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diff --git a/Engineering_Physics/chapter1_2.ipynb b/Engineering_Physics/chapter1_2.ipynb deleted file mode 100755 index bd2e1aac..00000000 --- a/Engineering_Physics/chapter1_2.ipynb +++ /dev/null @@ -1,1232 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:18ac31f959977ef2080ed3a1b1a6990ce93e604dcfb0f72ab45c0c28a2428e0e" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Quantum Mechanics and Quantum Computing" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example number 1.1, Page number 41" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "#Variable declaration\n", - "c=3*10**8 #velocity of light in m/s\n", - "h=6.626*10**-34 #planks constant \n", - "m=1.67*10**-27 #mass of proton\n", - "\n", - "#Calculation\n", - "v=c/10 #velocity of proton\n", - "lamda=h/(m*v) #de Broglie wave length\n", - "\n", - "#Result\n", - "print(\"the de Broglie wavelength in m is \",lamda);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "('the de Broglie wavelength in m is ', 1.3225548902195607e-14)\n" - ] - } - ], - "prompt_number": 17 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example number 1.2, Page number 42" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - " \n", - "#importing modules\n", - "import math\n", - "\n", - "#Variable declaration\n", - "V=400; #potential in Volts\n", - "\n", - "#Calculation\n", - "lamda=12.56/math.sqrt(V); #de Broglie wavelength\n", - "\n", - "#Result\n", - "print(\"The de Broglie wavelength in Armstrong is\",lamda);\n", - "\n", - "#answer given in the book is wrong" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "('The de Broglie wavelength in Armstrong is', 0.628)\n" - ] - } - ], - "prompt_number": 19 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example number 1.3, Page number 42\n" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - " \n", - "#importing modules\n", - "import math\n", - "\n", - "#Variable declaration\n", - "m=1.674*10**(-27); #mass of neutron in kg\n", - "h=6.626*10**(-34);\n", - "E=0.025; #kinetic energy in eV\n", - "\n", - "#Calculation\n", - "Ej=E*1.6*10**-19; #kinetic energy in J\n", - "lamda=h/math.sqrt(2*m*Ej); #de Broglie wavelength\n", - "lamdaA=lamda*10**10; #converting wavelength from m to Armstrong\n", - "lamdaA=math.ceil(lamdaA*10**3)/10**3; #rounding off to 3 decimals\n", - "\n", - "#Result\n", - "print(\"The de Broglie wavelength in metres is\",lamda);\n", - "print(\"The de Broglie wavelength in Armstrong is\",lamdaA);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "('The de Broglie wavelength in metres is', 1.81062582829353e-10)\n", - "('The de Broglie wavelength in Armstrong is', 1.811)\n" - ] - } - ], - "prompt_number": 20 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example number 1.4, Page number 42" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - " \n", - "#importing modules\n", - "import math\n", - "\n", - "#Variable declaration\n", - "V=1600; #potential in Volts\n", - "\n", - "#Calculation\n", - "lamda=12.56/math.sqrt(V); #de Broglie wavelength\n", - "lamda=math.ceil(lamda*10**2)/10**2; #rounding off to 2 decimals\n", - "\n", - "#Result\n", - "print(\"The de Broglie wavelength in Armstrong is\",lamda);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "('The de Broglie wavelength in Armstrong is', 0.32)\n" - ] - } - ], - "prompt_number": 21 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example number 1.5, Page number 42" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - " \n", - "#importing modules\n", - "import math\n", - "\n", - "#Variable declaration\n", - "deltax=0.2; #distance in armstrong\n", - "h=6.626*10**(-34);\n", - "\n", - "#Calculation\n", - "delta_xm=deltax*10**-10; #distance in m\n", - "delta_p=h/(2*math.pi*delta_xm);\n", - "\n", - "#Result\n", - "print(\"The uncertainity in momentum of electron in kg m/sec is\",delta_p);" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "('The uncertainity in momentum of electron in kg m/sec is', 5.2728032646344916e-24)\n" - ] - } - ], - "prompt_number": 22 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example number 1.6, Page number 43" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - " \n", - "#importing