutron is= %.2e m/sec\"%v\n", + "print\"Kinetic Energy of Neutron is= \",round(KE,3),\"eV\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Velocity of neutron is= 3.96e+03 m/sec\n", + "Kinetic Energy of Neutron is= 0.082 eV\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg6:pg-13" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "h=6.6*10**-34 #planck constant in joule-sec\n", + "m=9.1*10**-31 #mass of electron in kg\n", + "q=1.6*10**-19 #charge of electron in coulombs\n", + "E=(1.25*10**3)*(1.6*10**-19) #Kinetic energy in joule\n", + "lamda=h/sqrt(2*m*E)\n", + "print\"wavelength is =\",\"{:.2e}\".format(lamda),\"m\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "wavelength is = 3.46e-11 m\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg7:pg-13" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "h=6.63*10**-34 #planck constant in joule-sec\n", + "mo=1.67*10**-27 #mass of proton in Kg\n", + "v=2.0*10**8 #speed of proton in m/sec\n", + "c=3*10**8 #speed of light in m/sec\n", + "p=(mo*v)/sqrt(1-(v/c)**2) #momentum of proton\n", + "lamda=h/p\n", + "print\"wavelength is =\",\"{:.2e}\".format(lamda*10**10),\"Angstrom\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "wavelength is = 1.48e-5 Angstrom\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg8:pg-14" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "h=6.62*10**-34 #planck constant in joule-sec\n", + "mp=1.67*10**-27 #mass of proton in Kg\n", + "c=3*10**8 #speed of light in m/sec\n", + "v=c/20 #speed of proton in m/sec\n", + "lamda=h/(mp*v)\n", + "print\"de-Broglie wavelength =\",\"{:.3e}\".format(lamda),\"m\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "de-Broglie wavelength = 2.643e-14 m\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg9:pg-14" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "h=6.62*10**-34 #planck constant in joule-sec\n", + "m=9.1*10**-31 #mass of electron in kg\n", + "c=3*10**8 #speed of light in m/sec\n", + "lamda=2.0*10**-10#wavelength of electron and photon in meter\n", + "#(a) momenta\n", + "Pe=h/lamda #momentum of electron in Kg-m/s\n", + "Pp=h/lamda #momentum of photon in kg-m/s\n", + "print\"(a)Since wavelength of photon is same as that of an electron, therefore their moments is also same i.e \",Pe,\"Kg-m/s\"\n", + "\n", + "#(b) total energies\n", + "me=m\n", + "mo=m\n", + "KE=(Pe**2)/(2*me) #Kinetic energy of electron in joule\n", + "Re=mo*c**2 #rest energy of electron in joule\n", + "Re=Re/(1.6*10**-19) #in eV\n", + "Ee=Re/10**6 #total energy of electron in Mev(since K.E. of electron is negligible compared to its rest energy so total energy is equal to the rest energy) \n", + "Ep=Pp*c #total energy of photon in joule(since rest energy of photon is zero so its total energy is same as its K.E.) \n", + "Ep=(Ep/(1.6*10**-19))*10**-3 #in KeV\n", + "print\"(b)Total energy of electron is\",round(Ee,2),\"MeV\",\"\\n Total energy of photon is\",round(Ep,2),\"KeV\"\n", + "\n", + "#(c) ratio of kinetic energies\n", + "Ke=round((KE/(1.6*10**-19)),1)#Kinetic energy of electron in eV\n", + "Kp=round(Ep,2)*(10**3) #Kinetic energy of photon in eV\n", + "ratio=Ke/Kp\n", + "print\"(c)Ratio of kinetic energies=\",\"{:.2e}\".format(ratio)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a)Since wavelength of photon is same as that of an electron, therefore their moments is also same i.e 3.31e-24 Kg-m/s\n", + "(b)Total energy of electron is 0.51 MeV \n", + " Total energy of photon is 6.21 KeV\n", + "(c)Ratio of kinetic energies= 6.05e-03\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg10:pg-15" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "h=6.62*10**-34 #planck constant in joule-sec\n", + "m=1.67*10**-27 #mass of neutron in Kg\n", + "c=3*10**8 #speed of light in m/sec\n", + "E=28.8 #Kinetic energy of neutron in eV(given)\n", + "E=28.8*1.6*(10**-19)#in joule\n", + "Rn=m*c**2 #Rest mass energy of neutron in joule\n", + "Rn=(Rn/(1.6*10**-19))/10**6 #in MeV\n", + "#since Kinetic energy of neutron under consideration is very small compared to its rest mass energy,the relativistic consideration may be ignored. \n", + "lamda=h/sqrt(2*m*E)\n", + "print\"de-Broglie wavelength=\",round(lamda*10**10,5),\"Angstrom\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "de-Broglie wavelength= 0.05336 Angstrom\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg11:pg-15" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "h=6.63*10**-34 #planck constant in joule-sec\n", + "m=4*1.67*10**-27 #mass of alpha particle Kg=4*mass of proton\n", + "q=2*1.6*10**-19 #for alpha particle q=2*e coulomb\n", + "V=200 #potential difference in volts \n", + "lamda=h/sqrt(2*m*q*V)\n", + "print\"de-Broglie wavelength=\",round(lamda*10**10,5),\"Angstrom\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "de-Broglie wavelength= 0.00717 Angstrom\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg12:pg-15" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "h=6.63*10**-34 #planck constant in joule-sec\n", + "m=9.1*10**-31 #mass of electron in kg\n", + "mo=m #for velocity much less than the velocity of light, m=mo\n", + "#(a)wavelength for a ball of mass 1.0Kg and v=1.0m/s\n", + "v=1.0 #speed of ball in m/sec\n", + "mass=1.0 #mass of ball in Kg\n", + "lamda=h/(mass*v)\n", + "print\"de-Broglie wavelength for a ball of mass 1.0Kg and v=1.0m/s =\",\"{:.2e}\".format(lamda*10**10),\"Angstrom\"\n", + "\n", + "#(b)wavelength for an electron of mass 9.1*10**-31 Kg and v=10**6 m/sec\n", + "v=10**6 #speed of electron in m/sec\n", + "lamda=h/(mo*v)\n", + "print\"de-Broglie wavelength for an electron of mass 9.1*10**-31Kg and v=10**6m/sec =\",round(lamda*10**10,2),\"Angstrom\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "de-Broglie wavelength for a ball of mass 1.0Kg and v=1.0m/s = 6.63e-24 Angstrom\n", + "de-Broglie wavelength for an electron of mass 9.1*10**-31Kg and v=10**6m/sec = 7.29 Angstrom\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg13:pg-16" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "h=6.63*10**-34 #planck constant in joule-sec\n", + "m=1.67*10**-27 #mass of neutron in Kg\n", + "c=3*10**8 #speed of light in m/sec\n", + "E=1. #Kinetic energy of neutron in eV\n", + "Rn=m*c**2 #rest mass energy of neutron in joule\n", + "Rn=(Rn/(1.6*10**-19))/10**6 #in MeV\n", + "#Kinetic energy of given neutron 1eV is very small as compared to its rest mass energy,therefore the relativistic consideration may be ignored \n", + "E=1*1.6*10**-19 #in joule\n", + "lamda=h/sqrt(2*m*E)\n", + "print\"de-Broglie wavelength=\",round(lamda*10**10,3),\"Angstrom\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "de-Broglie wavelength= 0.287 Angstrom\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg14:pg-16" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "h=6.62*10**-34 #planck constant in joule-sec\n", + "m=1.675*10**-27 #mass of neutron in Kg\n", + "c=3*10**8 #speed of light in m/sec\n", + "E=12.8 #energy of neutron in MeV\n", + "Rn=m*c**2 #rest mass energy of neutron in joule\n", + "Rn=(Rn/(1.6*10**-19))/10**6 # in MeV\n", + "#since the given energy 12.8MeV is very small as compared to the rest mass energy,therefore the relativistic consideration may be ignored \n", + "E=E*(10**6)*(1.6*10**-19) # in eV\n", + "lamda=h/sqrt(2*m*E)\n", + "print\"de-Broglie wavelength=\",\"{:.1e}\".format(lamda*10**10),\"Angstrom\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "de-Broglie wavelength= 8.0e-5 Angstrom\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg15:pg-17" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "h=6.6*10**-34 #planck constant joule-sec\n", + "m=9.1*10**-31 #mass of electron kg\n", + "e=1.6*10**-19 #charge of electron in coulomb\n", + "lamda=0.40*10**-10#wavelength in meter\n", + "V=(h**2)/round(((lamda**2)*2*m*e),72)#lamda=h/sqrt(2mE)=h/sqrt(2meV)\n", + "print\"applied voltage=\",round(V,2),\"Volt\" " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "applied voltage= 934.76 Volt\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg17:pg-17" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "h=6.6*10**-34 #planck constant in joule-sec\n", + "m=1.67*10**-27 #mass of neutron in Kg\n", + "K=8.6*10**-5 #Boltzmann constant in eV/degree\n", + "K=K*1.6*10**-19 #in J/K\n", + "T=27+273 #temperature in Kelvin\n", + "E=K*T #energy of particle\n", + "lamda=h/round(sqrt(2*m*E),26)\n", + "print\"wavelength=\",round(lamda*10**10,3),\"Angstrom\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "wavelength= 1.779 Angstrom\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg18:pg-18" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "h=6.62*10**-34 #planck constant in joule-sec\n", + "m=4.65*10**-26 #mass of nitrogen atom in Kg\n", + "T=27+273 #Temperature in Kelvin\n", + "K=1.38*10**-23 #Boltzmann constant in J/K\n", + "E=(3./2)*K*T #for nitrogen atom E=(3/2)*K*T\n", + "lamda=h/sqrt(2*m*E) \n", + "print\"de-Broglie wavelength=\",round(lamda*10**10,4),\"Angstrom\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "de-Broglie wavelength= 0.2755 Angstrom\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg19:pg-18" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "h=6.62*10**-34 #planck constant in joule-sec\n", + "m=6.7*10**-27 #mass of helium atom in Kg\n", + "K=1.38*10**-23 #Boltzmann constant in J/K\n", + "T=400 #Temperature in Kelvin\n", + "E=(3./2)*K*T #for helium atom E=(3/2)*K*T\n", + "lamda=h/sqrt(2*m*E)\n", + "print\"de-Broglie wavelength=\",round(lamda*10**10,3),\"Angstrom\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "de-Broglie wavelength= 0.628 Angstrom\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg20:pg-18" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "E=100*1.6*10**-19 #kinetic energy of electron in joule\n", + "m=9.0*10**-31 #mass of electron in Kg\n", + "h=6.62*10**-34 #planck constant joule-sec\n", + "D=20 #distance of screen from foil in cm\n", + "diameter=2.44 #diameter of ring in cm\n", + "r=diameter/2 #radius of ring in cm\n", + "lamda=h/sqrt(2*m*E)\n", + "tan_theta=r/D\n", + "#for small value of theta tan(theta)=sin(theta)\n", + "#According to Bragg's law, 2d(sin(theta))=n*lamda\n", + "n=1\n", + "sin_theta=tan_theta\n", + "d=(n*lamda)/(2*sin_theta)\n", + "print\"spacing of the related lattice planes in the metal=\",round(d*10**10,2),\"Angstrom\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "spacing of the related lattice planes in the metal= 10.11 Angstrom\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg21:pg-19" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "r=0.53*10**-10 #radius of first Bohr orbit in hydrogen atom in Meter\n", + "h=6.6*10**-34 #planck constant in joule-sec\n", + "m=9.1*10**-31 #mass of electron in Kg\n", + "lamda=2*math.pi*r#since 2*pi*r=n*lamda where n=1 for the first Bohr orbit, So lamda=2*pi*r=h/(m*v)\n", + "v=h/(lamda*m)\n", + "print\"velocity of electron=\",\"{:.2e}\".format(v),\"m/s\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "velocity of electron= 2.18e+06 m/s\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg22:pg-19" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "h=6.63*10**-34 #planck constant in joule-sec\n", + "mo=9.1*10**-31 #rest mass of electron in Kg\n", + "lamda=5896*10**-10#wavelength in meter\n", + "#Since lamda=h/(mo*v) and Kinetic energy=(1/2)*mo*v**2 \n", + "#therefore on putting v=h/(mo*lamda) in equation of Kinetic energy\n", + "K=((h/lamda)**2)/(2*mo) #kinetic energy of electron in joule\n", + "K=K/(1.6*10**-19) #in eV\n", + "print\"Kinetic energy of electron=\",\"{:.2e}\".format(K),\"eV\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Kinetic energy of electron= 4.34e-06 eV\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg23:pg-20" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "h=6.63*10**-34 #planck constant in joule-sec\n", + "mo=1.67*10**-27 #rest mass of neutron in Kg\n", + "lamda=10**-10 #de-broglie wavelength in meter\n", + "v=h/(mo*lamda) #velocity of neutron in m/s (since lamda=h/(mo*v))\n", + "print\"velocity of neutron=\",\"{:.2e}\".format(v),\"m/s\"\n", + "K=(mo*v**2)/2 #kinetic energy of neutron in joule\n", + "K=K/(1.6*10**-19) #in eV\n", + "print\"kinetic energy of neutron=\",round(K,3),\"eV\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "velocity of neutron= 3.97e+03 m/s\n", + "kinetic energy of neutron= 0.082 eV\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg25:pg-21" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "h=6.62*10**-34 #planck constant in joule-sec\n", + "mo=9.1*10**-31 #rest mass of electron in Kg\n", + "c=3e8 #speed of light in m/sec\n", + "K=1 #Kinetic energy in MeV\n", + "Re=(mo*c**2/(1.6*10**-19))/10**6 #rest mass energy of electron in Mev\n", + "lamda=h*c/sqrt(K*(K+(2*Re)))\n", + "print\"Wavelength is %.2e Angstrom\"%(lamda*1e10/(1.6e-19*1e6))\n", + "#answer is wrong in book because of calculation mistake" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Wavelength is 8.73e-03 Angstrom\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg26:pg-22" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "h=6.63*10**-34 #planck constant in joule-sec\n", + "mo=9.1*10**-31 #rest mass of electron in Kg(in book it is given wrong in question)\n", + "c=3*10**8 #speed of light in m/sec\n", + "\n", + "#(a)wavelength associated with 1MeV electron\n", + "K=1 #kinetic energy of electron in MeV\n", + "Re=(mo*c**2/(1.6*10**-19))/10**6 #rest mass energy of electron in Mev\n", + "#since given K.E(1MeV)of electron is comparable with its rest mass energy \n", + "#therefore relativistic variation of mass with velocity is taken in to account\n", + "d=round(sqrt(K*(K+(2*Re))),2)*1.6*10**-13 #value of sqrt(K*(K+(2*mo*c**2))) in volt\n", + "lamda=h*c/d\n", + "print\"wavelength associated with 1MeV electron=\",\"{:.2e}\".format(lamda*10**10),\"Angstrom\"\n", + "\n", + "#(b)wavelength associated with 1MeV proton\n", + "K=1*1.6*10**-13 #kinetic energy of electron in volt\n", + "mo=1.67*10**-27 #rest mass of proton in Kg\n", + "Rp=(mo*c**2/(1.6*10**-19))/10**6 #rest mass energy of proton in Mev\n", + "#since given K.E(1MeV)of proton is much less than its rest mass energy \n", + "#therefore relativistic effect can be ignored\n", + "lamda=h/sqrt(2*mo*K)\n", + "print\"wavelength associated with 1MeV proton=\",\"{:.2e}\".format(lamda*10**10),\"Angstrom\"\n", + "\n", + "#(c)wavelength associated with 1MeV photon\n", + "#since rest mass of photon is zero so its rest mass energy is also zero\n", + "E=K #Energy of photon is entirely kinetic energy in volt\n", + "lamda=h*c/E\n", + "print\"wavelength associated with 1MeV photon=\",\"{:.2e}\".format(lamda*10**10),\"Angstrom\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "wavelength associated with 1MeV electron= 8.75e-03 Angstrom\n", + "wavelength associated with 1MeV proton= 2.87e-4 Angstrom\n", + "wavelength associated with 1MeV photon= 1.24e-02 Angstrom\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg27:pg-23" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "V=54. #potential difference in volt\n", + "lamda=12.28/sqrt(V) #de-Broglie wavelength of electron\n", + "phi=50 #angle of scattering in degree\n", + "sin_phi=sin(math.radians(phi))\n", + "D=lamda/sin_phi #according to Bragg's law for normal incidence lamda=D*(sin phi),where D is the distance between two consecutive atoms in the surface layer \n", + "print\"Distance between two neighbouring atoms in the surface of Ni-crystal=\",round(D,2),\"Angstrom\"\n", + "theta=90-(phi/2) #glancing angle in degree\n", + "sin_theta=sin(math.radians(theta))\n", + "d=lamda/(2*sin_theta) \n", + "print\"Distance between successive Bragg's planes of Ni-crystal=\",round(d,2),\"Angstrom\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Distance between two neighbouring atoms in the surface of Ni-crystal= 2.18 Angstrom\n", + "Distance between successive Bragg's planes of Ni-crystal= 0.92 Angstrom\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg28:pg-24" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "V=54. #potential difference in volt\n", + "lamda=12.28/sqrt(V) #de-Broglie wavelength of electron\n", + "D=2.15 #distance between successive atoms of crystal plane in angstrom\n", + "\n", + "#for first order diffraction\n", + "n=1 \n", + "phi=math.asin(n*lamda/D)#Bragg's equation, D*sin(phi)=n*lamda,where phi is angle of scattering \n", + "phi=math.degrees(phi)# in degree\n", + "#for second order diffraction\n", + "n=2\n", + "sin_phi_2=n*lamda/D\n", + "print\"since sin_phi_2=\",round(sin_phi_2,2),\">1 which is impossible because sin(phi) can never exceed 1.Hence second and third orders can't occur.