modules\n", - "import math\n", - "\n", - "#Variable declaration\n", - "n1=1;\n", - "n2=1;\n", - "n3=1; #values in lowest energy\n", - "h=6.62*10**(-34);\n", - "M=9.1*10**-31; #mass in kg\n", - "L=0.1; #side in nm\n", - "\n", - "#Calculation\n", - "L=L*10**-9; #side in m\n", - "n=(n1**2)+(n2**2)+(n3**2);\n", - "E1=(n*h**2)/(8*M*L**2); #energy in j\n", - "E1eV=E1/(1.6*10**-19); #energy in eV\n", - "E1eV=math.ceil(E1eV*10)/10; #rounding off to 1 decimals\n", - "\n", - "#Result\n", - "print(\"lowest energy of electron in Joule is\",E1);\n", - "print(\"lowest energy of electron is eV\",E1eV);\n", - "\n", - "#answer for lowest energy in eV given in the book is wrong" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "('lowest energy of electron in Joule is', 1.8059505494505486e-17)\n", - "('lowest energy of electron is eV', 112.9)\n" - ] - } - ], - "prompt_number": 23 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example number 1.7, Page number 43" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - " \n", - "#importing modules\n", - "import math\n", - "\n", - "#Variable declaration\n", - "M=9.1*10**(-31); #mass of electron in kg\n", - "h=6.66*10**(-34);\n", - "E=2000; #kinetic energy in eV\n", - "\n", - "#Calculation\n", - "Ej=E*1.6*10**-19; #kinetic energy in J\n", - "lamda=h/math.sqrt(2*M*Ej); #de Broglie wavelength\n", - "lamdaA=lamda*10**9; #converting wavelength from m to nm\n", - "lamdaA=math.ceil(lamdaA*10**3)/10**3; #rounding off to 3 decimals\n", - "\n", - "#Result\n", - "print(\"The de Broglie wavelength in nm is\",lamdaA);" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "('The de Broglie wavelength in nm is', 0.028)\n" - ] - } - ], - "prompt_number": 24 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example number 1.8, Page number 43" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - " \n", - "#importing modules\n", - "import math\n", - "\n", - "#Variable declaration\n", - "n=1; #for minimum energy\n", - "h=6.626*10**(-34);\n", - "m=9.1*10**-31; #mass in kg\n", - "L=4*10**-10; #size in m\n", - "\n", - "#Calculation\n", - "E1=(n*h**2)/(8*m*L**2); #energy in j\n", - "\n", - "#Result\n", - "print(\"lowest energy of electron in Joule is\",E1);\n", - "\n", - "#answer given in the book is wrong" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "('lowest energy of electron in Joule is', 3.7692201236263733e-19)\n" - ] - } - ], - "prompt_number": 23 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example number 1.9, Page number 44" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - " \n", - "#importing modules\n", - "import math\n", - "from __future__ import division\n", - "\n", - "#Variable declaration\n", - "h=6.626*10**(-34);\n", - "m=9.1*10**-31; #mass in kg\n", - "lamda=1.66*10**-10; #wavelength in m\n", - "\n", - "#Calculation\n", - "v=h/(m*lamda); #velocity in m/sec\n", - "v_km=v*10**-3; #velocity in km/sec\n", - "E=(1/2)*m*v**2; #kinetic energy in joule\n", - "EeV=E/(1.6*10**-19); #energy in eV\n", - "EeV=math.ceil(EeV*10**3)/10**3; #rounding off to 3 decimals\n", - "\n", - "#Result\n", - "print(\"velocity of electron in m/sec is\",round(v));\n", - "print(\"velocity of electron in km/sec is\",round(v_km));\n", - "print(\"kinetic energy of electron in Joule is\",E);\n", - "print(\"kinetic energy of electron in eV is\",EeV);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "('velocity of electron in m/sec is', 4386337.0)\n", - "('velocity of electron in km/sec is', 4386.0)\n", - "('kinetic energy of electron in Joule is', 8.754176510091736e-18)\n", - "('kinetic energy of electron in eV is', 54.714)\n" - ] - } - ], - "prompt_number": 26 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example number 1.10, Page number 44" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - " \n", - "#importing modules\n", - "import math\n", - "\n", - "#Variable decleration\n", - "V=15; #potential in kV\n", - "\n", - "#Calculation\n", - "v=V*10**3; #potential in V\n", - "lamda=12.26/math.sqrt(v); #de Broglie wavelength\n", - "lamda=math.ceil(lamda*10**2)/10**2 #rounding off to 2 decimals\n", - "\n", - "#result\n", - "print(\"The de Broglie wavelength in Armstrong is\",lamda);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "('The de Broglie wavelength in Armstrong is', 0.11)\n" - ] - } - ], - "prompt_number": 27 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example number 1.11, Page number 44\n" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - " \n", - "#importing modules\n", - "import math\n", - "\n", - "#Calculation\n", - "m=1.675*10**-27; #mass of neutron in kg\n", - "h=6.626*10**-34;\n", - "E=10; #kinetic energy in keV\n", - "\n", - "#Calculation\n", - "EeV=E*10**3; #Energy in eV\n", - "Ej=EeV*1.6*10**-19; #kinetic energy in J\n", - "v=math.sqrt(2*Ej/m); #velocity in m/s\n", - "lamda=h/(m*v); #de broglie wavelength in m\n", - "lamda_A=lamda*10**10; #de broglie wavelength in armstrong\n", - "lamda_A=math.ceil(lamda_A*10**4)/10**4 #rounding off to 4 decimals\n", - "\n", - "#Result\n", - "print(\"The velocity in m/sec is\",round(v));\n", - "print(\"The de Broglie wavelength in metres is\",lamda);\n", - "print(\"The de Broglie wavelength in Armstrong is\",lamda_A);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "('The velocity in m/sec is', 1382189.0)\n", - "('The de Broglie wavelength in metres is', 2.861996093951046e-13)\n", - "('The de Broglie wavelength in Armstrong is', 0.0029)\n" - ] - } - ], - "prompt_number": 28 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example number 1.12, Page number 45" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - " \n", - "#importing modules\n", - "import math\n", - "\n", - "#Variable decleration\n", - "m=9.1*10**-31; #mass of electron in kg\n", - "h=6.6*10**-34;\n", - "E=2; #kinetic energy in keV\n", - "\n", - "#Calculation\n", - "EeV=E*10**3; #Energy in eV\n", - "Ej=EeV*1.6*10**-19; #kinetic energy in J\n", - "p=math.sqrt(2*m*Ej); #momentum\n", - "lamda=h/p; #de broglie wavelength in m\n", - "lamda_A=lamda*10**10; #de broglie wavelength in armstrong\n", - "lamda_A=math.ceil(lamda_A*10**4)/10**4 #rounding off to 4 decimals\n", - "\n", - "#Result\n", - "print(\"The de Broglie wavelength in metres is\",lamda);\n", - "print(\"The de Broglie wavelength in Armstrong is\",lamda_A);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "('The de Broglie wavelength in metres is', 2.7348483695436575e-11)\n", - "('The de Broglie wavelength in Armstrong is', 0.2735)\n" - ] - } - ], - "prompt_number": 29 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example number 1.13, Page number 45" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - " \n", - "#import module\n", - "import math\n", - "\n", - "#Variable decleration\n", - "m=1.676*10**-27; #mass of neutron in kg\n", - "h=6.62*10**-34;\n", - "E=0.025; #kinetic energy in eV\n", - "\n", - "#Calculation\n", - "Ej=E*1.6*10**-19; #kinetic energy in J\n", - "v=math.sqrt(2*Ej/m); #velocity in m/s\n", - "lamda=h/(m*v); #wavelength in m\n", - "lamda_A=lamda*10**10; #de broglie wavelength in armstrong\n", - "lamda_A=math.ceil(lamda_A*10**5)/10**5 #rounding off to 5 decimals\n", - "\n", - "#Result\n", - "print(\"The neutrons wavelength in metres is\",lamda);\n", - "print(\"The wavelength in Armstrong is\",lamda_A);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "('The neutrons wavelength in metres is', 1.8079065940980725e-10)\n", - "('The wavelength in Armstrong is', 1.80791)\n" - ] - } - ], - "prompt_number": 30 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example number 1.14, Page number 45" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - " \n", - "#import module\n", - "import math\n", - "\n", - "#Variable decleration\n", - "V=10; #potential in kV\n", - "\n", - "#Calculation\n", - "V=V*10**3; #potential in V\n", - "lamda=12.26/math.sqrt(V); #wavelength\n", - "\n", - "#Result\n", - "print(\"The wavelength in Armstrong is\",lamda);" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "('The wavelength in Armstrong is', 0.1226)\n" - ] - } - ], - "prompt_number": 31 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example number 1.15, Page number 45" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - " \n", - "#import module\n", - "import math\n", - "\n", - "#Varialble decleration\n", - "h=6.626*10**-34;\n", - "m=9.1*10**-31; #mass in kg\n", - "l=1; #width