\"\n", + "\n", + "#when V is increased from 54volt to 60 volt\n", + "V=60.\n", + "n=1\n", + "lamda=12.28/sqrt(V)#de-Broglie wavelength of electron\n", + "phi=math.asin(n*lamda/D)\n", + "phi=math.degrees(phi)\n", + "print\"When accelerating potential were changed from 54volt to 60volt, first order diffracted beams occur at=\",int(round(phi)),\"degree\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "since sin_phi_2= 1.55 >1 which is impossible because sin(phi) can never exceed 1.Hence second and third orders can't occur.\n", + "When accelerating potential were changed from 54volt to 60volt, first order diffracted beams occur at= 48 degree\n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg30:pg-28" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from sympy import *\n", + "h=6.63*10**-34 #planck constant in joule-sec\n", + "mo=9.1*10**-31 #rest mass of electron in kg\n", + "C=3*10**8 #speed of light in m/sec\n", + "lamda=10**-12 #wavelength in meter\n", + "pc=(h*C/lamda)/(1.6*10**-19)\n", + "Eo=(mo*C**2)/(1.6*10**-19)#rest energy of electron in eV\n", + "E=sqrt((pc**2)+(Eo**2)) #total energy of electron eV\n", + "c=Symbol('c')\n", + "v=c*round(sqrt(1-(Eo/E)**2),3)\n", + "print\"Group velocity of the de-Broglie waves,Vg = v =\",v\n", + "Vp=round((1/sqrt(1-(Eo/E)**2)),2)*(c**2/c)#Vp=(c**2)/Vg\n", + "print\"Phase velocity of the de-Broglie waves,Vp=\",Vp" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Group velocity of the de-Broglie waves,Vg = v = 0.925*c\n", + "Phase velocity of the de-Broglie waves,Vp= 1.08*c\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg31:pg-29" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from sympy import *\n", + "h=6.63*10**-34 #planck constant in joule-sec\n", + "mo=9.1*10**-31 #rest mass of electron in kg\n", + "C=3*10**8 #speed of light in m/sec\n", + "lamda=2.0*10**-12 #wavelength in meter \n", + "pc=round((h*C/lamda)/(1.6*10**-16),2)#in KeV\n", + "Eo=int((mo*C**2)/(1.6*10**-16))#rest energy of electron in KeV\n", + "E=round(sqrt((pc**2)+(Eo**2)),2)#total energy of electron KeV\n", + "KE=E-Eo\n", + "print\"Kinetic energy of electron=\",KE,\"KeV\"\n", + "c=Symbol('c')\n", + "v=c*round(sqrt(1-(Eo/E)**2),4)\n", + "print\"Group velocity of the de-Broglie waves,Vg = v =\",v\n", + "Vp=round((1/sqrt(1-(Eo/E)**2)),2)*(c**2/c)#Vp=(c**2)/Vg\n", + "print\"Phase velocity of the de-Broglie waves,Vp=\",Vp" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Kinetic energy of electron= 293.65 KeV\n", + "Group velocity of the de-Broglie waves,Vg = v = 0.7725*c\n", + "Phase velocity of the de-Broglie waves,Vp= 1.29*c\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg33:pg-39" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#From Heisenberg uncertainty relation (delta_x*delta_p)>=h or delta_p>=h/delta_x\n", + "h=6.63*10**-34 #planck constant joule-sec\n", + "delta_x=0.2*10**-10 #uncertainty in position in meter\n", + "delta_p=h/(2*math.pi*delta_x) #uncertainty in momentum \n", + "print\"Uncertainty in momentum =\",\"{:.2e}\".format(delta_p),\"Kg-ms-1\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Uncertainty in momentum = 5.28e-24 Kg-ms-1\n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg34:pg-39" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#From Heisenberg uncertainty relation (delta_x*delta_p)>=h or delta_x>=h/delta_p\n", + "h=6.62*10**-34 #planck constant in joule-sec\n", + "mo=9.0*10**-31 #rest mass of electron in kg\n", + "c=3*10**8 #speed of light in m/sec\n", + "v=3.0*10**7 #velocity of electron in m/sec\n", + "delta_p=mo*v/sqrt(1-(v/c)**2)#maximum uncertainty in momentum\n", + "delta_x=h/(2*math.pi*delta_p)#smallest uncertainty in position \n", + "print\"Smallest possible uncertainty in position of an electron=\",round(delta_x*10**10,4),\"Angstrom\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Smallest possible uncertainty in position of an electron= 0.0388 Angstrom\n" + ] + } + ], + "prompt_number": 29 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg35:pg-40" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#From Heisenberg uncertainty relation (delta_x*delta_p)>=h or delta_p>=h/delta_x\n", + "h=1.05*10**-34 #planck constant in joule-sec\n", + "m=9.1*10**-31 # mass of electron in kg\n", + "delta_x=1.1*10**-8#uncertainty in position in meter\n", + "delta_p=h/delta_x #uncertainty in momentum\n", + "delta_v=delta_p/m #minimum uncertainty in velocity\n", + "print\"minimum uncertainty in velocity of an electron=\",\"{:.2e}\".format(delta_v),\"m/s\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "minimum uncertainty in velocity of an electron= 1.05e+04 m/s\n" + ] + } + ], + "prompt_number": 30 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg36:pg-40" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#From Heisenberg uncertainty relation (delta_x*delta_p)>=h or delta_p>=h/delta_x\n", + "h=6.62*10**-34 #planck constant in joule-sec\n", + "m=9.1*10**-31 #mass of electron in kg\n", + "delta_x=10*10**-10#uncertainty in position in meter\n", + "delta_p=h/(2*math.pi*delta_x)#uncertainty in momentum in Kg-m/s\n", + "delta_v=delta_p/m #uncertainty in velocity of an electron\n", + "print\"uncertainty in velocity of an electron=\",\"{:.2e}\".format(delta_v),\"m/s\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "uncertainty in velocity of an electron= 1.16e+05 m/s\n" + ] + } + ], + "prompt_number": 31 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg37:pg-40" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#From Heisenberg uncertainty relation (delta_x*delta_p)>=h or delta_x>=h/delta_p\n", + "h=6.62*10**-34 #planck constant in joule-sec\n", + "m=9.0*10**-31 #mass of electron in kg\n", + "v=1.05*10**4 #speed of electron in m/s\n", + "p=m*v #momentum of electron in Kg-m/s\n", + "delta_p=(0.01/100)*p#uncertainty in momentum(since uncertainty in value of p is 0.01% of its value)\n", + "delta_x=h/(2*math.pi*delta_p)#uncertainty in position of electron\n", + "print\"uncertainty in the position of electron=\",\"{:.3e}\".format(delta_x),\"m\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "uncertainty in the position of electron= 1.115e-04 m\n" + ] + } + ], + "prompt_number": 32 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg38:pg-40" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#From Heisenberg uncertainty relation (delta_x*delta_p)>=h or delta_x>=h/delta_p\n", + "h=6.63*10**-34 #planck constant in joule-sec\n", + "m=9.0*10**-31 #mass of electron in kg\n", + "v=600 #speed of electron in m/s\n", + "delta_v=(.005/100)*v#uncertainty in velocity of an electron in m/s\n", + "delta_p=m*delta_v#uncertainty in momentum of an electron in Kg-m/s\n", + "#value of delta_p is wrong in book\n", + "delta_x=h/(2*math.pi*delta_p)#uncertainty in position of electron\n", + "print\"uncertainty in position of electron=\",round(delta_x,4),\"m\"#answer is wrong in book" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "uncertainty in position of electron= 0.0039 m\n" + ] + } + ], + "prompt_number": 33 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg39:pg-41" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "h=6.63*10**-34 #planck constant in joule-sec\n", + "m=9*10**-31 #mass of electron in kg\n", + "v=6.6*10**4 #speed of electron in m/s\n", + "p=m*v #momentum of electron in Kg-m/s\n", + "delta_p=(0.01/100)*p#uncertainty in momentum(since uncertainty in value of p is 0.01% of its value)\n", + "delta_x=h/(2*math.pi*delta_p)#uncertainty in position of electron(From Heisenberg uncertainty relation)\n", + "print\"uncertainty in the position of electron=\",\"{:.2e}\".format(delta_x),\"m\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "uncertainty in the position of electron= 1.78e-05 m\n" + ] + } + ], + "prompt_number": 34 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg40:pg-41" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "h=6.63*10**-34 #planck constant in joule-sec\n", + "m=9.1*10**-31 #mass of electron in kg\n", + "r=0.5*10**-10 #radius of hydrogen atom in meter\n", + "delta_x=r #uncertainty in position of electron meter\n", + "delta_Px=round(h/(2*math.pi*delta_x),25)#uncertainty in momentum of electron in Kg-m/s(From Heisenberg uncertainty relation)\n", + "p=delta_Px #momentum of electron(since magnitude of momentum can't be less than that of uncertainty)\n", + "KE=((p**2)/(2*m))/(1.6*10**-19)#Kinetic energy in eV\n", + "print\"Kinetic energy=\",round(KE,1),\"eV\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Kinetic energy= 15.1 eV\n" + ] + } + ], + "prompt_number": 35 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg41:pg-41" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "h=6.63*10**-34 #planck constant in joule-sec\n", + "m=9.0*10**-31 #mass of electron in kg\n", + "v=5.00*10**3 #speed of electron in m/s\n", + "p=m*v #momentum of electron in Kg-m/s\n", + "delta_p=(0.003/100)*p#uncertainty in momentum(since uncertainty in value of p is 0.003% of its value)\n", + "delta_x=h/(2*math.pi*delta_p)#uncertainty in position of electron(From Heisenberg uncertainty relation)\n", + "print\"uncertainty in the position of electron=\",\"{:.2e}\".format(delta_x),\"m\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "uncertainty in the position of electron= 7.82e-04 m\n" + ] + } + ], + "prompt_number": 36 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg42:pg-41" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "h=6.63*10**-34 #planck constant in joule-sec\n", + "m=9.1*10**-31 #mass of electron in kg\n", + "r=0.53*10**-10 #radius of hydrogen atom in meter\n", + "delta_x=r #uncertainty in position of electron in meter\n", + "delta_Px=h/(2*math.pi*delta_x)#uncertainty in momentum of electron in Kg-m/s(From Heisenberg uncertainty relation)\n", + "p=delta_Px #momentum of electron(since magnitude of momentum can't be less than that of uncertainty)\n", + "KE=((p**2)/(2*m))/(1.6*10**-19)#Kinetic energy in eV\n", + "print\"Minimum energy=\",round(KE,1),\"eV\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Minimum energy= 13.6 eV\n" + ] + } + ], + "prompt_number": 37 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg43:pg-42" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "c=3*10**8 #speed of light in m/sec\n", + "dlamda=(10**-4)*(10**-10) #width of spectral line in meter\n", + "lamda=5000*10**-10 #wavelength of spectral line in meter\n", + "\n", + "#From Heisenberg uncertainty relation (delta_E*delta_t)>=h or delta_t>=h/(2*pi*delta_E)\n", + "#since E=h*c/lamda so delta_E=(h*c/lamda**2)*dlamda\n", + "#putting value of delta_E in Heisenberg uncertainty relation,delta_t=lamda**2/(2*pi*c*dlamda)\n", + "delta_t=(lamda**2)/(2*math.pi*c*dlamda)\n", + "print\"minimum time required=\",\"{:.3e}\".format(delta_t),\"sec\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "minimum time required= 1.326e-08 sec\n" + ] + } + ], + "prompt_number": 38 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg44:pg-42" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "h=6.62*10**-34 #planck constant joule-sec\n", + "m=9.0*10**-31 #mass of electron kg\n", + "delta_x=.01*10**-2#uncertainty in position of a particle in meter\n", + "#(a)uncertainty in momentum of particle\n", + "delta_p=h/(2*math.pi*delta_x)#From Heisenberg uncertainty relation\n", + "print\"uncertainty in momentum of particle=\",\"{:.3e}\".format(delta_p),\"Kg-m/s\"\n", + "\n", + "#(b)uncertainty in velocity of electron\n", + "delta_x=5*10**-10 #uncertainty in position of a electron in meter\n", + "delta_p=h/(2*math.pi*delta_x)\n", + "delta_v=delta_p/m\n", + "print\"uncertainty in velocity of electron=\",\"{:.2e}\".format(delta_v),\"m/s\"\n", + "\n", + "#(c)uncertainty in velocity of alpha-particle\n", + "mp=1.67*10**-27 #mass of proton in Kg\n", + "m=4*mp #mass of alpha=particle in Kg\n", + "delta_x=5*10**-10 #uncertainty in position of alpha-particle in meter\n", + "delta_p=h/(2*math.pi*delta_x)\n", + "delta_v=delta_p/m\n", + "print\"uncertainty in velocity of alpha-particle=\",round(delta_v,2),\"m/s\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "uncertainty in momentum of particle= 1.054e-30 Kg-m/s\n", + "uncertainty in velocity of electron= 2.34e+05 m/s\n", + "uncertainty in velocity of alpha-particle= 31.55 m/s\n" + ] + } + ], + "prompt_number": 39 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg45:pg-43" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#From Heisenberg uncertainty relation for energy E and time t, (delta_E*delta_t)>=h\n", + "#Also E=h*v or delta_E=h*(delta_v)\n", + "#putting this value in uncertainty relation, delta_v>=1/(2*pi*delta_t)\n", + "delta_t=10**-8 #uncertainty in time in sec\n", + "delta_v=1/(2*math.pi*delta_t) \n", + "print\"minimum uncertainty in the frequency=\",\"{:.3e}\".format(delta_v),\"sec-1\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "minimum uncertainty in the frequency= 1.592e+07 sec-1\n" + ] + } + ], + "prompt_number": 40 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg46:pg-43" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#From Heisenberg uncertainty relation for energy E and time t, (delta_E*delta_t)>=h\n", + "#Also E=h*v or delta_E=h*(delta_v)\n", + "#putting this value in uncertainty relation, delta_v>=1/(2*pi*delta_t)\n", + "delta_t=10**-8 #uncertainty in time in sec\n", + "delta_v=1/(2*math.pi*delta_t) \n", + "print\"minimum uncertainty in the frequency=\",\"{:.3e}\".format(delta_v),\"sec-1\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "minimum uncertainty in the frequency= 1.592e+07 sec-1\n" + ] + } + ], + "prompt_number": 41 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg47:pg-43" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "h=6.62*10**-34 #planck constant joule-sec\n", + "m=9.0*10**-31 #mass of electron kg\n", + "delta_x=1*10**-10#uncertainty in position of a electron in meter\n", + "delta_p=h/(2*math.pi*delta_x)#uncertainty in momentum of electron in Kg-m/s(From Heisenberg uncertainty relation) \n", + "E=(1*10**3)*(1.6*10**-19)#energy in joule\n", + "p=sqrt(2*m*E) #momentum in Kg-m/s\n", + "percentage=(delta_p/p)*100\n", + "print\"Percentage of uncertainty in its momentum=\",round(percentage,1),\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Percentage of uncertainty in its momentum= 6.2 %\n" + ] + } + ], + "prompt_number": 42 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg48:pg-43" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "h=6.63*10**-34 #planck constant joule-sec\n", + "delta_t=2.5*10**-20 #uncertainty in time in sec\n", + "delta_E=h/(2*math.pi*delta_t)#From Heisenberg uncertainty relation for energy E and time t\n", + "delta_E=delta_E/(1.6*10**-19)\n", + "print\"minimum error=\",round(delta_E/10**3,3),\"KeV\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "minimum error= 26.38 KeV\n" + ] + } + ], + "prompt_number": 43 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg49:pg-44" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "h=1.054*10**-34 #planck constant joule-sec\n", + "delta_t=10**-12 #uncertainty in time in sec\n", + "delta_E=h/(2*delta_t)#uncertainty in energy(From Heisenberg uncertainty relation)\n", + "delta_E=delta_E/(1.6*10**-19)\n", + "print\"uncertainty in energy of gamma-ray photon emitted=\",\"{:.1e}\".format(delta_E),\"eV\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "uncertainty in energy of gamma-ray photon emitted= 3.3e-04 eV\n" + ] + } + ], + "prompt_number": 44 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg50:pg-44" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "h=6.62*10**-34 #planck constant joule-sec\n", + "m=9.1*10**-31 #mass of electron kg\n", + "e=1.6*10**-19 #charge of electron in coulomb\n", + "V=1000. #potential difference in volt\n", + "delta_V=12 #in volt\n", + "#delta_P=m*e*delta_V/sqrt(2*m*e*V)\n", + "#From Heisenberg uncertainty relation delta_x*delta_p)>=h or delta_x>=h/delta_p\n", + "delta_x=((h*sqrt(2.0))/sqrt(m*e))*(sqrt(V)/delta_V)\n", + "print\"uncertainty in position of an electron=\",round(delta_x,11),\"m =\",round(delta_x*1e10,1),\"Angstrom\"\n", + "#answer in book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "uncertainty in position of an electron= 6.47e-09 m = 64.7 Angstrom\n" + ] + } + ], + "prompt_number": 45 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg51:pg-44" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#According