in armstrong\n", - "\n", - "#Calculation\n", - "L=l*10**-10; #width in m\n", - "#permitted electron energies En=(n**2*h**2)/(8*m*L**2)\n", - "#let X = h**2/(8*m*L**2)\n", - "X = h**2/(8*m*L**2); #energy in J\n", - "XeV=X/(1.6*10**-19); #energy in eV\n", - "#in the 1st level n1=1\n", - "n1=1;\n", - "E1=(n1**2)*XeV; #energy in eV\n", - "\n", - "#in second level n2=2\n", - "n2=2;\n", - "E2=(n2**2)*XeV; #energy in eV\n", - "#in third level n3=\n", - "n3=3;\n", - "E3=(n3**2)*XeV; #energy in eV\n", - "\n", - "#Result\n", - "print(\"minimum energy the electron can have in eV is\",round(E1));\n", - "print(\"other values of energy are in eV and in eV\",round(E2),round(E3));\n", - "\n", - "#answers given in the book are wrong" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "('minimum energy the electron can have in eV is', 38.0)\n", - "('other values of energy are in eV and in eV', 151.0, 339.0)\n" - ] - } - ], - "prompt_number": 32 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example number 1.16, Page number 46\n" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - " \n", - "#import module\n", - "import math\n", - "\n", - "#Variable decleration\n", - "n=1; #lowest state\n", - "L=10; #width in armstrong\n", - "\n", - "#Calculation\n", - "L=L*10**-10; #width in m\n", - "x=L/2;\n", - "delta_x=1; #interval in armstrong\n", - "delta_x=delta_x*10**-10; #interval in m\n", - "psi1=(math.sqrt(2/L))*math.sin(math.pi*x/L);\n", - "A=psi1**2;\n", - "p=A*delta_x;\n", - "p=math.ceil(p*10)/10; #de broglie wavelength in armstrong\n", - "\n", - "#Result\n", - "print(\"probability of finding the particle is \",p);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "('probability of finding the particle is ', 0.2)\n" - ] - } - ], - "prompt_number": 33 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example number 1.17, Page number 46" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - " \n", - "#import module\n", - "import math\n", - "from __future__ import division\n", - "\n", - "#Variable decleration\n", - "d=970; #density of Na in kg/m^3\n", - "n=6.02*10**26;\n", - "h=6.62*10**(-34);\n", - "m=9.1*10**-31; #mass in kg\n", - "w=23; #atomic weight\n", - "\n", - "#Calculation\n", - "N=(d*n)/w; #number of atoms per m^3\n", - "A=(h**2)/(8*m);\n", - "B=(3*N)/math.pi;\n", - "Ef=A*B**(2/3);\n", - "EfeV=Ef/(1.6*10**-19);\n", - "EfeV=math.ceil(EfeV*10**2)/10**2 #rounding of to 2 decimals\n", - "\n", - "#Result\n", - "print(\"fermi energy of Na in eV is\",EfeV);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "('fermi energy of Na in eV is', 3.16)\n" - ] - } - ], - "prompt_number": 34 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example number 1.18, Page number 46" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - " \n", - "#import module\n", - "import math\n", - "from __future__ import division\n", - "\n", - "#Variable decleration\n", - "n1=1;\n", - "n2=1;\n", - "n3=1; #values in lowest energy\n", - "h=6.62*10**(-34);\n", - "m=9.1*10**-31; #mass in kg\n", - "L=0.1; #side in nm\n", - "\n", - "#Calculation\n", - "L=L*10**-9; #side in m\n", - "n=(n1**2)+(n2**2)+(n3**2);\n", - "E1=(n*h**2)/(8*m*L**2); #energy in j\n", - "E1eV=E1/(1.6*10**-19); #energy in eV\n", - "E1eV=math.ceil(E1eV*10**1)/10**1 #rounding off to 2 decimals\n", - "\n", - "#Result\n", - "print(\"lowest energy of electron in Joule is\",E1);\n", - "print(\"lowest energy of electron in eV is\",E1eV);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "('lowest energy of electron in Joule is', 1.8059505494505486e-17)\n", - "('lowest energy of electron in eV is', 112.9)\n" - ] - } - ], - "prompt_number": 35 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example number 1.19, Page number 47" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - " \n", - "#import module\n", - "import math\n", - "from __future__ import division\n", - "\n", - "#Variable decleration\n", - "mn=1.676*10**-27; #mass of neutron in kg\n", - "me=9.1*10**-31; #mass of electron in kg\n", - "h=6.62*10**-34;\n", - "c=3*10**8; #velocity of light in m/sec\n", - "\n", - "#Calculation\n", - "En=2*me*c**2;\n", - "lamda=h/math.sqrt(2*mn*En); #wavelength in m\n", - "lamda_A=lamda*10**10; #converting lamda from m to A\n", - "lamda_A=math.ceil(lamda_A*10**6)/10**6 #rounding off to 6 decimals\n", - "\n", - "#Result\n", - "print(\"The de broglie wavelength in Angstrom is\",lamda_A);" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "('The de broglie wavelength in Angstrom is', 0.000283)\n" - ] - } - ], - "prompt_number": 36 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example number 1.20, Page number 47 ***************************************************************************" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "#import module\n", - "import math\n", - "from __future__ import division\n", - "\n", - "#Variable decleration\n", - "n2=2; #second quantum state\n", - "n4=4; #fourth quantum state\n", - "h=6.626*10**-34;\n", - "m=9.1*10**-31; #mass in kg\n", - "a=2; #potential box length in armstrong\n", - "\n", - "#Calculation\n", - "a=a*10**-10; #length in m\n", - "A=n2**2*h**2;\n", - "B=8*m*a**2;\n", - "E2=A/B; #energy in j\n", - "E2eV=E2/(1.6*10**-19); #energy in eV\n", - "C=n4**2*h**2;\n", - "E4=C/B; #energy in j\n", - "E4eV=E4/(1.6*10**-19); #energy in eV\n", - "\n", - "#Result\n", - "print(\"energy corresponding to second quantum state in Joule is\",E2);\n", - "print(\"energy corresponding to second quantum state in eV is\",E2eV);\n", - "print(\"energy corresponding to fourth quantum state in Joule is\",E4);\n", - "print(\"energy corresponding to fourth quantum state in eV is\",E4eV);\n", - "\n", - "\n", - "#answers given in the book are wrong" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "('energy corresponding to second quantum state in Joule is', 6.030752197802197e-18)\n", - "('energy corresponding to second quantum state in eV is', 37.69220123626373)\n", - "('energy corresponding to fourth quantum state in Joule is', 2.412300879120879e-17)\n", - "('energy corresponding to fourth quantum state in eV is', 150.7688049450549)\n" - ] - } - ], - "prompt_number": 13 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example number 1.21, Page number 48 ***********" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - " \n", - "#import module\n", - "import math\n", - "from __future__ import division\n", - "\n", - "#Variable decleration\n", - "V=344; #accelerated voltage in V\n", - "n=1; #first reflection\n", - "theta=60; #glancing angle in degrees\n", - "\n", - "#Calculation\n", - "lamda=12.27/math.sqrt(V);\n", - "d=(n*lamda)/(2*math.sin(theta));\n", - "\n", - "#Result\n", - "print(\"The spacing of the crystal in Angstrom is\",lamda);" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "('The spacing of the crystal in Angstrom is', 0.6615540636030947)\n" - ] - } - ], - "prompt_number": 38 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example number 1.22, Page number 49 *************" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - " \n", - "#import module\n", - "import math\n", - "from __future__ import division\n", - "\n", - "#Variable decleration\n", - "n2=2; #second quantum state\n", - "n3=3; #fourth quantum state\n", - "h=6.626*10**-34;\n", - "m=9.1*10**-31; #mass in kg\n", - "a=1*10**-10; #width of potential well in m\n", - "\n", - "#Calculation\n", - "B=8*m*a**2;\n", - "E1=h**2/B; #ground state energy\n", - "E1eV=E1/(1.6*10**-19); #energy in eV\n", - "A=n2**2*h**2;\n", - "E2=A/B; #energy in j\n", - "E2eV=E2/(1.6*10**-19); #energy in eV\n", - "C=n3**2*h**2;\n", - "E3=C/B; #energy in j\n", - "E3eV=E3/(1.6*10**-19); #energy in eV\n", - "E1=math.ceil(E1*10**3)/10**3 #rounding off to 3 decimals\n", - "E1eV=math.ceil(E1eV*10**3)/10**3 #rounding off to 3 decimals\n", - "E2eV=math.ceil(E2eV*10**3)/10**3 #rounding off to 3 decimals\n", - "E3eV=math.ceil(E3eV*10**3)/10**3 #rounding off to 3 decimals\n", - "\n", - "#Result\n", - "print(\"ground state energy in Joule is\",E1);\n", - "print(\"ground state energy in eV is\",E1eV);\n", - "print(\"first energy state in eV is\",E2eV);\n", - "print(\"second energy state in eV is\",E3eV);\n", - "\n", - "#answers given in the book are wrong by one decimal" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "('ground state energy in Joule is', 0.001)\n", - "('ground state energy in eV is', 37.693)\n", - "('first energy