to uncertainty principle (delta_x*delta_p)>=h ......eq.1\n", + "#momentum,p=h/lamda or p*lamda=h ........eq.2\n", + "#on differentiating eq.2 and solving we get delta_p=(h*delta_lamda)/lamda**2,put this value in eq.1\n", + "#we get (delta_x*delta_lamda)>=(lamda**2)/2*pi\n", + "lamda=10**-10 #wavelength in meter\n", + "delta_lamda=lamda*10**-6#since uncertainty in wavelength is given to be in ratio 1/10**6\n", + "delta_x=lamda/(2*math.pi*10**-6)#since (delta_x)=(lamda**2)/(2*pi*delta_lamda)\n", + "print\"uncertainty in position=\"\"{:.2e}\".format(delta_x),\"m\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "uncertainty in position=1.59e-05 m\n" + ] + } + ], + "prompt_number": 46 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg52:pg-54" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "h=6.63*10**-34 #planck constant in joule-sec\n", + "m=9.11*10**-31 #mass of electron in kg\n", + "L=10**-10 #width of box in meter\n", + "n=1 #minimum energy of the particle is obtained for n=1\n", + "E=(n*h)**2/(8*m*L**2) #in joule\n", + "E=round(E,20)/(1.6*10**-19)#in eV\n", + "print\"energy of an electron=\",round(E,2),\"eV\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "energy of an electron= 37.69 eV\n" + ] + } + ], + "prompt_number": 47 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg54:pg-54" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "delta_x=5*10**-10 #uncertainty in position in meter\n", + "L=25*10**-10 #width of box in meter\n", + "x=L/2 #at the center of the box x=L/2\n", + "n=1 #since the particle is in the state of least energy so n=1\n", + "def si(x): #si(x) is the wave function of particle moving in an infinite potential well\n", + " return sqrt(2/L)*sin(n*math.pi*x/L)\n", + "p=(abs(si(x))**2)*delta_x\n", + "print\"probability =\",round(p,1)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "probability = 0.4\n" + ] + } + ], + "prompt_number": 48 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg57:pg-56" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "h=6.63*10**-34 #planck constant in joule-sec\n", + "m=9.1*10**-31 #mass of electron in kg\n", + "L=2.5*10**-10 #width of box in meter\n", + "#first lowest permitted energy value\n", + "n=1\n", + "E=(n*h)**2/(8*m*L**2) #in joule\n", + "E=E/(1.6*10**-19) #in eV\n", + "print\"first lowest permitted energy value=\",round(E,2),\"eV\"\n", + "#second lowest permitted energy value\n", + "n=2\n", + "E=round(E,2)*n**2 #in eV\n", + "print\"second lowest permitted energy value=\",round(E,2),\"eV\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "first lowest permitted energy value= 6.04 eV\n", + "second lowest permitted energy value= 24.16 eV\n" + ] + } + ], + "prompt_number": 49 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg58:pg-56" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#five antinodes signify that particle is in 5th quantum state i.e n=5\n", + "E5=5**2 #energy of the 5th quantum state\n", + "E1=230*1.6*10**-19 #energy of the 1st quantum state in joule\n", + "E1=round(E1/E5,20)\n", + "h=6.62*10**-34 #planck constant in joule-sec\n", + "L=0.2*10**-9 #width of well in meter\n", + "#for n=1\n", + "m=h**2/(8*E1*L**2) #in Kg(since En=(n*h)**2/(8*m*L**2))\n", + "print\"mass of the particle=\",\"{:.2e}\".format(m),\"Kg\"\n", + "En=(1*10**3)*(1.6*10**-19)#in joule(given)\n", + "n=sqrt(En/E1)\n", + "print\"since n=\",round(n,2),\" is not an integer. Hence,En=1KeV is not permitted value of energy.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "mass of the particle= 9.32e-31 Kg\n", + "since n= 10.43 is not an integer. Hence,En=1KeV is not permitted value of energy.\n" + ] + } + ], + "prompt_number": 50 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg59:pg-57" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from scipy import integrate\n", + "x1=0.35 #lower limit\n", + "x2=0.45 #upper limit\n", + "#Wn=a*x,wave function\n", + "p=integrate.quad(lambda x: x**2,x1,x2)\n", + "p=round(p[0],4)\n", + "print\"Probability=%s*a**2\"%p\n", + "X=integrate.quad(lambda x: x**3,0,1)\n", + "X=(X[0])\n", + "print\"Expectation value of particle's position=%s*a**2\"%X" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Probability=0.0161*a**2\n", + "Expectation value of particle's position=0.25*a**2\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg60:pg-58" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "L = 1 # let unit length\n", + "x1=0.45*L #lower limit\n", + "x2=0.55*L #upper limit\n", + "n=1 #for ground state\n", + "p = (1/L)*((x2-(L/(2*math.pi*n) *math.sin(2*x2*math.pi*n/L)))- (x1-(L/(2*math.pi*n) *math.sin(2*x1*math.pi*n/L))))\n", + "p_per = p*100 # probability of finding particle in percentage\n", + "print(\"Probability of finding particle(ground state)=\"),round(p_per,1),\"%\"\n", + "n=2 #for first excited state\n", + "p = (1/L)*((x2-(L/(2*math.pi*n) *math.sin(2*x2*math.pi*n/L)))- (x1-(L/(2*math.pi*n) *math.sin(2*x1*math.pi*n/L))))\n", + "p_per = p*100 # probability of finding particle in percentage\n", + "print(\"Probability of finding particle(first excited state)=\"),round(p_per,2),\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Probability of finding particle(ground state)= 19.8 %\n", + "Probability of finding particle(first excited state)= 0.65 %\n" + ] + } + ], + "prompt_number": 53 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg61:pg-58" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "L = 1e-8 # length of box in cm\n", + "h = 6.626e-34 # Plank constant in joule-sec\n", + "m = 9.1e-31 # mass of electron in Kg\n", + "\n", + "E1 = (h)**2/(8*m*(L*1e-2)**2) # Calculation of energy of ground state in Joule\n", + "E1_eV = round(E1/1.6e-19 )# Calculation of energy in eV\n", + "E2_eV =2**2*E1_eV # Calculation of energy of first excited state in eV\n", + "del_E = E2_eV - E1_eV # calculation of difference between first state and ground state\n", + "print(\"Energy difference between ground state and first excited state =\"),int(del_E),\"eV\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Energy difference between ground state and first excited state = 114 eV\n" + ] + } + ], + "prompt_number": 54 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg62:pg-59" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "a=1 #let\n", + "n=2 #for second energy state\n", + "x=a/2 #at the center of the box\n", + "W2=sqrt(2/a)*round(sin(n*math.pi*x/a))#wave function of particle in second energy state\n", + "print\"probability of finding particle in interval del_x, p=del_x*(W2)**2= \",W2\n", + "#probability of finding particle in interval del_x is, p=del_x*(W2)**2" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "probability of finding particle in interval del_x, p=del_x*(W2)**2= 0\n" + ] + } + ], + "prompt_number": 55 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg63:pg-59" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "L = 2. # length of box in Angstrom\n", + "x1=1.6000 #lower limit in Angstrom\n", + "x2=1.6001 #upper limit in Angstrom\n", + "n=1 #given\n", + "p = (1/L)*((x2-(L/(2*math.pi*n) *math.sin(2*x2*math.pi*n/L)))- (x1-(L/(2*math.pi*n) *math.sin(2*x1*math.pi*n/L))))\n", + "print\"Probability of finding particle=\",\"{:.2e}\".format(p)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Probability of finding particle= 3.45e-05\n" + ] + } + ], + "prompt_number": 56 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg64:pg-60" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "lamda=1.21*10**-10 #de-Broglie wavelength in meter\n", + "L=lamda/2 #length of a loop in meter\n", + "#since there are 7 loops between the walls of the box \n", + "a=7*L\n", + "print\"Distance between the walls=\", a,\"m\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Distance between the walls= 4.235e-10 m\n" + ] + } + ], + "prompt_number": 57 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg65:pg-60" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "L = 10**-9 # width of potential well in meter\n", + "h = 6.63e-34 # Plank constant in joule-sec\n", + "m = 9.1e-31 # mass of electron in Kg\n", + "n1=1\n", + "n2=2\n", + "n3=3\n", + "lamda1 = 2*L/n1 # Calculation of wavelength\n", + "lamda2 = 2*L/n2 # Calculation of wavelength\n", + "lamda3 = 2*L/n3 # Calculation of wavelength\n", + "E=h**2/(8*m*L**2) # Calculation of energy in Joule\n", + "E=round(E/(1.6*10**-19),2) # Calculation of energy in eV\n", + "E1_eV = n1**2*E # Calculation of energy in eV\n", + "E2_eV = n2**2*E # Calculation of energy in eV\n", + "E3_eV = n3**2*E # Calculation of energy in eV\n", + "print\"For first energy state: wavelength in angstrom & Energy in eV=\",int(lamda1*10**10),\",\",round(E1_eV,2)\n", + "print\"For second energy state: wavelength in angstrom & Energy in eV=\",int(lamda2*10**10),\",\",round(E2_eV,2)\n", + "print\"For third energy state: wavelength in angstrom & Energy in eV=\",round(lamda3*10**10,1),\",\",round(E3_eV,2)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "For first energy state: wavelength in angstrom & Energy in eV= 20 , 0.38\n", + "For second energy state: wavelength in angstrom & Energy in eV= 10 , 1.52\n", + "For third energy state: wavelength in angstrom & Energy in eV= 6.7 , 3.42\n" + ] + } + ], + "prompt_number": 58 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg66:pg-60" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "c=3*10**10 #speed of light in cm/sec\n", + "del_t=2*10**-7#uncertainty in time in sec\n", + "#x is the distance between source and reflecting object\n", + "del_x=(c/2)*del_t\n", + "print\"Uncertainty in distance=\",int(del_x),\"cm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Uncertainty in distance= 3000 cm\n" + ] + } + ], + "prompt_number": 59 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Engineering_Physics_(Volume-2)_by_S.K._Gupta/chapter2_1.ipynb b/Engineering_Physics_(Volume-2)_by_S.K._Gupta/chapter2_1.ipynb new file mode 100644 index 00000000..de4b7a8a --- /dev/null +++ b/Engineering_Physics_(Volume-2)_by_S.K._Gupta/chapter2_1.ipynb @@ -0,0 +1,1050 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:ac6310227f42f13ba4b3bd9417ff991ecd0dd863f2cb872dda0e8a0feb18d473" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter2:X-RAY DIFFRACTION" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg1:pg-70" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "V=25*10**3 #potential difference in Volt\n", + "h=6.63*10**-34 #planck constant in joule-sec\n", + "c=3*10**8 #speed of light in m/sec\n", + "e=1.6*10**-19 #charge of electron in coulomb\n", + "theta=radians(15.8) #glancing angle for NaCl crystal for CuKa line\n", + "d=2.82 #for NaCl\n", + "lamda=2*d*sin(theta) \n", + "print \"wavelength of CuKa line=\",round(lamda,4),\"Angstrom\"\n", + "lamda_min=(h*c/(e*V))*10**10\n", + "print \"wavelength of X-Ray photon at shortest limit=\",round(lamda_min,4),\"Angstrom\"\n", + "theta_1=degrees(math.asin(lamda_min/(2*d)))\n", + "print \"glancing angle for photons at the shortest wavelength limit=\",round(theta_1,2),\"degree\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "wavelength of CuKa line= 1.5357 Angstrom\n", + "wavelength of X-Ray photon at shortest limit= 0.4972 Angstrom\n", + "glancing angle for photons at the shortest wavelength limit= 5.06 degree\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg2:pg-70" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "theta=radians(30) #glancing angle in radians\n", + "d=1.87 #spacing between lattice planes in angstrom\n", + "n=2 #for second order reflection\n", + "lamda=2*d*sin(theta)/n\n", + "print \"wavelength of X-Rays=\",lamda,\"Angstrom\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "wavelength of X-Rays= 0.935 Angstrom\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg3:pg-70" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "lamda=0.36*10**-8 #wavelength in cm\n", + "theta=radians(4.8)#glancing angle in radians\n", + "n=1 #for first order diffraction\n", + "d=n*lamda/(2*sin(theta))\n", + "print \"interplanar separation of atomic planes in crystal=\",\"{:.2e}\".format(d),\"cm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "interplanar separation of atomic planes in crystal= 2.15e-08 cm\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg4:pg-71" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "lamda=2.6*10**-10 #wavelength in meter\n", + "theta=radians(20) #in radians\n", + "n=2 #for second order diffraction\n", + "d=n*lamda/(2*sin(theta))\n", + "print \"spacing constant of the crystal=\",round(d*10**10,2),\"Angstrom\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "spacing constant of the crystal= 7.6 Angstrom\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg5:pg-71" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "d=2.82*10**-10 #spacing in meter\n", + "n=2 #for second order\n", + "sin_theta=1 #maximum value of sin(theta)\n", + "lamda_max=2*d*sin_theta/n\n", + "print \"longest wavelength=\",lamda_max*10**10,\"Angstrom\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "longest wavelength= 2.82 Angstrom\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg6:pg-71" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "lamda=0.842 #wavelength in angstrom\n", + "theta_1=8+(35./60) #1' = (1/60)\u00ba = 0.01666667\u00ba\n", + "theta_3=math.asin(round(3*sin(radians(theta_1)),2))\n", + "print \"glancing angle for 3rd order reflection=\",round(math.degrees(theta_3),1),\"degrees\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "glancing angle for 3rd order reflection= 26.7 degrees\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg7:pg-71" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "lamda=0.97 #wavelength of first X-ray beam in angstrom\n", + "theta=radians(60) #angle of reflection in radians\n", + "n=3 #for third order reflection\n", + "d=n*lamda/(2*sin(theta))\n", + "n_1=1 #for first order reflection\n", + "theta_1=radians(30) #angle of reflection in radians\n", + "lamda_1=2*d*sin(theta_1)\n", + "print \"wavelength of the second X-ray beam=\",round(lamda_1,2),\"Angstrom\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "wavelength of the second X-ray beam= 1.68 Angstrom\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg8:pg-72" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "lamda=0.30 #wavelength in angstrom\n", + "d=0.5 #lattice spacing in angstrom\n", + "n=2 #for second order diffraction\n", + "theta=math.asin(n*lamda/(2*d))\n", + "print \"For second order maxima, angle=\",round(math.degrees(theta),2),\"degrees\"\n", + "n=3 #for third order diffraction\n", + "theta=math.asin(n*lamda/(2*d))\n", + "print \"For third order maxima, angle=\",round(math.degrees(theta),2),\"degrees\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "For second order maxima, angle= 36.87 degrees\n", + "For third order maxima, angle= 64.16 degrees\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg9:pg-72" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "d=2.82*10**-8 #lattice spacing in cm \n", + "c=3*10**10 #speed of light in cm/sec\n", + "e=1.6*10**-19 #charge on electron in coulomb\n", + "v=9045 #voltage in volt\n", + "theta=radians(14)#angle in radians\n", + "n=1 #first order\n", + "lamda=2*d*sin(theta)/n\n", + "h=(e*v*lamda/c)*10**7 #since 1 joule=10**7 erg\n", + "print \"h=\",\"{:.2e}\".format(h),\"erg-sec\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "h= 6.58e-27 erg-sec\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg10:pg-72" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "do=2.82 #lattice spacing in angstrom\n", + "theta=radians(10) #angle in radians\n", + "lamda=2*do*round(sin(theta),4)\n", + "print \"wavelength=\",round(lamda,4),\"Angstrom\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "wavelength= 0.9791 Angstrom\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg11:pg-72" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "d=0.4086*10**-10 #lattice spacing in meter\n", + "h=6.6*10**-34 #planck constant in joule-sec\n", + "m=9.1*10**-31 #mass of electron in Kg\n", + "n=1 #first order\n", + "theta=radians(65) #glancing angle in radians\n", + "lamda=2*d*sin(theta)/n\n", + "print \"wavelength=\",\"{:.3e}\".format(lamda),\"m\"\n", + "v=h/(m*lamda)\n", + "print \"velocity of electron=\",\"{:.3e}\".format(v),\"m/sec\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "wavelength= 7.406e-11 m\n", + "velocity of electron= 9.793e+06 m/sec\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg12:pg-73" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "h=6.62*10**-34 #planck constant in joule-sec\n", + "e=1.6*10**-19 #charge on electron in coulomb\n", + "m=9*10**-31 #mass of electron in Kg\n", + "v=344 #voltage in volt\n", + "n=1 #first order\n", + "theta=radians(60)#glancing angle in radians\n", + "lamda=h/sqrt(2*m*e*v)\n", + "d=n*lamda/(2*sin(theta))\n", + "print \"spacing of the crystal=\",round(d*10**10,2),\"Angstrom\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "spacing of the crystal= 0.38 Angstrom\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg13:pg-73" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "#given that\n", + "lamda=1.32*10**-10 #wavelength in meter\n", + "theta_deg=9 #angle fraction in degree\n", + "theta_min=30 #angle fraction in minute\n", + "theta =theta_deg+(theta_min/60.) # Total angle\n", + "for n in range(1,5):\n", + " d = lamda/(n*2*math.sin(theta*math.pi/180)) # Inter layer spacing\n", + " print \"If order is %d then spacing is\"%(n),\"{:.2e}\".format(d),\"meter\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "If order is 1 then spacing is 4.00e-10 meter\n", + "If order is 2 then spacing is 2.00e-10 meter\n", + "If order is 3 then spacing is 1.33e-10 meter\n", + "If order is 4 then spacing is 1.00e-10 meter\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg14:pg-74" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "# given that\n", + "theta1_deg = 5 # Absolut degree part of angle for first angle\n", + "theta1_min = 23# remainder minute part of angle for first angle\n", + "theta2_deg = 7 # Absolut degree part of angle for second angle\n", + "theta2_min = 37# remainder minute part of angle for second angle\n", + "theta3_deg = 9 # Absolut degree part of angle for third angle\n", + "theta3_min = 22# remainder minute part of angle for third angle\n", + "\n", + "val1 = math.sin((theta1_deg+ theta1_min/60.)*math.pi/180)# Sin value for first angle\n", + "val2 = math.sin((theta2_deg+ theta2_min/60.)*math.pi/180) #Sin value for second angle\n", + "val3 = math.sin((theta3_deg+ theta3_min/60.)*math.pi/180)#Sin value for third angle\n", + "ratio_21 = val2/val1\n", + "ratio_31 = val3/val1\n", + "print \"Interatomic layer separation ratios in crystal are as 1 : %f : %f\"%(ratio_21,ratio_31)\n", + "print \"Above relation shows that crystal has a simple cubic crystal structure.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Interatomic layer separation ratios in crystal are as 1 : 1.412775 : 1.734750\n", + "Above relation shows that crystal has a simple cubic crystal structure.\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg15:pg-82" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "h=6.63*10**-34 #planck constant in joule-sec\n", + "c=3*10**8 #speed of light in m/sec\n", + "mo=9.1*10**-31 #mass of electron in Kg\n", + "theta=radians(180)#scattering angle in radians\n", + "d_lamda=h*(1-math.cos(theta))/(mo*c)\n", + "print \"change in wavelength of photon=\",round(d_lamda*10**10,4),\"Angstrom\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "change in wavelength of photon= 0.0486 Angstrom\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg16:pg-82" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "#given that\n", + "E=100. # Energy of X ray beam in KeV\n", + "theta=30 # Scattering angle in degree\n", + "mo=9.1*10**-31 # mass of electron in kg\n", + "c=3*10**8 # Speed of light in m/s\n", + "E_rest=(mo*c**2)/(1.6e-19*1e3) # Rest mass energy in KeV\n", + "k=(1/E)+ ((1-math.cos(radians(theta)))/(E_rest))\n", + "k=int(k*10000)*10**-4\n", + "del_e=E-1/k # Energy of recoiled electron\n", + "print \"Energy of recoiled electrons is \",round(del_e,2),\"KeV\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Energy of recoiled electrons is 1.96 KeV\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg17:pg-82" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "#given that\n", + "lamda=1 # wavelength in angstrom\n", + "h=6.63*10**-34 # Planck's constant in joule-sec\n", + "mo=9.1*10**-31 # mass of electron in kg\n", + "c=3*10**8 # speed of light in m/sec\n", + "theta=90 # scattering angle in degree\n", + "d_lambda=h*(1-math.cos(radians(90)))/(mo*c) # calculation of compton shift \n", + "print \"compton shift is \",round(d_lambda*1e10,4),\"Angstrom\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "compton shift is 0.0243 Angstrom\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg18:pg-83" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "#given that\n", + "lamda=0.015 #wavelength in angstrom\n", + "h=6.63*10**-34 #Planks constant in joule-sec\n", + "mo=9.1*10**-31 #mass of electron in kg\n", + "c=3*10**8 #speed of light in m/sec\n", + "theta=60 #scattering angle in degree\n", + "d_lambda=h*(1-math.cos(theta*math.pi/180))*1e10/(mo*c) \n", + "lambda_n=lamda+d_lambda\n", + "print \"Wavelength of the scattered X-ray is \",round(lambda_n,3),\"Angstrom\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Wavelength of the scattered X-ray is 0.027 Angstrom\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg19:pg-83" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "#given that\n", + "lamda=1 # wavelength in angstrom\n", + "h=6.63*10**-34 # Planck's constant in joule-sec\n", + "mo=9.1*10**-31 # mass of electron in kg\n", + "c=3*10**8 # speed of light in m/sec\n", + "theta=90 # scattering angle in degree\n", + "d_lambda= h*(1-math.cos(radians(90)))*1e10/(mo*c) # calculation of wavelength shift in angstrom\n", + "lambda_n=lamda+d_lambda # Calculation of wavelength of scattered beam in angstrom\n", + "K_E=h*c*(lambda_n-lamda)*1e10/(1.6e-19*lambda_n*lamda)# Calculation of K.E of recoiled electron in eV\n", + "phi=math.atan(round((lamda/lambda_n),2))# calculation of Direction of the recoiled electron\n", + "print \"Wavelength of the scattered beam is \",round(lambda_n,4),\"Angstrom\"\n", + "print \"Kinetic Energy imparted to the recoiled electron is \",round(K_E),\"eV\"\n", + "print \"Direction of the recoiled electron is \",round(degrees(phi),1),\"degree\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Wavelength of the scattered beam is 1.0243 Angstrom\n", + "Kinetic Energy imparted to the recoiled electron is 295.0 eV\n", + "Direction of the recoiled electron is 44.4 degree\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg20:pg-84" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "#given that\n", + "lamda=1 # wavelength in angstrom\n", + "h=6.63*10**-34 # Planck's constant in joule-sec\n", + "mo=9.1*10**-31 # mass of electron in kg\n", + "c=3*10**8 # speed of light in m/sec\n", + "theta=90 # scattering angle in degree\n", + "d_lambda= h*(1-math.cos(radians(90)))*1e10/(mo*c) # calculation of compton shift in angstrom\n", + "lambda_n=lamda+d_lambda # Calculation of wavelength of scattered beam in angstrom\n", + "K_E=h*c*(lambda_n-lamda)*1e10/(1.6e-19*lambda_n*lamda)# Calculation of K.E of recoiled electron in eV\n", + "print \"Compton shift is \",round(d_lambda,4),\"Angstrom\"\n", + "print \"Kinetic Energy imparted to the recoiled electron is \",round(K_E),\"eV\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Compton shift is 0.0243 Angstrom\n", + "Kinetic Energy imparted to the recoiled electron is 295.0 eV\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg21:pg-84" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "h=6.63*10**-34 # Planck's constant in joule-sec\n", + "mo=9.1*10**-31 # mass of electron in kg\n", + "c=3*10**8 # speed of light in m/sec\n", + "E=0.88*10**6 #energy of gamma-rays in eV\n", + "theta=180 #scattering angle in degree for maximum energy of recoiled electron\n", + "lamda=h*c*10**10/(E*1.6*10**-19)\n", + "d_lamda_max=h*(1-math.cos(radians(theta)))*1e10/(mo*c)\n", + "lamda_n=lamda+d_lamda_max\n", + "K_E_max=h*c*d_lamda_max*1e10/(1.6e-19*lamda_n*lamda)\n", + "print \"Maximum energy of compton recoil electrons is \",round(K_E_max*10**-6,3),\"MeV\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Maximum energy of compton recoil electrons is 0.682 MeV\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg22:pg-85" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "h=6.62*10**-34 # Planck's constant in joule-sec\n", + "mo=9.0*10**-31 # mass of electron in kg\n", + "c=3*10**8 # speed of light in m/sec\n", + "theta=90 # scattering angle in degree \n", + "lamda=h*(1-math.cos(radians(theta)))*1e10/(mo*c)\n", + "d_lamda=lamda # compton shift \n", + "E=h*c/(round(lamda,4)*1e-10)\n", + "print \"Wavelength of incident photon is \",round(lamda,4),\"Angstrom\"\n", + "print \"Energy of incident photon is \",\"{:.3e}\".format(E),\"joule\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Wavelength of incident photon is 0.0245 Angstrom\n", + "Energy of incident photon is 8.106e-14 joule\n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg23:pg-85" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "h=6.63*10**-34 # Planck's constant in joule-sec\n", + "mo=9.1*10**-31 # mass of electron in kg\n", + "c=3*10**8 # speed of light in m/sec\n", + "theta=90 # scattering angle in degree \n", + "d_lamda=h*(1-math.cos(radians(theta)))*1e10/(mo*c)\n", + "print \"Percentage change in energy when photon is:\"\n", + "#(a) for microwave photon\n", + "lamda=3*10**8 #wavelength of microwave photon in Angstrom\n", + "energy_change=d_lamda*100/(lamda+d_lamda)\n", + "print \"A microwave photon= \",\"{:.1e}\".format(energy_change),\"%\"\n", + "\n", + "#(b) for visible light photon\n", + "lamda=5000 #wavelength of visible light photon in Angstrom\n", + "energy_change=d_lamda*100/(lamda+d_lamda)\n", + "print \"A visible light photon= \",\"{:.2e}\".format(energy_change),\"%\"\n", + "\n", + "#(c) for X-ray photon\n", + "lamda=1 #wavelength of X-ray photon in Angstrom\n", + "energy_change=d_lamda*100/(lamda+d_lamda)\n", + "print \"An X-ray photon= \",round(energy_change,1),\"%\"\n", + "\n", + "#(d) for gamma-ray photon\n", + "lamda=0.0124 #wavelength of gamma-ray photon in Angstrom\n", + "energy_change=d_lamda*100/(lamda+d_lamda)\n", + "print \"A gamma-ray photon= \",int(energy_change),\"%\"\n", + "print \"Hence, the compton effect is dominant only in the gamma-ray region and shorter X-ray region.It is not observable in the visible region and microwave region\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Percentage change in energy when photon is:\n", + "A microwave photon= 8.1e-09 %\n", + "A visible light photon= 4.86e-04 %\n", + "An X-ray photon= 2.4 %\n", + "A gamma-ray photon= 66 %\n", + "Hence, the compton effect is dominant only in the gamma-ray region and shorter X-ray region.It is not observable in the visible region and microwave region\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg24:pg-86" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "lamda=2 # wavelength in angstrom\n", + "h=6.62*10**-34 # Planck's constant in joule-sec\n", + "mo=9.1*10**-31 # mass of electron in kg\n", + "c=3*10**8 # speed of light in m/sec\n", + "theta=45 # scattering angle in degree\n", + "d_lamda=h*(1-math.cos(radians(theta)))*1e10/(mo*c) \n", + "lamda_n=lamda+d_lamda \n", + "f=d_lamda/lamda_n # Calculation of fraction of energy lost by photon \n", + "print \"Fraction of energy lost by photon is \",round(f,4)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Fraction of energy lost by photon is 0.0035\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg25:pg-87" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "C_W=0.0242 #compton wavelength of electron in Angstrom\n", + "theta=45 # scattering angle in degree\n", + "d_lamda=C_W*(1-math.cos(radians(theta)))\n", + "lamda= d_lamda\n", + "print \"Wavelength= \",round(lamda,3),\"Angstrom\"\n", + "#answer is incomplete in book as only wavelength is calculated and no region is specified\n", + "print \"Hence, such a photon lie in the Gamma-ray region of electromagnetic spectrum.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Wavelength= 0.007 Angstrom\n", + "Hence, such a photon lie in the Gamma-ray region of electromagnetic spectrum.\n" + ] + } + ], + "prompt_number": 29 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg26:pg-87" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "h=6.6*10**-34 # Planck's constant in joule-sec\n", + "mo=9.1*10**-31 # mass of electron in kg\n", + "c=3*10**8 # speed of light in m/sec\n", + "E=510*10**3 # energy of gamma-rays in eV\n", + "theta=90 # scattering angle in degree \n", + "lamda=h*c/(E*1.6*10**-19)\n", + "d_lamda=h*(1-math.cos(radians(theta)))/(mo*c)\n", + "lamda_n=lamda+d_lamda\n", + "Er=h*c*d_lamda/(lamda_n*lamda)\n", + "phi=math.atan(lamda/lamda_n)\n", + "print \"Wavelength of scattered radiation is \",\"{:.3e}\".format(lamda_n),\"meter\"\n", + "print \"Energy of recoil electron is \",\"{:.3e}\".format(Er),\"joule\"\n", + "print \"Direction of the recoil electron is \",round(degrees(phi),2),\"degree\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Wavelength of scattered radiation is 4.844e-12 meter\n", + "Energy of recoil electron is 4.073e-14 joule\n", + "Direction of the recoil electron is 26.61 degree\n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg27:pg-88" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "h=6.63*10**-34 # Planck's constant in joule-sec\n", + "mo=9.1*10**-31 # mass of electron in kg\n", + "c=3*10**8 # speed of light in m/sec\n", + "E=510*10**3 # energy of gamma-rays in eV\n", + "theta=90 # scattering angle in degree \n", + "lamda=h*c/(E*1.6*10**-19)\n", + "d_lamda=h*(1-math.cos(radians(theta)))/(mo*c)\n", + "lamda_n=lamda+d_lamda\n", + "print \"Wavelength of scattered radiation is \",round(lamda_n*10**10,4),\"Angstrom\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Wavelength of scattered radiation is 0.0487 Angstrom\n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg28:pg-88" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "h=6.62*10**-34 # Planck's constant in joule-sec\n", + "mo=9.1*10**-31 # mass of electron in kg\n", + "c=3*10**8 # speed of light in m/sec\n", + "theta=180 # scattering angle in degree for minimum energy of incident photon\n", + "lamda_max=h*(1-math.cos(radians(theta)))/(mo*c)\n", + "E_min=h*c/lamda_max\n", + "print \"Minimum energy of incident photon is \",int(round(E_min/(1.6*10**-16))),\"KeV\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Minimum energy of incident photon is 256 KeV\n" + ] + } + ], + "prompt_number": 26 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Engineering_Physics_(Volume-2)_by_S.K._Gupta/chapter3_1.ipynb b/Engineering_Physics_(Volume-2)_by_S.K._Gupta/chapter3_1.ipynb new file mode 100644 index 00000000..210e8592 --- /dev/null +++ b/Engineering_Physics_(Volume-2)_by_S.K._Gupta/chapter3_1.ipynb @@ -0,0 +1,685 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:b69efe67845471267465a642fdefaa06de6ab058dfd36d1fd2799087ad02e07e" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter3:DIELECTRIC PROPERTIES OF MATERIALS" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg1:pg-119" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "Eg=6.0 #dielectric constant of glass plate\n", + "dg=0.25 #thickness of glass plate in mm\n", + "Ep=3.0 #dielectric constant of plastic film\n", + "dp=0.1 #thickness of plastic film in mm\n", + "Eo=8.85e-12 #permittivity of free space in F/m \n", + "A=1 #let surface area be 1\n", + "Cg=Eg*Eo*A/dg\n", + "Cp=Ep*Eo*A/dp\n", + "ratio=Cg/Cp\n", + "print\"Cg = \",ratio,\"Cp\"\n", + "print\"Since Cp>Cg,the plastic film filled capacitor holds more charge than the glass plate filled capacitor\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Cg = 0.8 Cp\n", + "Since Cp>Cg,the plastic film filled capacitor holds more charge than the glass plate filled capacitor\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg2:pg-120" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "Er=2.8 #dielectric constant of a dielectric material\n", + "D=3e-8 #magnitude of electric displacement vector in C/m**2\n", + "p=(Er-1)*D/Er\n", + "print\"Polarization is \",round(p,10),\"C/m**2\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Polarization is 1.93e-08 C/m**2\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg3:pg-120" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "E=1000 #electric field in V/m\n", + "p=4.3e-8 #polarization in C/m**2\n", + "Eo=8.85e-12#permittivity of free space in F/m \n", + "Er=1+(p/(Eo*E))\n", + "print\"Relative permittivity of NaCl is \",round(Er,2)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Relative