state in eV is', 150.769)\n", - "('second energy state in eV is', 339.23)\n" - ] - } - ], - "prompt_number": 39 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example number 1.23, Page number 49" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - " \n", - "#import module\n", - "import math\n", - "\n", - "#Variable decleration\n", - "n3=3; #fourth quantum state\n", - "h=6.626*10**-34;\n", - "m=9.1*10**-31; #mass in kg\n", - "\n", - "\n", - "#ground state energy E1 = h**2/(8*m*a**2)\n", - "#second excited state E3 = (9*h**2)/(8*m*a**2)\n", - "#required energy E = E3-E1\n", - "#E = (9*h**2)/(8*m*a**2) - h**2/(8*m*a**2)\n", - "#E = (h**2/(8*m*a**2))*(9-1)\n", - "#therefore E = (8*h**2)/(8*m*a**2)\n", - "#hence E = (h**2)/(m*a**2)\n", - "\n", - "#Result \n", - "# the required energy is E = (h**2)/(m*a**2)" - ], - "language": "python", - "metadata": {}, - "outputs": [], - "prompt_number": 4 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example number 1.24, Page number 50" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - " \n", - "#import module\n", - "import math\n", - "from __future__ import division\n", - "\n", - "#Variable decleration\n", - "delta_x=10**-8; #length of box in m\n", - "h=6.626*10**-34;\n", - "m=9.1*10**-31; #mass in kg\n", - "\n", - "#Calculation\n", - "delta_v=h/(m*delta_x); #uncertainity in m/sec\n", - "delta_vk=delta_v*10**-3; #uncertainity in km/sec\n", - "delta_vk=math.ceil(delta_vk*10**2)/10**2 #rounding off to 2 decimals\n", - "\n", - "#Result\n", - "print(\"minimum uncertainity in velocity in m/sec is\",round(delta_v));\n", - "print(\"minimum uncertainity in velocity in km/sec is\",delta_vk);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "('minimum uncertainity in velocity in m/sec is', 72813.0)\n", - "('minimum uncertainity in velocity in km/sec is', 72.82)\n" - ] - } - ], - "prompt_number": 40 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example number 1.25, Page number 50" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - " \n", - "#import module\n", - "import math\n", - "from __future__ import division\n", - "\n", - "#Variable decleration\n", - "mp=1.6*10**-27; #mass of proton in kg\n", - "me=9.1*10**-31; #mass of electron in kg\n", - "h=6.626*10**(-34);\n", - "c=3*10**10; #velocity of light in m/sec\n", - "\n", - "#Calculation\n", - "Ep=me*c**2;\n", - "lamda=h/math.sqrt(2*mp*Ep); #wavelength in m\n", - "lamda_A=lamda*10**10; #converting lamda from m to A\n", - "\n", - "#Result\n", - "print(\"The de broglie wavelength in Angstrom is\",lamda_A);" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "('The de broglie wavelength in Angstrom is', 4.092931643497047e-06)\n" - ] - } - ], - "prompt_number": 41 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example number 1.26, Page number 51 *************************************************" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - " \n", - "#import module\n", - "import math\n", - "from __future__ import division\n", - "\n", - "#Variable decleration\n", - "m=1.675*10**(-27); #mass of neutron in kg\n", - "h=6.626*10**(-34);\n", - "n=1; #diffractive order\n", - "d=0.314; #spacing in nm\n", - "E=0.04; #kinetic energy in eV\n", - "\n", - "#Calculation\n", - "d=d*10**-9; #spacing in m\n", - "Ej=E*1.6*10**-19; #kinetic energy in J\n", - "lamda=h/math.sqrt(2*m*Ej); #de Broglie wavelength\n", - "lamdaA=lamda*10**9; #converting wavelength from m to nm\n", - "theta=math.asin((n*lamda)/(2*d));\n", - "print(\"The de Broglie wavelength in metres is\",lamda);\n", - "print(\"The de Broglie wavelength in nm is\",lamdaA);\n", - "print(\"glancing angle in degrees is\",theta);\n", - "\n", - "#answer given in the book is wrong" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "('The de Broglie wavelength in metres is', 1.4309980469755228e-10)\n", - "('The de Broglie wavelength in nm is', 0.1430998046975523)\n", - "('glancing angle in degrees is', 0.2298853909391574)\n" - ] - } - ], - "prompt_number": 16 - }, - { - "cell_type": "code", - "collapsed": false, - "input": [], - "language": "python", - "metadata": {}, - "outputs": [] - } - ], - "metadata": {} - } - ] -}
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