permittivity of NaCl is 5.86\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg4:pg-120" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "Er=1.000074 #dielectric constant of helium \n", + "Eo=8.85e-12 #permittivity of free space in F/m (in book F/m2 is printed which is wrong)\n", + "E=100 #electric field in V/m\n", + "Na=6e23 #Avogadro number\n", + "V=22.4 #volume occupied by 1gm atom of gas at NTP in litres\n", + "N=Na/(V*1e-3)\n", + "p=Eo*(Er-1)*E\n", + "P=p/N\n", + "print\"Induced dipole moment is \",round(P,42),\"Cm\"#answer in book is in different form as 24.42e-40 Cm" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Induced dipole moment is 2.445e-39 Cm\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg5:pg-121" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "Epsilon=1.46e-10 #permittivity of diamond in C**2/Nm**2\n", + "Eo=8.86e-12 #permittivity of free space in C**2/Nm**2\n", + "Er=Epsilon/Eo \n", + "X=Eo*(Er-1)\n", + "print\"Dielectric constant is \",round(Er,2)\n", + "print\"Electrical susceptibility is \",round(X,12),\"C**2/Nm**2\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Dielectric constant is 16.48\n", + "Electrical susceptibility is 1.37e-10 C**2/Nm**2\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg6:pg-121" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "Xe=35.4e-12 #electrical susceptibility in C**2/Nm**2\n", + "Eo=8.85e-12 #permittivity of free space in C**2/Nm**2 \n", + "K=1+(Xe/Eo)\n", + "Epsilon=Eo*K\n", + "print\"Dielectric constant is \",int(K)\n", + "print\"Permittivity of the material is \",Epsilon,\"C**2/Nm**2\"\n", + "#answer in book is in different form as 44.25e-12 C**2/Nm**2" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Dielectric constant is 5\n", + "Permittivity of the material is 4.425e-11 C**2/Nm**2\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg7:pg-121" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "Vo=60 #applied potential difference in volt\n", + "V=30 #reduced potential difference in volt\n", + "K=Vo/V\n", + "print\"Dielectric constant of the liquid is \",K" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Dielectric constant of the liquid is 2\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg8:pg-121" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "Vo=100 #potential difference in volts\n", + "t=0.3 #thickness of insulator in cm\n", + "A=100 #area in cm**2\n", + "d=1 #separation between plates in cm\n", + "K=7 #dielectric constant \n", + "Eo=8.9e-12 #permittivity of free space in C**2/Nm**2\n", + "E_o=Vo/(d*1e-2)\n", + "E=E_o/K\n", + "D=K*Eo*E\n", + "p=(K-1)*Eo*E\n", + "print\"E = \",\"{:.2e}\".format(E),\"Volt/m\"\n", + "print\"D = \",D,\"C/m**2\"\n", + "print\"p = \",round(p,9),\"C/m**2\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "E = 1.43e+03 Volt/m\n", + "D = 8.9e-08 C/m**2\n", + "p = 7.6e-08 C/m**2\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg9:pg-122" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "E=6e4 #electric field in V/m\n", + "K=1.000134 #dielectric constant of neon\n", + "Eo=8.9e-12 #permittivity of free space in F/m\n", + "Na=6e23 #Avogadro number\n", + "V=22.4 #volume occupied by 1gm atom of gas at NTP in litres\n", + "p=Eo*(K-1)*E\n", + "N=Na/(V*1e-3)\n", + "P=p/N\n", + "alpha=P/(Eo*E)\n", + "print\"Induced dipole moment is\",round(P,38),\"Cm\"\n", + "print\"Atomic polarizability of neon is \",round(alpha,32),\"m**3\"\n", + "#answer in book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Induced dipole moment is 2.67e-36 Cm\n", + "Atomic polarizability of neon is 5e-30 m**3\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg11:pg-123" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "Er=1.0024 #dielectric constant of argon atom\n", + "N=2.7e25 #number of atoms per cubic meter\n", + "Eo=8.85e-12 #permittivity of free space in F/m\n", + "alpha_e=Eo*(Er-1)/N\n", + "print\"Electronic polarizability is \",round(alpha_e,41),\"Fm**2\"\n", + "#answer is wrong in book" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Electronic polarizability is 7.9e-40 Fm**2\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg12:pg-123" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "N=9.8e26 #number of atoms in volume of one cubic meter of hydrogen gas\n", + "Eo=8.85e-12 #permittivity of free space in F/m\n", + "ao=0.53e-10 #radius of hydrogen atom in meter\n", + "alpha=4*math.pi*Eo*ao**3\n", + "Er=1+(4*math.pi*N*ao**3)\n", + "print\"Polarizability is \",round(alpha,43),\"Fm**2\"\n", + "print\"Relative permittivity is \",round(Er,4)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Polarizability is 1.66e-41 Fm**2\n", + "Relative permittivity is 1.0018\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg13:pg-124" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "alpha_300=2.5e-39 # total polarizability in C**2m/N at 300 K\n", + "alpha_400=2.0e-39 # total polarizability in C**2m/N at 400 K\n", + "T1 =300 # temperature in Kelvin\n", + "T2 =400 # temperature in Kelvin\n", + "beta=(alpha_300-alpha_400)*(T1*T2/(T2-T1))\n", + "alpha_def_300=alpha_300 - beta/300\n", + "alpha_oriant_300=beta/300\n", + "alpha_oriant_400=beta/400\n", + "print\"Deformational Polarizability is \",alpha_def_300,\"C**2mN**-1\"\n", + "print\"Orientational Polarizability at %d K is \"%T1,alpha_oriant_300,\"C**2mN**-1\"\n", + "print\"Orientational Polarizability at %d K is \"%T2,alpha_oriant_400,\"C**2mN**-1\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Deformational Polarizability is 5e-40 C**2mN**-1\n", + "Orientational Polarizability at 300 K is 2e-39 C**2mN**-1\n", + "Orientational Polarizability at 400 K is 1.5e-39 C**2mN**-1\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg14:pg-132" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "m=32 # Atomic weight of sulphur\n", + "d=2.08 # Density in g/cm**3\n", + "alpha_e=3.28e-40 # Electronic polarizability in Fm**2\n", + "Na=6.023e23 # Avogadro Number\n", + "Eo=8.85e-12 # Permittivity of free space in F/m\n", + "N=Na*d*1e6/m \n", + "k=N*alpha_e/(3*Eo)\n", + "epsilon_r = (1+ k*2)/(1-k)# Calculation of relative permittivity\n", + "print\"Relative dielectric constant is \",round(epsilon_r,1)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Relative dielectric constant is 3.8\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg15:pg-132" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "n=1.5 # Refractive index\n", + "Er=5.6 # Static dielectric constant\n", + "per=(1-((n**2-1)/(n**2+2))*(Er+2)/(Er-1))*100 # Pecentage of ionic polarizability\n", + "print\"Percentage of ionic polarizability is \",round(per,1),\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Percentage of ionic polarizability is 51.4 %\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg16:pg-133" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "n=sqrt(2.69) # Refractive index\n", + "Er=4.94 # Static dielectric constant\n", + "k1=(Er-1)/(Er+2)\n", + "k2=(n**2-1)/(n**2+2)\n", + "ratio=1/round(((k1/k2)-1),3) \n", + "print\"Ratio of electronic to ionic polarizability is \",round(ratio,3)\n", + "#in book ai/ae is mentioned instead of ae/ai in final answer which is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Ratio of electronic to ionic polarizability is 1.736\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg17:pg-133" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "Er=6.75 #dielectric constant of glass\n", + "n=1.5 #refractive index of glass\n", + "f=1e9 #frequency in Hz\n", + "per=(Er-n**2)*100/(Er-1)\n", + "print\"Percentage attributed to ionic polarizability is \",round(per,1),\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Percentage attributed to ionic polarizability is 78.3 %\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg18:pg-142" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "t=5.5e-3 #thickness of quartz crystal plate in meter\n", + "p=2.65e3 #density of quartz crystal in Kg/m**3\n", + "Y=8e10 #Young's modulus of quartz in N/m**2 (value is wrong in question in book)\n", + "m=1 \n", + "f=m*sqrt(Y/p)/(2*t)\n", + "print\"Frequency is \",int(f*1e-3),\"KHz\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Frequency is 499 KHz\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg19:pg-148" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import cmath\n", + "Er=4.36 #real part of dielectric constant of bakelite\n", + "N=4e28 #number of atoms per cubic meter\n", + "tan_d=2.8e-2#loss tangent at 1 MHz freuqency\n", + "Eo=8.853e-12#permittivity of free space in F/m\n", + "alpha=(3*Eo/N)*(Er*(1-(1j*tan_d))-1)/(Er*(1-(1j*tan_d))+2)\n", + "x=round(alpha.real*1e40,1)\n", + "y=round(alpha.imag*1e40,2)\n", + "alpha=complex(x,y)\n", + "print\"Complex polarizability is \",alpha*1e-40,\"Fm**2\"\n", + "#in book answer is in different form and as (3.5-0.06i)*10**-40\n", + "#in book unit of answer is not mentioned" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Complex polarizability is (3.5e-40-6e-42j) Fm**2\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg20:pg-149" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "t=18e-6 # Relaxation time in second\n", + "Er_d=1 # let real part of dielectric constant be 1\n", + "Er_dd=1 # let imaginary part of dielectric constant be 1\n", + "f=1/(2*math.pi*t) # Calculation of frequency\n", + "delta=math.atan(Er_dd/Er_d)\n", + "phi=90-(delta*180/math.pi) # Calculation of phase difference\n", + "print\"Frequency is \",round(f/1e3,1),\"KHz\"\n", + "print\"Phase difference between current and voltage is %d degree\"%(phi)\n", + "print\"Current leads the voltage \"#this part is not mentioned in answer in book" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Frequency is 8.8 KHz\n", + "Phase difference between current and voltage is 45 degree\n", + "Current leads the voltage \n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Engineering_Physics_(Volume-2)_by_S.K._Gupta/chapter4_1.ipynb b/Engineering_Physics_(Volume-2)_by_S.K._Gupta/chapter4_1.ipynb new file mode 100644 index 00000000..3b5fccb8 --- /dev/null +++ b/Engineering_Physics_(Volume-2)_by_S.K._Gupta/chapter4_1.ipynb @@ -0,0 +1,441 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:3eb552fb3c1a7764bbfbba83330f0d6a18ee201068631d5f4e4579961984a5eb" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter4:MAGNETIC PROPERTIES OF MATERIALS" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg1:pg-153" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "phi_B=2.4e-5 #magnetic flux in weber\n", + "A=0.2 #cross-sectional area in cm**2\n", + "H=1600 #magnetising field in A/m\n", + "mu_o=4*round(math.pi,2)*1e-7 #absolute permeability of air in N/A**2\n", + "B=phi_B/(A*1e-4)\n", + "mu=B/H\n", + "Xm=mu/mu_o-1\n", + "print\"Magnetic permeability of iron bar is \",\"{:.1e}\".format(mu),\"N/A**2\"\n", + "print\"Magnetic susceptibility of iron bar is \",round(Xm,2)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Magnetic permeability of iron bar is 7.5e-04 N/A**2\n", + "Magnetic susceptibility of iron bar is 596.13\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg2:pg-154" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from sympy import *\n", + "Xm=948e-11 #magnetic susceptibility of a medium\n", + "pi=Symbol('pi')\n", + "mu_o=4*pi*1e-7 #absolute permeability of air in H/m\n", + "mu_r=1+Xm\n", + "mu=int(mu_r)*mu_o\n", + "print\"Relative Permeability is =\",mu_r,\"=\",int(mu_r),\"or >\",int(mu_r)\n", + "print\" Relative permeability is slightly greater than one.\"\n", + "print\"Permeability is =\",mu,\"H/m\" #answer in book is 4*(pi)*1e-7 H/m" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Relative Permeability is = 1.00000000948 = 1 or > 1\n", + " Relative permeability is slightly greater than one.\n", + "Permeability is = 4.0e-7*pi H/m\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg3:pg-154" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "mu_r=1200 #relative permeability of iron rod\n", + "n=5 #number of turns per cm\n", + "i=0.5 #current in ampere\n", + "V=1e-3 #volume of iron rod in m**3\n", + "I=(mu_r-1)*(n*1e2)*i\n", + "M=I*V\n", + "print\"Magnetic moment is \",\"{:.0e}\".format(M),\"Am**2\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Magnetic moment is 3e+02 Am**2\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg4:pg-155" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "mu_r=100 #relative permeability of iron rod\n", + "n=300 #number of turns per meter\n", + "i=0.5 #current in ampere\n", + "D=10 #diameter of iron rod in mm\n", + "r=D/2 #radius of iron rod in mm\n", + "l=2 #length of iron rod in meter\n", + "I=(mu_r-1)*n*i\n", + "V=round(math.pi,2)*(r*1e-3)**2*l\n", + "M=I*V\n", + "print\"Magnetic moment is \",round(M,3),\"Am**2\"\n", + "#answer in book is wrong as the value of l is taken wrong in calcultion. " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Magnetic moment is 2.331 Am**2\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg5:pg-163" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "n=1e29 #number of atoms per unit volume in atoms/m**3\n", + "Pm=1.8e-23 #magnetic moment of one atom in A-m**2\n", + "K=1.38e-23 #Boltzmann's constant in J/K\n", + "T=300 #temperature in Kelvin\n", + "mu_o=4*round(math.pi,2)*10**-7 #absolute permeability of air in N/A**2\n", + "B=0.1 #magnetic flux density in weber/m**2\n", + "A=1 #cross-section area in cm**2\n", + "l=10 #length of iron bar in cm\n", + "Xm=mu_o*n*Pm**2/(3*K*T) #magnetic susceptibility of iron bar\n", + "P_m=Pm**2*B/(3*K*T) #mean dipole moment of an iron atom in A-m**2\n", + "V=(A*1e-4)*(l*1e-2) #volume of iron bar in m**3\n", + "n_o_a=V*n \n", + "dm=n_o_a*P_m #dipole moment of the iron bar \n", + "I=Pm*n \n", + "m=I*V\n", + "print\"Magnetic Susceptibility is \",\"{:.3e}\".format(Xm)\n", + "print\"Dipole moment is \",\"{:.3e}\".format(dm),\"Am**2\"\n", + "print\"Magnetisation is \",\"{:.1e}\".format(I),\"A/m\"\n", + "print\"Magnetic moment is \",int(m),\"Am**2\"#this answer is wrong in book" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Magnetic Susceptibility is 3.277e-03\n", + "Dipole moment is 2.609e-03 Am**2\n", + "Magnetisation is 1.8e+06 A/m\n", + "Magnetic moment is 18 Am**2\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg6:pg-169" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "H=5e3 #Coercivity of bar magnet in ampere/m \n", + "l=10 #length of solenoid in cm\n", + "n=50 #number of turns in solenoid\n", + "i=H*(l*1e-2)/n\n", + "print\"Current is \",int(i),\"amp\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Current is 10 amp\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg8:pg-170" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "a=250 #area of B-H loop in J/m**3\n", + "f=50 #frequency of a.c. in Hz\n", + "m=9.0 #mass of iron core in Kg\n", + "p=7500 #density of iron in Kg/m**3\n", + "V=m/p\n", + "n=50*60*60\n", + "E=n*V*a\n", + "print\"Hysteresis loss of energy E per hour is \",\"{:.1e}\".format(E),\"J\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Hysteresis loss of energy E per hour is 5.4e+04 J\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg10:pg-170" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "A=0.785e-4 #cross-sectional area of rowland ring in m**2\n", + "Ri=5.0 #inner radius in cm\n", + "Ro=6.0 #outer radius in cm\n", + "N=400 #number of turns of wire\n", + "Bo=2e-4 #magnetic flux density in weber/m**2\n", + "mu_o=4*math.pi*10**-7 #absolute permeability of air in N/A**2\n", + "Ns=50 #number of turns in secondary coil\n", + "R=8.0 #resistance in ohm\n", + "B1=800*Bo #magnetic flux density in weber/m**2\n", + "l=2*math.pi*(Ri+Ro)*1e-2/2\n", + "i=Bo*l/(mu_o*N)\n", + "q=Ns*B1*A/R\n", + "print\"Required Current is \",i,\"amp\"\n", + "print\"Charge passed is \",q,\"coulomb\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Required Current is 0.1375 amp\n", + "Charge passed is 7.85e-05 coulomb\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg11:pg-171" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "mu_r=400 #relative permeability of iron ring\n", + "r=0.1 #mean radius of iron ring in meter\n", + "A=5e-4 #cross-sectional area of iron ring in m**2\n", + "n=1000 #number of turns of wire\n", + "i=4 #current in ampere\n", + "mu_o=4*math.pi*10**-7 #absolute permeability of air in N/A**2\n", + "B=mu_o*mu_r*n*i/(2*math.pi*r)\n", + "phi=B*A \n", + "print\"Flux in the ring is \",\"{:.2e}\".format(phi),\"weber\"\n", + "n_o=500 #number of turns in secondary coil per meter\n", + "R=10 #resistance in ohm\n", + "q=2*n_o*A*B/R\n", + "print\"Electricity discharged through the secondary coil is \",q,\"coulomb\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Flux in the ring is 1.60e-03 weber\n", + "Electricity discharged through the secondary coil is 0.16 coulomb\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg12:pg-171" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "m=12 #weight of the iron core in Kg\n", + "p=7.5 #density of iron core in gm/cc\n", + "f=50 #frequency in cycles/sec\n", + "a=3000 #area of hysteresis loop in ergs/cm**3 (unit is misprinted in question in book)\n", + "V=(m*1e3)/p\n", + "n=f*60*60\n", + "E=n*V*a\n", + "print\"Hourly loss of energy is \",E,\"erg\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Hourly loss of energy is 8.64e+11 erg\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg13:pg-172" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "a=0.5 #area of B-H loop in cm**2\n", + "H=1e3 #value of 1 cm on X-axis in A/m\n", + "B=1 #value of 1 cm on Y-axis in Tesla\n", + "V=1e-3 #volume of specimen in m**3\n", + "n=50 #frequency of a.c. in Hz\n", + "area=a*H*B #area of B-H loop in J/m**3 (this is misprinted in solution in book)\n", + "p=n*V*area\n", + "print\"Hysteresis power loss is \",int(p),\"Watt\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Hysteresis power loss is 25 Watt\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_(Volume-2)_by_S.K._Gupta/chapter5_1.ipynb b/Engineering_Physics_(Volume-2)_by_S.K._Gupta/chapter5_1.ipynb new file mode 100644 index 00000000..62847c30 --- /dev/null +++ b/Engineering_Physics_(Volume-2)_by_S.K._Gupta/chapter5_1.ipynb @@ -0,0 +1,129 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:cc6c5a4a0e0adf2cdedcb0781ed844438a49a2dfa38fdca65079fafeb0f6b0cf" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter5:ULTRASONIC" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg1:pg-177" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "v=5760 #velocity of propagation of ultrasonic sound wave along X-direction in m/s\n", + "t=1.0*10**-3 #thickness of a piezo-electric quartz plate in meter\n", + "lamda=2*t #wavelength in meter(since t=lamda/2 corresponding to fundamental frequency)\n", + "V=v/lamda #fundamental frequency of the crystal\n", + "print\"fundamental frequency of the crystal=\",V/10**6,\"MHz\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "fundamental frequency of the crystal= 2.88 MHz\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg2:pg-177" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "Y=7.9*10**10 #Young's modulus in newton/m**2\n", + "p=2650 #density for Quartz in Kg/m**3\n", + "t=0.005 #thickness of a Quartz crystal in meter\n", + "v=sqrt(Y/p) #velocity for longitudinal vibrations in m/sec\n", + "lamda=2*t #wavelength in meter(since t=lamda/2 corresponding to fundamental frequency)\n", + "V=v/lamda #fundamental frequency of the crystal\n", + "print\"fundamental frequency of the crystal=\",int(round(V/10**3)),\"KHz\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "fundamental frequency of the crystal= 546 KHz\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg3:pg-177" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "f=200.*10**3 #frequency of ultrasonic sound in Hz\n", + "S_a=340 #speed of sound in air in m/s\n", + "S_w=1486 #speed of sound in water in m/s\n", + "lamda_r=S_a/f#wavelength of reflected sound in metre\n", + "print\"Wavelength of reflected sound=\",\"{:.2e}\".format(lamda_r),\"m\"\n", + "lamda_t=S_w/f#wavelength of transmitted sound in metre\n", + "print\"Wavelength of transmitted sound=\",\"{:.2e}\".format(lamda_t),\"m\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Wavelength of reflected sound= 1.70e-03 m\n", + "Wavelength of transmitted sound= 7.43e-03 m\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Engineering_Physics_(Volume-2)_by_S.K._Gupta/chapter6_1.ipynb b/Engineering_Physics_(Volume-2)_by_S.K._Gupta/chapter6_1.ipynb new file mode 100644 index 00000000..08c4435e --- /dev/null +++ b/Engineering_Physics_(Volume-2)_by_S.K._Gupta/chapter6_1.ipynb @@ -0,0 +1,865 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:4df436839f5993306536c3970f88a0cacfc571f6977bdd36882e86f5929b935c" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter6:ELECTROMAGNETICS" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg1:pg-206" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from sympy import *\n", + "r=1 #radius in meter\n", + "H=2 #magnitude of field vector in amp/meter\n", + "pi=Symbol('pi')\n", + "I=H*2*pi*r \n", + "print\"Current in the wire is \",I,\"amp\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Current in the wire is 4*pi amp\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg5:pg-212" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "from sympy import *\n", + "sigma=1e-4 #conductivity in siemen/m\n", + "Er=2.25 #relative permittivity \n", + "E0=1/(4*math.pi*9e9) #permittivity of free space\n", + "E=5e-6*sin(Symbol('9e+09*t')) #electric field in the material volt/m\n", + "J=sigma*E\n", + "diff_E=5e-6*9e9*cos(Symbol('9e+09*t'))\n", + "Jd=E0*Er*diff_E\n", + "print\"Conduction current density is \",J,\"A/m**2\"\n", + "print\"Displacement current density is \",Jd,\"A/m**2\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Conduction current density is 5.0e-10*sin(9e+09*t) A/m**2\n", + "Displacement current density is 8.95246554891911e-7*cos(9e+09*t) A/m**2\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg13:pg-236" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "H0=1 #magnitude of field vector in amp/meter\n", + "mu_0=4*round(math.pi,2)*1e-7 #permeability of free space in H/m\n", + "e0=8.85e-12 #permittivity of free space in F/m\n", + "E0=H0*sqrt(mu_0/e0)\n", + "print\"Magnitude of electric field for plane wave in free space is \",round(E0,2),\"V/m\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Magnitude of electric field for plane wave in free space is 376.72 V/m\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg14:pg-236" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "E0=1e2 #maximum electric field in plane electromagnetic wave in Newton/coul.\n", + "c=3e8 #speed of light in m/sec\n", + "B0=E0/c \n", + "print\"Maximum magnetic field is \",round(B0,9),\"Tesla\"\n", + "print\"Maximum magnetic field will be in Z-direction.\"#this part is not printed in answer in book" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Maximum magnetic field is 3.33e-07 Tesla\n", + "Maximum magnetic field will be in Z-direction.\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg15:pg-236" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "S=2*4.2e4/60 #energy flux per unit area per second at the earth surface\n", + "mu_0=4*round(math.pi,2)*1e-7 #permeability of free space in H/m\n", + "e0=8.85e-12 #permittivity of free space in F/m\n", + "EH=S\n", + "E_div_H=sqrt(mu_0/e0)\n", + "E=sqrt(E_div_H*EH)\n", + "H=EH/E\n", + "E0=round(E,1)*round(sqrt(2.),3)\n", + "H0=H*sqrt(2.)\n", + "print\"Amplitude of electric field is \",round(E0,1),\"V/m\"\n", + "print\"Amplitude of magnetic field is \",round(H0,3),\"A-turn m-1\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Amplitude of electric field is 1026.8 V/m\n", + "Amplitude of magnetic field is 2.726 A-turn m-1\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg16:pg-236" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "P0=1000 #power in watt\n", + "r=2 #distance in meter\n", + "Sav=P0/(4*round(math.pi,2)*r**2)\n", + "mu_0=4*round(math.pi,2)*1e-7 #permeability of free space in H/m\n", + "e0=8.85e-12 #permittivity of free space in F/m\n", + "EH=Sav\n", + "E_div_H=sqrt(mu_0/e0)\n", + "E=sqrt(E_div_H*EH)\n", + "H=EH/E\n", + "print\"Average value of electric field intensity is \",round(E,2),\"V/m\"\n", + "print\"Average value of magnetic field intensity is \",round(H,2),\"A-turn m-1\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Average value of electric field intensity is 86.59 V/m\n", + "Average value of magnetic field intensity is 0.23 A-turn m-1\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg17:pg-237" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "S=1.38 #energy flux in KW/m**2\n", + "c=3e8 #speed of light in m/sec\n", + "mu_0=4*math.pi*1e-7 #permeability of free space in H/m\n", + "E0=sqrt(2*mu_0*c*S*1e3)\n", + "B0=E0/c\n", + "print\"Peak value of electric field is \",round(E0*1e-3,2),\"KV/m\"\n", + "print\"Peak value of magnetic field is \",round(B0,7),\"Wb/m**2\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Peak value of electric field is 1.02 KV/m\n", + "Peak value of magnetic field is 3.4e-06 Wb/m**2\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg18:pg-237" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "E0=100 #in Newton/coul.\n", + "A=1e-3 #area in m**2\n", + "l=100 #length in cm\n", + "e0=8.85e-12 #permittivity of free space in F/m\n", + "V=A*l*1e-2\n", + "U=e0*E0**2*V/2\n", + "print\"Energy contained in cylinder is \",U,\"Joule\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Energy contained in cylinder is 4.425e-11 Joule\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg19:pg-238" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "E0=0.05 #amplitude of electric field strength in V/m\n", + "v=6 #frequency in MHz\n", + "c=3e8 #speed of light in m/sec\n", + "mu_0=4*math.pi*1e-7 #permeability of free space in H/m\n", + "e0=8.85e-12 #permittivity of free space in F/m\n", + "T=round(1/(v*1e6),9)\n", + "lamda=c/(v*1e6)\n", + "H0=E0/sqrt(mu_0/e0)\n", + "Sx_av=E0*round(H0,6)/2\n", + "print\"E=\",E0,\"sin(\",\"{:.2e}\".format(2*math.pi/T),\"t -\",(2*round(math.pi,2)/lamda),\"x) V/m\"\n", + "print\"H=\",\"{:.2e}\".format(H0),\"sin(\",\"{:.2e}\".format(2*math.pi/T),\"t -\",(2*round(math.pi,2)/lamda),\"x) A/m\" \n", + "print\"B=\",round(E0/c,12),\"sin(\",\"{:.2e}\".format(2*math.pi/T),\"t -\",(2*round(math.pi,2)/lamda),\"x) Wb/m**2\" \n", + "print\"Average poynting vector S=\",Sx_av,\"Wb/m**2\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "E= 0.05 sin( 3.76e+07 t - 0.1256 x) V/m\n", + "H= 1.33e-4 sin( 3.76e+07 t - 0.1256 x) A/m\n", + "B= 1.67e-10 sin( 3.76e+07 t - 0.1256 x) Wb/m**2\n", + "Average poynting vector S= 3.325e-06 Wb/m**2\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg20:pg-239" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "lamda=7 #wavelength in mm\n", + "E0=42 #maximum magnitude of electric field in V/m\n", + "c=3e8 #speed of light in m/sec\n", + "print\"E=\",E0,\"sin(2*pi*(ct-x)/\",lamda,\") V/m\"\n", + "print\"B=\",E0/c,\"sin(2*pi*(ct-x)/\",lamda,\") Wb/m**2 \\nThe magnetic field is along Z-axis.\"\n", + "#unit is not mentioned in answer in book" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "E= 42 sin(2*pi*(ct-x)/ 7 ) V/m\n", + "B= 1.4e-07 sin(2*pi*(ct-x)/ 7 ) Wb/m**2 \n", + "The magnetic field is along Z-axis.\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg21:pg-239" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "er=81 #relative permittivity of distilled water\n", + "e0=1 #let, permittivity of free space\n", + "mu_0=1 #let, permeability of free space\n", + "e=e0*er\n", + "c=3e8 #speed of light in m/sec\n", + "mu=mu_0#for distilled water\n", + "MU=sqrt((mu*e)/(mu_0*e0))\n", + "v=c/MU\n", + "print\"Refractive index is \",MU\n", + "print\"Velocity of light in distilled water is \",\"{:.2e}\".format(v),\"m/s\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Refractive index is 9\n", + "Velocity of light in distilled water is 3.33e+7 m/s\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg23:pg-241" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "E0=7.5 #electric field intensity in KV/m\n", + "w=2e9 #angular frequency in rad/sec\n", + "c=3e8 #speed of light in m/sec\n", + "mu_0=4*round(math.pi,2)*1e-7 #permeability of free space in H/m\n", + "e0=8.85e-12 #permittivity of free space in F/m\n", + "f=w/(2*round(math.pi,2))\n", + "lamda=c/f\n", + "T=1/f\n", + "H0=E0*1e3/sqrt(mu_0/e0)\n", + "print\"Wavelength is \",lamda,\"m\"\n", + "print\"Frequency is \",round(f*1e-6,1),\"MHz\"\n", + "print\"Time period is \",T,\"sec\"\n", + "print\"Amplitude of magnetic field intensity is \",round(H0,2),\"A/m\"\n", + "print\"Therefore, Hz=\",round(H0,2),\"cos( (%.e*t)-(beta*x)) A/m\"%w#unit is not printed in book" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Wavelength is 0.942 m\n", + "Frequency is 318.5 MHz\n", + "Time period is 3.14e-09 sec\n", + "Amplitude of magnetic field intensity is 19.91 A/m\n", + "Therefore, Hz= 19.91 cos( (2e+09*t)-(beta*x)) A/m\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg24:pg-241" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from sympy import *\n", + "mu_0=4*math.pi*1e-7 #permeability of free space in H/m\n", + "e0=8.854e-12 #permittivity of free space in F/m\n", + "E=array([0, 45*sin(Symbol('6*pi*1e8*t-(2*pi*x)')),15*cos(Symbol('6*pi*1e8*t-(2*pi*x)'))]) \n", + "#E=Ey*sin((w*t)-(beta*x))j + Ez*cos((w*t)-(beta*x))k\n", + "#compairing given equation with above equation\n", + "beta=2*Symbol('pi')\n", + "w=6e8*Symbol('pi')\n", + "f=w/(2*Symbol('pi'))\n", + "n0=sqrt(mu_0/e0)\n", + "Hx=0\n", + "Hy=round(15/n0,4)*cos(Symbol('6*pi*10**8*t-(2*pi*x)'))\n", + "Hz=round(45/n0,3)*sin(Symbol('6*pi*10**8*t-(2*pi*x)'))\n", + "H=array([ Hx , Hy , Hz ])\n", + "print\"Phase constant is \",beta,\"rad/s\"\n", + "print\"Angular frequency is %.e*pi rad/s\"%(w/Symbol('pi'))\n", + "print\"Frequency is \",\"{:.0e}\".format(f),\"Hz\"\n", + "print\"Intrinsic impedance is \",int(round(n0)),\"Ohm\"\n", + "print\"Magnetic field is \",H,\"A/m\" #unit is not printed in book" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Phase constant is 2*pi rad/s\n", + "Angular frequency is 6e+08*pi rad/s\n", + "Frequency is 3e+8 Hz\n", + "Intrinsic impedance is 377 Ohm\n", + "Magnetic field is [0 0.0398*cos(6*pi*10**8*t-(2*pi*x)) 0.119*sin(6*pi*10**8*t-(2*pi*x))] A/m\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg25:pg-242" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from sympy import *\n", + "x, Beta, y, Y = symbols('x Beta y Y')\n", + "Hz=(6*x*cos(Beta))+(12*y*sin(Y))\n", + "a=diff((6*x*cos(Beta))+(12*y*sin(Y)),y)\n", + "b=diff(-((6*x*cos(Beta))+(12*y*sin(Y))),x)\n", + "c=0\n", + "J=array([ a, b, c])\n", + "print\"Current density is \",J,\"A/m**2\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Current density is [12*sin(Y) -6*cos(Beta) 0] A/m**2\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg26:pg-244" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "A=1.3 #area in m**2\n", + "t=3 #time in hours\n", + "S=1.1 #intensity of sun rays in KW/m**2\n", + "c=3e8 #speed of light in m/sec\n", + "p=A*(t*3600)*(S*1000)/c\n", + "print\"Momentum is %se-4 Kg-m/s\"%(p*10000)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Momentum is 514.8e-4 Kg-m/s\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg27:pg-245" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "S=10 #energy flux in watt/m**2\n", + "A=1 #area in m**2\n", + "t=1 #time in hour\n", + "c=3e8 #speed of light in m/sec\n", + "p=2*S*A*(t*3600)/c\n", + "F=2*S*A/c\n", + "print\"Momentum is %.1e Kg-m/s\"%p\n", + "print\"Force is %.2e N\"%F" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Momentum is 2.4e-04 Kg-m/s\n", + "Force is 6.67e-08 N\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg29:pg-251" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "mu=4*math.pi*1e-7 #permeability in H/m\n", + "f=71.6 #frequency in MHz\n", + "sigma=3.54e7 #conductivity in siemens/m\n", + "d=1/sqrt(math.pi*f*1e6*mu*sigma)\n", + "print\"Depth of penetration is \",int(round(d*1e6)),\"micro meter\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Depth of penetration is 10 micro meter\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg30:pg-251" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "f=3e6 #frequency in Hz\n", + "mu_r=1 \n", + "mu_0=4*round(math.pi,2)*1e-7 # in H/m\n", + "sigma=38e6 # in S/m\n", + "mu=mu_r*mu_0\n", + "d=1/sqrt(round(math.pi,2)*f*mu*sigma)\n", + "alpha=1/(d)\n", + "beta=alpha\n", + "magnitude=sqrt(alpha**2+beta**2)\n", + "angle=degrees(math.atan(beta/alpha))\n", + "v=2*round(math.pi,2)*f/round(beta)\n", + "print\"Skin depth is \",round(d*1e3,5),\"mm\"\n", + "print\"Propagation constant =[ %.4e , %s degree] m**-1\"%(magnitude,int(angle)) #in polar form\n", + "print\"Wave velocity is \",round(v,2),\"m/s\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Skin depth is 0.04716 mm\n", + "Propagation constant =[ 2.9987e+04 , 45 degree] m**-1\n", + "Wave velocity is 888.51 m/s\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg31:pg-252" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "mu=4*math.pi*1e-7 # in H/m\n", + "e0=8.854e-12 # in F/m\n", + "e=70*e0\n", + "sigma=5\n", + "d=(2./sigma)*sqrt(e/mu)\n", + "alpha=1/round(d,4)\n", + "print\"skin depth is \",round(d,4),\"m\"\n", + "print\"Attenuation constant is \",round(alpha,2),\"Np/m\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "skin depth is 0.0089 m\n", + "Attenuation constant is 112.36 Np/m\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg32:pg-253" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "sigma=2e-3 #in S/m\n", + "e0=8.854e-12 #in F/m\n", + "e=80*e0\n", + "f=10 #in KHz\n", + "mu=4*math.pi*1e-7 #in H/m\n", + "ratio=sigma/(2*round(math.pi,2)*f*1e3*e)\n", + "\n", + "#since ratio= sigma/(w*e) = 44.96 >>1,therefore, medium is a good conductor.\n", + "#So calculations will be done considering medium as a good conductor.\n", + "\n", + "alpha=sqrt(2*math.pi*f*1e3*mu*sigma/2)\n", + "beta=int(alpha*1e5)*1e-5\n", + "magnitude=sqrt(alpha**2+beta**2)\n", + "angle=degrees(math.atan(beta/alpha))\n", + "ni=round(round(sqrt(2*math.pi*f*1e3*mu/sigma),2)/round(sqrt(2),2),3)*(1+1j)\n", + "lamda=2*round(math.pi,2)/beta\n", + "v=2*math.pi*f*1e3/beta\n", + "print\"Attenuation constant is %.2e neper/m\"%(int(alpha*1e5)*1e-5)\n", + "print\"Phase constant is %.2e rad/m\"%beta\n", + "print\"Propagation constant = [ %.3e , %.f degree] m**-1\"%(magnitude,angle)#in polar form(unit is not printed in book) \n", + "print\"Intrinsic impedance is \",ni,\"ohm\"\n", + "print\"Wavelength is %.2f m\"%lamda\n", + "print\"Velocity of wave is %.2e m/s\"%v" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Attenuation constant is 8.88e-03 neper/m\n", + "Phase constant is 8.88e-03 rad/m\n", + "Propagation constant = [ 1.256e-02 , 45 degree] m**-1\n", + "Intrinsic impedance is (4.454+4.454j) ohm\n", + "Wavelength is 707.21 m\n", + "Velocity of wave is 7.08e+06 m/s\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg33:pg-254" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "f=100 #in MHz\n", + "mu_r=1\n", + "mu_0=4*round(math.pi,2)*1e-7 #in H/m\n", + "mu=mu_0*mu_r\n", + "sigma=58e6 #in S/m\n", + "alpha=sqrt(round(math.pi,2)*f*1e6*mu*sigma)\n", + "alpha=int(alpha/10)*10\n", + "beta=alpha\n", + "magnitude=sqrt(alpha**2+beta**2)\n", + "angle=degrees(math.atan(beta/alpha))\n", + "sqrt_j=45\n", + "ni=sqrt(2*round(math.pi,2)*f*1e6*mu/sigma)\n", + "v=2*round(math.pi,2)*f*1e6/beta\n", + "print\"Attenuation constant is %.4e neper/m\"%(int(alpha*1e5)*1e-5)\n", + "print\"Phase constant is %.4e rad/m\"%beta\n", + "print\"Propagation constant = [ %.4e , %.f degree] m**-1\"%(magnitude,angle)#in polar form(unit is not printed in book) \n", + "print\"Intrinsic impedance = [ %.3e , %s degree ] ohm\"%(ni,sqrt_j)#in polar form(unit is not printed in book)\n", + "print\"Velocity of wave is %.3f Km/s\"%(v/1e3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Attenuation constant is 1.5124e+05 neper/m\n", + "Phase constant is 1.5124e+05 rad/m\n", + "Propagation constant = [ 2.1389e+05 , 45 degree] m**-1\n", + "Intrinsic impedance = [ 3.688e-03 , 45 degree ] ohm\n", + "Velocity of wave is 4.152 Km/s\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg34:pg-255" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "mu=4*math.pi*1e-7 #in H/m\n", + "sigma=3.54e7 #in S/m\n", + "d=0.0664 #penetration depth in mm\n", + "f=1/(math.pi*mu*sigma*(d*1e-3)**2)\n", + "print\"Frequency is %.2f MHz\"%(f/1e6)\n", + "#answer is wrong in book because d=0.0644 is taken in calculation which is wrong(given d=0.0664 mm)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Frequency is 1.62 MHz\n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Engineering_Physics_(Volume-2)_by_S.K._Gupta/chapter7_1.ipynb b/Engineering_Physics_(Volume-2)_by_S.K._Gupta/chapter7_1.ipynb new file mode 100644 index 00000000..871de61f --- /dev/null +++ b/Engineering_Physics_(Volume-2)_by_S.K._Gupta/chapter7_1.ipynb @@ -0,0 +1,503 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:6f3332c2cf8866e9257d8f3e11193b59b2e8f54645aa9a7c58a49b95bb6e6cc9" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter7:SUPERCONDUCTIVITY" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg1:pg-272" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "Tc=3.7 #critical temperature in K\n", + "Hc_0=0.0306 #critical magnetic field in Tesla at 0K\n", + "T=2 #temperature in K\n", + "Hc=Hc_0*(1-(T/Tc)**2)\n", + "print\"Critical field at 2 K is \",round(Hc,4),\"Tesla\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Critical field at 2 K is 0.0217 Tesla\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg2:pg-272" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "Tc=7.2 #transition temperature in K\n", + "T=5 #temperature in K\n", + "Hc_T=3.3e4 #critical magnetic field at 5K in A/m\n", + "Hc_0=Hc_T/(1-(T/Tc)**2)\n", + "print\"Maximum value of H at 0 K is \",\"{:.2e}\".format(Hc_0),\"A/m\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Maximum value of H at 0 K is 6.37e+04 A/m\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg3:pg-273" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "Tc=7.2 #critical temperature in K\n", + "Hc_0=1 #let,critical magnetic field at 0K\n", + "Hc_T=0.1*Hc_0 #critical magnetic field at T Kelvin\n", + "T=sqrt(1-Hc_T/Hc_0)*Tc\n", + "print\"Temperature is \",round(T,2),\"K\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Temperature is 6.83 K\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg4:pg-273" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "T=4.2 #temperature in K\n", + "Hc_0=0.0803 #critical magnetic field at 0K in Wb/m**2\n", + "Tc=7.2 #critical temperature for Pb in K\n", + "Hc_T=Hc_0*(1-(T/Tc)**2)\n", + "print\"Critical field at 4.2 K is \",round(Hc_T,5),\"Tesla\"#answer is wrong in book" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Critical field at 4.2 K is 0.05298 Tesla\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg5:pg-273" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "Hc_T=105e3 #critical magnetic field at T Kelvin in A/m\n", + "Hc_0=150e3 #critical magnetic field at 0K in A/m\n", + "Tc=9.2 #critical temperature in K\n", + "T=sqrt(1-Hc_T/Hc_0)*Tc\n", + "print\"Temperature is \",round(T,2),\"K\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Temperature is 5.04 K\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg6:pg-274" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "Hc_T=1e5 #critical magnetic field at 8K in A/m\n", + "T=8 #temperature in K\n", + "Hc_0=2e5 #critical magnetic field at 0K in A/m\n", + "Tc=T/sqrt(1-Hc_T/Hc_0)\n", + "print\"Transition temperature is \",round(Tc,1),\"K\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Transition temperature is 11.3 K\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg7:pg-274" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "Tc=7.26 #critical temperature in K\n", + "Hc_0=8e5 #critical magnetic field at 0K in A/m\n", + "Hc_T=4e4 #critical magnetic field at T kelvin in A/m\n", + "T=sqrt(1-Hc_T/Hc_0)*Tc\n", + "print\"T =\",round(T,2),\"K\",\"\\nThe temperature of the metal should be held below\",round(T,2),\"K\" " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "T = 7.08 K \n", + "The temperature of the metal should be held below 7.08 K\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg8:pg-275" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "T1=14 #temperature in K\n", + "T2=12 #temperature in K\n", + "T=4.2 #temperature in K\n", + "Hc_T1=0.176 #critical magnetic field at temperature T1\n", + "Hc_T2=0.528 #critical magnetic field at temperature T2\n", + "Tc=sqrt((Hc_T2*T1**2-Hc_T1*T2**2)/(Hc_T2-Hc_T1))\n", + "Tc=int(Tc*10)/10. #rounding off\n", + "Hc_0=Hc_T1/(1-(T1/Tc)**2)\n", + "Hc_T=Hc_0*(1-(T/Tc)**2)\n", + "print\"Transition temperature is \",Tc,\"K\"\n", + "print\"Critical field at 0 K is \",round(Hc_0,3),\"Tesla\"\n", + "print\"Critical field at 4.2 K is \",round(Hc_T,2),\"Tesla\"\n", + "#answers in book are wrong because value of T2 is taken as 13K in calculation which is wrong." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Transition temperature is 14.8 K\n", + "Critical field at 0 K is 1.673 Tesla\n", + "Critical field at 4.2 K is 1.54 Tesla\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg9:pg-275" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "D=1.0 #diameter of Pb wire in mm\n", + "Bc=0.0548 #in Tesla\n", + "mu_0=4*math.pi*1e-7 #absolute permeability of air in N/A**2\n", + "Ic=math.pi*D*1e-3*Bc/mu_0\n", + "print\"Current is \",int(Ic),\"amp\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Current is 137 amp\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg10:pg-276" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "Hc_0=6.5e3 #critical magnetic field at 0K in A/m\n", + "Tc=7.18 #critical temperature in K\n", + "Hc_T=4.5e3 #critical magnetic field at T Kelvin in A/m\n", + "T=sqrt(1-Hc_T/Hc_0)*Tc\n", + "print\"Temperature is \",round(T,2),\"K\"\n", + "D=2 #diameter of the lead wire in mm\n", + "r=D/2 \n", + "Ic=2*math.pi*r*1e-3*Hc_T\n", + "Jc=Ic/(math.pi*(r*1e-3)**2)\n", + "print\"Critical current density is \",\"{:.1e}\".format(Jc),\"A/m**2\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Temperature is 3.98 K\n", + "Critical current density is 9.0e+06 A/m**2\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg11:pg-281" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "T=3.5 #temperature in K\n", + "lamda_T=750 #penetration depth of Hg at 3.5K in Angstrom\n", + "Tc=4.153 #critical temperature in K\n", + "lamda_0=lamda_T*sqrt(round(1-(T/Tc)**4,3))\n", + "print\"Penetration depth at 0 K is\",round(lamda_0,1),\"Angstrom\"#answer is wrong in book because of calculation mistake " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Penetration depth at 0 K is 528.2 Angstrom\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg12:pg-281" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "m=9.1e-31 #mass of electron kg\n", + "mu_0=12.56e-7 #absolute permeability of air in N/A**2\n", + "e=1.6e-19 #charge of electron in coulomb\n", + "ns=1e28 #number of super electrons per meter cube\n", + "lamda_0=sqrt(m/(mu_0*ns*e**2))\n", + "lamda_0=round(lamda_0,9)*1e10\n", + "print\"Penetration depth at 0 K is \",int(lamda_0),\"Angstrom\"\n", + "Tc=3 #critical temperature in K\n", + "T=1. #temperature in K\n", + "lamda_T=lamda_0/sqrt(1-(T/Tc)**4)\n", + "print\"Penetration depth at 1 K is \",int(lamda_T),\"Angstrom\"\n", + "#in book lamda(at 3K) is printed,which is wrong. Correct notation is lamda(at 1K)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Penetration depth at 0 K is 530 Angstrom\n", + "Penetration depth at 1 K is 533 Angstrom\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg13:pg-286" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "Tc=9.25 #critical temperature in K\n", + "T=0 #temperature in K\n", + "Kb=1.38e-23 #Boltzmann's constant in J/K\n", + "Eg=3.53*Kb*Tc/(1.6e-19)\n", + "h=6.63e-34 #planck constant joule-sec\n", + "c=3e8 #speed of light in m/sec\n", + "print\"Energy gap Eg is \",round(Eg*1e3,2),\"meV\"\n", + "lamda_min=h*c/round(Eg*1.6e-19,23)\n", + "print\"Minimum photon wavelength is \",\"{:.2e}\".format(lamda_min),\"m\"\n", + "print\" This wavelength lie in the far-infrared region of electromagnetic radiations.\"\n", + "v=round(Eg*1.6e-19,23)/h\n", + "print\"Frequency needed is \",\"{:.2e}\".format(v),\"s**-1\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Energy gap Eg is 2.82 meV\n", + "Minimum photon wavelength is 4.42e-04 m\n", + " This wavelength lie in the far-infrared region of electromagnetic radiations.\n", + "Frequency needed is 6.79e+11 s**-1\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Eg14:pg-286" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "M=200.59 #average atomic mass of Hg in amu\n", + "m=204 #mass of isotope in amu\n", + "T=4.153 #temperature in K\n", + "t=4.118 #temperature in K\n", + "dM=m-M\n", + "dTc=t-T\n", + "alpha=-(M*dTc/(dM*T))\n", + "print\"Isotope effect coefficient is \",round(alpha,3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Isotope effect coefficient is 0.496\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +} \ No newline at end of file diff --git a/Engineering_Physics_(Volume-2)_by_S.K._Gupta/screenshots/X-ray_diffraction_1.png b/Engineering_Physics_(Volume-2)_by_S.K._Gupta/screenshots/X-ray_diffraction_1.png new file mode 100644 index 00000000..958cc24e Binary files /dev/null and b/Engineering_Physics_(Volume-2)_by_S.K._Gupta/screenshots/X-ray_diffraction_1.png differ diff --git a/Engineering_Physics_(Volume-2)_by_S.K._Gupta/screenshots/ultrasonics_1.png b/Engineering_Physics_(Volume-2)_by_S.K._Gupta/screenshots/ultrasonics_1.png new file mode 100644 index 00000000..094e7bc2 Binary files /dev/null and b/Engineering_Physics_(Volume-2)_by_S.K._Gupta/screenshots/ultrasonics_1.png differ diff --git a/Engineering_Physics_(Volume-2)_by_S.K._Gupta/screenshots/wave_mechanics_1.png b/Engineering_Physics_(Volume-2)_by_S.K._Gupta/screenshots/wave_mechanics_1.png new file mode 100644 index 00000000..1be3ed59 Binary files /dev/null and b/Engineering_Physics_(Volume-2)_by_S.K._Gupta/screenshots/wave_mechanics_1.png differ diff --git a/sample_notebooks/MohdAsif/Ch2.ipynb b/sample_notebooks/MohdAsif/Ch2.ipynb new file mode 100644 index 00000000..493ebc40 --- /dev/null +++ b/sample_notebooks/MohdAsif/Ch2.ipynb @@ -0,0 +1,1269 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:5a4681bcd32ee53961d617cc97d1b729b583a17da43dad40dee1f6a5e0b10f08" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 2 - Junctions & Interfaces" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex 2.6.1 Page 2-21" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log\n", + "#Given : \n", + "Ge=4.4*10**22 #atoms/cm**3\n", + "NA=Ge/10**8 #per cm**3\n", + "NA=NA*10**6 #per m**3\n", + "ND=NA*10**3 #per m**3\n", + "ni=2.5*10**13 #per cm**3\n", + "ni=ni*10**6 #per m**3\n", + "VT=26 #mV\n", + "Vj=VT*log(NA*ND/ni**2) #mV\n", + "print \"Junction potential = %0.1f mV\"%Vj" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Junction potential = 328.7 mV\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex 2.6.2 Page 2- 22" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "#Given : \n", + "ni=2.5*10**15 #per cm**3\n", + "Ge=4.4*10**22 #atoms/cm**3\n", + "NA=Ge/10**8 #per cm**3\n", + "NA=NA*10**6 #per m**3\n", + "ND=NA*10**3 #per m**3\n", + "ni=ni*10**6 #per m**3\n", + "T=27+273 #K\n", + "VT=T/11600 #V\n", + "Vo=VT*log(NA*ND/ni**2) #V\n", + "print \"Contact potential = %0.4f V \" %Vo" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Contact potential = 0.0888 V \n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex 2.6.3 Page 2-23" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given : \n", + "mu_n=1500*10**-4 #m**2/V-s\n", + "mu_p=475*10**-4 #m**2/V-s\n", + "ni=1.45*10**10*10**6 #per m**3\n", + "q=1.6*10**-19 #Coulomb\n", + "rho_p=10 #ohm-cm\n", + "rho_p=rho_p*10**-2 #ohm-m\n", + "rho_n=3.5 #ohm-cm\n", + "rho_n=rho_n*10**-2 #ohm-m\n", + "sigma_p=1/rho_p #(ohm-m)**-1\n", + "NA=sigma_p/q/mu_p #m**3\n", + "sigma_n=1/rho_n #(ohm-m)**-1\n", + "ND=sigma_p/q/mu_n #m**3\n", + "VT=26*10**-3 #V\n", + "Vj=VT*log(NA*ND/ni**2) #V\n", + "print \"Height of potential barrier = %0.3f V \"%Vj\n", + "#Anser in the book is wrong." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Height of potential barrier = 0.564 V \n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex 2.6.4 Page 2-24" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given : \n", + "rho_p=2 #ohm-cm\n", + "rho_p=rho_p*10**-2 #ohm-m\n", + "rho_n=1 #ohm-cm\n", + "rho_n=rho_n*10**-2 #ohm-m\n", + "mu_n=1500*10**-4 #m**2/V-s\n", + "mu_p=2100*10**-4 #m**2/V-s\n", + "ni=2.5*10**13 #per m**3\n", + "q=1.6*10**-19 #Coulomb\n", + "sigma_p=1/rho_p #(ohm-m)**-1\n", + "NA=sigma_p/q/mu_p #m**3\n", + "sigma_n=1/rho_n #(ohm-m)**-1\n", + "ND=sigma_p/q/mu_n #m**3\n", + "T=27+273 #K\n", + "VT=T/11600 #V\n", + "Vj=VT*log(NA*ND/ni**2) #V\n", + "print \"Height of potential barrier = %0.4f V\"%Vj\n", + "#Anser in the book is wrong." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Height of potential barrier = 0.9347 V\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex 2.7.1 Page 2-27" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given : \n", + "Vgamma=0.6 #Volt\n", + "rf=12 #ohm\n", + "V=5 #Volts\n", + "R=1 #kohm\n", + "IF=(V-Vgamma)/(R*1000+rf) #A\n", + "print \"Diode current = %0.1f mA\"%(IF*1000)\n", + "VF=Vgamma+IF*rf #volts\n", + "print \"Diode voltage = %0.2f Volts\"%VF" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Diode current = 4.3 mA\n", + "Diode voltage = 0.65 Volts\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex 2.7.2 Page 2-35" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from sympy import symbols, sin, N\n", + "#Given : \n", + "Vgamma=0.6 #Volt\n", + "Rf=10 #ohm\n", + "Eta=2 \n", + "Vm=0.2 #Volts\n", + "Vdc=10 #Volts\n", + "RL=1 #kohm\n", + "IDQ=(Vdc-Vgamma)/(RL*1000+Rf) #A\n", + "VT=25*10**-3 #Volts\n", + "rd=Eta*VT/IDQ #ohm\n", + "omega,t = symbols('omega t')\n", + "Vs = Vm*sin(omega*t)\n", + "Vo_ac =(RL*1000)/(RL*1000+rd)*Vs\n", + "print \"Alternating component of voltage across RL, Vo(ac) =\",N(Vo_ac,4)\n", + "Vo_DC=IDQ*RL*1000 #Volts\n", + "Vo_DC_total=Vo_DC+(Vo_ac) #Volts\n", + "print \"\\nTotal load voltage =\",N(Vo_DC_total,4)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Alternating component of voltage across RL, Vo(ac) = 0.1989*sin(omega*t)\n", + "\n", + "Total load voltage = 0.1989*sin(omega*t) + 9.307\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex 2.7.3 Page 2-37" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import exp\n", + "#Given : \n", + "Eta=2 #for Si diode\n", + "T=300 #K\n", + "VT=T/11600 #V\n", + "IbyIo=90/100 \n", + "#I=Io*(exp(V/Eta/VT)-1)\n", + "V=log(IbyIo+1)*Eta*VT #V\n", + "print \"Saturation value of voltage = %0.2f mV\"%(V*1000)\n", + "VF=0.5 #Volts\n", + "VR=-0.5 #Volts\n", + "IFbyIR=(exp(VF/Eta/VT)-1)/(exp(VR/Eta/VT)-1) #ratio\n", + "print \"Ratio of forward to reverse current : %0.1f\" %IFbyIR\n", + "#Answer in the book is wrong." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Saturation value of voltage = 33.20 mV\n", + "Ratio of forward to reverse current : -15782.7\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex 2.7.4 Page 2-37" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given : \n", + "Eta=2 #for Si diode\n", + "T=300 #K\n", + "VT=T/11600 #V\n", + "IbyIo=90/100 \n", + "#I=Io*(exp(V/Eta/VT)-1)\n", + "V=log(IbyIo+1)*Eta*VT #V\n", + "print \"Saturation value of voltage = %0.2f mV\"%(V*1000)\n", + "VF=0.2 #Volts\n", + "VR=-0.2 #Volts\n", + "IFbyIR=(exp(VF/Eta/VT)-1)/(exp(VR/Eta/VT)-1) #ratio\n", + "print \"Ratio of forward to reverse current : %0.2f\" %IFbyIR\n", + "#Answer in the book is wrong." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Saturation value of voltage = 33.20 mV\n", + "Ratio of forward to reverse current : -47.78\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex 2.9.1 Page 2-61" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given : \n", + "IF=10 #mA\n", + "VF=0.75 #volts\n", + "T=27+273 #K\n", + "Eta=2 #for Si diode\n", + "VT=T/11600 #V\n", + "Io=IF/(exp(VF/Eta/VT)-1) #mA\n", + "print \"Reverse saturation current = %0.3f nA\" %(Io*10**6)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Reverse saturation current = 5.043 nA\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex 2.9.2 Page 2-61" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "#Given : \n", + "IF=10 #mA\n", + "VF=0.3 #Volts\n", + "T=27+273 #K\n", + "Eta=1 #for Ge diode\n", + "VT=T/11600 #V\n", + "Io=IF/(exp(VF/Eta/VT)-1) #mA\n", + "print \"Reverse saturation current = %0.2f nA \"%(Io*10**6) \n", + "# Answer wrong in the textbook." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Reverse saturation current = 91.66 nA \n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex 2.9.3 Page 2-61" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given : \n", + "Io=1*10**-9 #A\n", + "T=27+273 #K\n", + "VT=T/11600 #V\n", + "VF=0.3 #Volts\n", + "Eta=1 #for Ge diode\n", + "IF=Io*(exp(VF/Eta/VT)-1) #mA\n", + "print \"Forwad current = %0.4f mA \"%(IF*10**3) " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Forwad current = 0.1091 mA \n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex 2.9.4 Page 2-62" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given : \n", + "T=27+273 #K\n", + "V1=0.4 #V\n", + "V2=0.42 #V\n", + "I1=10 #mA\n", + "I2=20 #mA\n", + "VT=T/11600 #V\n", + "Eta=1/log(I1/I2)*(V1-V2)/VT\n", + "print \"Value of Eta : %0.2f\" %Eta\n", + "Io=I1/(exp(V1/Eta/VT)-1)*10**-3 #A\n", + "print \"Current, Io = %0.2f nA \"%(Io*10**9) \n", + "#Ans in the book is not accurate." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Value of Eta : 1.12\n", + "Current, Io = 9.54 nA \n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex 2.9.5 Page 2-62" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given : \n", + "Io1=10**-12 #A\n", + "Io2=10**-10 #A\n", + "I=2 #mA\n", + "Eta=1 #constant\n", + "T=27+273 #K\n", + "VT=26/1000 #V\n", + "#I=I1+I2\n", + "V=(log(I*10**-3/(Io1+Io2))+1)*Eta*VT #V\n", + "print \"Voltage across the diodes = %.4f V \"%V \n", + "#Ans in the book is not accurate." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Voltage across the diodes = 0.4628 V \n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex 2.9.6 Page 2-64" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given : \n", + "Io1=10*10**-9 #A\n", + "Io2=10*10**-9 #A\n", + "Eta=1.1 #constant\n", + "T=25+273 #K\n", + "V=0.2 #V(assumed)\n", + "VT=T/11600 #V\n", + "I1=Io1*(exp(V/Eta/VT)-1) #A\n", + "I2=Io2*(exp(V/Eta/VT)-1) #A\n", + "I=I1+I2 #A\n", + "print \"Source current = %0.2f micro Ampere \"%(I*10**6) " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Source current = 23.68 micro Ampere \n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex 2.9.7 Page 2-65" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given : \n", + "Io=10**-13 #A\n", + "T=27+273 #K\n", + "Eta=1 #constant\n", + "V=0.6 #V\n", + "VT=26/1000 #V\n", + "I3=Io*(exp(V/Eta/VT)-1) #A\n", + "R=1*1000 #ohm\n", + "Ir=V/R #A\n", + "Itotal=I3+Ir #A\n", + "VD1=log(Itotal/Io)*Eta*VT #V\n", + "VD2=VD1 #V\n", + "Vin=VD1+VD2+V #V\n", + "print \"Voltage Vin = %0.3f V \"%Vin " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Voltage Vin = 1.823 V \n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex 2.9.8 Page 2-66" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given : \n", + "Vs=10 #V\n", + "print \"Case(i) : Vb=9.8V\" \n", + "Vb=9.8 #V\n", + "#D1 forward & D2 reverse biased: Breakdown D2\n", + "VD2=Vb #V\n", + "VD1=Vs-Vb #V\n", + "print \"VD1 = %0.3f V \"%VD1 \n", + "print \"VD2 = %0.2f V \"%VD2\n", + "print \"Case(ii) : Vb=10.2V\" \n", + "Vb=10.2 #V\n", + "#D1 forward & D2 reverse biased: none will be breakdown\n", + "VD2=Vb #V\n", + "#I=I0 so exp(V1/Eta/VT)-1=1\n", + "Eta=1 #constant\n", + "VT=26/1000 #V\n", + "VD1=log(1+1)*Eta*VT #V\n", + "VD2=Vs-VD1 #V\n", + "print \"VD1 = %0.3f V \"%VD1 \n", + "print \"VD2 = %0.3f V \"%VD2" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Case(i) : Vb=9.8V\n", + "VD1 = 0.200 V \n", + "VD2 = 9.80 V \n", + "Case(ii) : Vb=10.2V\n", + "VD1 = 0.018 V \n", + "VD2 = 9.982 V \n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex 2.9.9 Page 2-67" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given : \n", + "Vs=5 #Volt\n", + "Eta=1 #constant\n", + "VT=26/1000 #V\n", + "#I=I0 so exp(V1/Eta/VT)-1=1\n", + "V1=log(1+1)*Eta*VT #Volt\n", + "V2=Vs-V1 #Volt\n", + "print \"Voltage across diode D1 = %0.3f V \"%V1\n", + "print \"Voltage across diode D2 = %0.3f V \"%V2 " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Voltage across diode D1 = 0.018 V \n", + "Voltage across diode D2 = 4.982 V \n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex 2.10.2 Page 2-70" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given : \n", + "rho_n=10 #ohm-cm\n", + "rho_p=3.5 #ohm-cm\n", + "ni=1.5*10**10 #per cm**3\n", + "Vj=0.56 #volt\n", + "q=1.6*10**-19 #Coulomb\n", + "mu_n=1500 #cm**2/V-s\n", + "mu_p=500 #cm**2/V-s\n", + "sigma_p=1/rho_p #(ohm-cm)**-1\n", + "NA=sigma_p/q/mu_p #per cm**3\n", + "sigma_n=1/rho_n #(ohm-cm)**-1\n", + "ND=sigma_n/q/mu_n #per cm**3\n", + "VT=Vj/log(NA*ND/ni**2) #V\n", + "T=11600*VT #K\n", + "print \"Temperature of junction = %0.2f degree K \" %T\n", + "t=T-273 #degree C\n", + "print \"Temperature of junction = %0.2f degree C : \" %t" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Temperature of junction = 287.28 degree K \n", + "Temperature of junction = 14.28 degree C : \n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex 2.11.1 Page 2-75" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given : \n", + "Io=10 #nA\n", + "T1=27+273 #K\n", + "T2=87+273 #K\n", + "VT=T1/11600 #V\n", + "Eta=2 #for Si\n", + "m=1.5 #for Si\n", + "VGO=-1.21 #volt\n", + "K=Io*10**-9/T1**m/exp(VGO/Eta/VT) #constant\n", + "VT=T2/11600 #V\n", + "Io2=K*T2**m*exp(VGO/Eta/VT) #A\n", + "print \"Reverse saturation current at 87 degree C = %0.2f nA \"%(Io2*10**9) " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Reverse saturation current at 87 degree C = 648.69 nA \n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex 2.11.2 Page 2-76" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given : \n", + "V=0.45 #volt\n", + "Eta=2 #for Si\n", + "T1=27+273 #K\n", + "T2=125+273 #K\n", + "VT1=T1/11600 #V\n", + "VT2=T2/11600 #V\n", + "I1BYIo1=exp(V/Eta/VT1) \n", + "I2BYIo2=exp(V/Eta/VT2) \n", + "m=1.5 #for Si\n", + "VGO=1.21 #volt\n", + "Io1BYIo2=(T1/T2)**m*exp(-VGO/Eta/VT1+VGO/Eta/VT2) #constant\n", + "I2BYI1=I2BYIo2/I1BYIo1/Io1BYIo2 \n", + "print \"Factor by which current increases : %0.2f\"%I2BYI1\n", + "#Answer is wrong in the textbook." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Factor by which current increases : 56.94\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex 2.11.3 Page 2-78" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given : \n", + "Io1=2 #nA\n", + "T1=10+273 #K\n", + "V=0.4 #volt\n", + "VT1=T1/11600 #V\n", + "m=1.5 #for Si\n", + "Eta=2 #for Si\n", + "VGO=-1.21 #volt\n", + "K=Io1*10**-9/T1**m/exp(VGO/Eta/VT1) #constant\n", + "I1=Io1*10**-9*(exp(V/Eta/VT1)-1) #nA\n", + "T2=70+273 #K\n", + "VT2=T2/11600 #V\n", + "Io2=K*T2**m*(exp(VGO/Eta/VT2)) #A\n", + "I2=Io2*(exp(V/Eta/VT2)-1) #nA\n", + "change=(I2-I1)/I1*100 #%\n", + "print \"%% change in diode current : %0.1f\" %change\n", + "#Answer is wrong in the textbook." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "% change in diode current : 2332.4\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex 2.11.4 Page 2-79" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given : \n", + "T=300 #K\n", + "m_Si=1.5 #for Si\n", + "m_Ge=1.5 #for Ge\n", + "EGO_Si=1.21 #Volt\n", + "EGO_Ge=0.785 #Volt\n", + "Eta_Si=2 \n", + "Eta_Ge=1 \n", + "VT=26/1000 #V\n", + "print \"Part(i) : \" \n", + "d_logIoBYdt_Ge=m_Ge/T+EGO_Ge/(Eta_Ge*T*VT) #per degree C\n", + "print \"d(log(Io))/dt for Ge = %0.2f per degree C \"%d_logIoBYdt_Ge \n", + "d_logIoBYdt_Si=m_Si/T+EGO_Si/(Eta_Si*T*VT) #per degree C\n", + "print \"d(log(Io))/dt for Si = %0.2f per degree C\"%d_logIoBYdt_Si\n", + "print \"\\nPart(ii) : \" \n", + "V=0.2 #Volt\n", + "dVBYdt_Ge=V/T-Eta_Ge*VT*d_logIoBYdt_Ge \n", + "print \"dV/dt for Si = %0.1f mV per degree C\"%(dVBYdt_Ge*1000)\n", + "V=0.6 #Volt\n", + "dVBYdt_Si=V/T-Eta_Si*VT*d_logIoBYdt_Si\n", + "print \"dV/dt for Si = %0.1f mV per degree C\"%(dVBYdt_Si*1000)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Part(i) : \n", + "d(log(Io))/dt for Ge = 0.11 per degree C \n", + "d(log(Io))/dt for Si = 0.08 per degree C\n", + "\n", + "Part(ii) : \n", + "dV/dt for Si = -2.1 mV per degree C\n", + "dV/dt for Si = -2.3 mV per degree C\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex 2.12.1 Page 2-85" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "#Given : \n", + "NA=4*10**20 #per m**3\n", + "Vj=0.2 #Volt\n", + "V1=-1 #Volts\n", + "V2=-5 #Volts\n", + "epsilon_r=16 #for Ge\n", + "epsilon_o=8.85*10**-12 #permitivity\n", + "q=1.6*10**-19 #Coulomb\n", + "W1=sqrt(2*epsilon_r*epsilon_o*(Vj-V1)/q/NA) #m\n", + "print \"Width of depletion region = %0.2f micro meter \"%(W1*10**6) \n", + "W2=sqrt(2*epsilon_r*epsilon_o*(Vj-V2)/q/NA) #m\n", + "print \"New value of Width of depletion region = %0.2f micro meter \"%(W2*10**6) " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Width of depletion region = 2.30 micro meter \n", + "New value of Width of depletion region = 4.80 micro meter \n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex 2.12.2 Page 2-86" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given : \n", + "NA=4*10**20 #per m**3\n", + "Vj=0.2 #Volt\n", + "V1=-1 #Volts\n", + "V2=-5 #Volts\n", + "A=0.8*10**-6 #m**2\n", + "epsilon_r=16 #for Ge\n", + "epsilon_o=8.85*10**-12 #permitivity\n", + "q=1.6*10**-19 #Coulomb\n", + "W1=sqrt(2*epsilon_r*epsilon_o*(Vj-V1)/q/NA) #m\n", + "CT1=epsilon_r*epsilon_o*A/W1 #\n", + "print \"Transition capacitance = %0.2f pF\"%(CT1*10**12)\n", + "W2=sqrt(2*epsilon_r*epsilon_o*(Vj-V2)/q/NA) #m\n", + "CT2=epsilon_r*epsilon_o*A/W2 #\n", + "print \"New value of Transition capacitance = %0.1f pF\"%(CT2*10**12)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Transition capacitance = 49.16 pF\n", + "New value of Transition capacitance = 23.6 pF\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex 2.12.3 Page 2-87" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given : \n", + "NA=3*10**20 #per m**3\n", + "Vj=0.2 #Volt\n", + "V=-10 #Volts\n", + "A=1*10**-6 #m**2\n", + "epsilon_r=16 #for Ge\n", + "epsilon_o=8.854*10**-12 #permitivity\n", + "q=1.6*10**-19 #Coulomb\n", + "W=sqrt(2*epsilon_r*epsilon_o*(Vj-V)/q/NA) #m\n", + "print \"Width of depletion region = %0.2f micro meter\"%(W*10**6)\n", + "CT=epsilon_r*epsilon_o*A/W #\n", + "print \"Transition capacitance = %0.2f pF\"%(CT*10**12)\n", + "#Answer is wrong in the textbook." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Width of depletion region = 7.76 micro meter\n", + "Transition capacitance = 18.26 pF\n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex 2.12.4 Page 2-88" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given : \n", + "W=2*10**-4*10**-2 #m\n", + "A=1*10**-6 #m**2\n", + "epsilon_r=16 #for Ge\n", + "epsilon_o=8.854*10**-12 #permitivity\n", + "q=1.6*10**-19 #Coulomb\n", + "CT=epsilon_r*epsilon_o*A/W #\n", + "print \"Barrier capacitance = %0.1f pF\"%(CT*10**12)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Barrier capacitance = 70.8 pF\n" + ] + } + ], + "prompt_number": 26 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex 2.12.5 Page 2-88" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "#Given : \n", + "Vj=0.5 #Volt\n", + "V=-4.5 #Volt\n", + "rho_p=5*10**-2 #ohm-m\n", + "epsilon_r=12 #for Si\n", + "epsilon_o=8.854*10**-12 #permitivity\n", + "q=1.6*10**-19 #Coulomb\n", + "CT=100*10**-12 #F\n", + "mu_p=500*10**-4 #m**2/V-s\n", + "sigma_p=1/rho_p #(ohm-m)**-1\n", + "NA=sigma_p/q/mu_p #per m**3\n", + "W=sqrt(2*epsilon_r*epsilon_o*(Vj-V)/q/NA) #m\n", + "A=CT*W/(epsilon_r*epsilon_o) #\n", + "r=sqrt(A/pi) #m\n", + "D=2*r #m\n", + "print \"Diameter = %0.2f micro meter \"%(D*10**6) \n", + "#Answer is wrong in the textbook. Sqrt is not taken while calculatng W value and also other mistakes." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Diameter = 1397.53 micro meter \n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Ex 2.12.6 Page 2-90" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given : \n", + "Eta=2 #for Si\n", + "T=300 #K\n", + "VT=26/1000 #V\n", + "IbyIo=0.9 \n", + "#part (i)\n", + "V=log(IbyIo+1)*Eta*VT #Volt\n", + "print \"Value of reverse voltage = %0.2f mV \" %(V*1000)\n", + "#part (ii)\n", + "VF=0.2 #Volt\n", + "VR=-0.2 #Volt\n", + "IFbyIR=(exp(VF/Eta/VT)-1)/(exp(VR/Eta/VT)-1) \n", + "print \"Ratio of forward bias current to reverse saturation current : %0.2f\" %IFbyIR\n", + "#Answer is wrong in the textbook." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Value of reverse voltage = 33.38 mV \n", + "Ratio of forward bias current to reverse saturation current : -46.81\n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex 2.12.7 Page 2-91" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import numpy as np\n", + "#Given : \n", + "Vs=100 #V\n", + "Rf1=20 #ohm\n", + "Vgamma1=0.2 #Volts\n", + "Rf2=15 #ohm\n", + "Vgamma2=0.6 #Volts\n", + "Vb_Ge=0.2 #Volts\n", + "Vb_Si=0.6 #Volts\n", + "R1=10*10**3 #ohm\n", + "R2=1*10**3 #ohm\n", + "#Case(i)\n", + "Imax=Vs/R1 #A\n", + "#D1 ON & D2 off\n", + "V=Vb_Ge+Rf1*Imax #Volt\n", + "#D2 off as V