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diff --git a/Engineering_Physics/Chapter_4.ipynb b/Engineering_Physics/Chapter_4.ipynb deleted file mode 100755 index d93ccbff..00000000 --- a/Engineering_Physics/Chapter_4.ipynb +++ /dev/null @@ -1,743 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:2a55c0c681215b0dc959ddeda0187458e8ed07320f22e00a7385acd5044d2ee9" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Quantum Physics" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example number 4.1, Page number 133 " - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "import math\n", - "\n", - "#Variable declaration\n", - "h=6.63*10**-34; #plancks constant in Js\n", - "m0=9.1*10**-31; #mass of the electron in kg\n", - "c=3*10**8; #velocity of light in m/s\n", - "phi=135; #angle of scattering in degrees\n", - "phi=phi*0.0174532925 #converting degrees to radians \n", - "\n", - "#Calculation\n", - "delta_lamda=(h*(1-math.cos(phi)))/(m0*c);\n", - "\n", - "#Result\n", - "print(\"change in wavelength in metres is\",delta_lamda);" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "('change in wavelength in metres is', 4.1458307496867315e-12)\n" - ] - } - ], - "prompt_number": 6 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example number 4.2, Page number 134 " - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - " \n", - "import math\n", - "\n", - "#Variable declaration\n", - "h=6.63*10**-34; #plancks constant in Js\n", - "m0=9.1*10**-31; #mass of the electron in kg\n", - "c=3*10**8; #velocity of light in m/s\n", - "lamda=2; #wavelength in angstrom\n", - "lamdaA=lamda*10**-10; #converting lamda from Angstrom to m\n", - "phi=90; #angle of scattering in degrees\n", - "phi=phi*0.0174532925 #converting degrees to radians \n", - "\n", - "#Calculation\n", - "delta_lamda=(h*(1-math.cos(phi)))/(m0*c);\n", - "delta_lamda=delta_lamda*10**10; #converting delta_lamda from m to Angstrom\n", - "delta_lamda=math.ceil(delta_lamda*10**5)/10**5; #rounding off to 5 decimals\n", - "lamda_dash=delta_lamda+lamda;\n", - "lamdaA_dash=lamda_dash*10**-10; #converting lamda_dash from Angstrom to m\n", - "#energy E=h*new-h*new_dash\n", - "E=h*c*((1/lamdaA)-(1/lamdaA_dash));\n", - "EeV=E/(1.602176565*10**-19); #converting J to eV\n", - "EeV=math.ceil(EeV*10**3)/10**3; #rounding off to 3 decimals\n", - "new=c/lamda;\n", - "new_dash=c/lamda_dash;\n", - "theta=math.atan((h*new*math.sin(phi))/((h*new)-(h*new_dash*math.cos(phi))));\n", - "theta=theta*57.2957795; #converting radians to degrees\n", - "\n", - "#Result\n", - "print(\"change in compton shift in Angstrom is\",delta_lamda);\n", - "print(\"wavelength of scattered photons in Angstrom is\",lamda_dash);\n", - "print(\"energy of recoiling electron in J is\",E);\n", - "print(\"energy of recoiling electron in eV is\",EeV);\n", - "print(\"angle at which recoiling electron appears in degrees is\",int(theta));\n", - "\n", - "#answers given in the book are wrong" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "('change in compton shift in Angstrom is', 0.02429)\n", - "('wavelength of scattered photons in Angstrom is', 2.02429)\n", - "('energy of recoiling electron in J is', 1.1933272900621974e-17)\n", - "('energy of recoiling electron in eV is', 74.482)\n", - "('angle at which recoiling electron appears in degrees is', 45)\n" - ] - } - ], - "prompt_number": 10 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example number 4.3, Page number 135" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - " \n", - "import math\n", - "\n", - "#Variable declaration\n", - "h=6.626*10**-34; #plancks constant in Js\n", - "m0=9.1*10**-31; #mass of the electron in kg\n", - "c=3*10**8; #velocity of light in m/s\n", - "phi=60; #angle of scattering in degrees\n", - "phi=phi*0.0174532925; #converting degrees to radians\n", - "E=10**6; #energy of photon in eV\n", - "E=E*1.6*10**-19; #converting eV into J\n", - "\n", - "#Calculation\n", - "delta_lamda=(h*(1-math.cos(phi)))/(m0*c);\n", - "delta_lamda=delta_lamda*10**10; #converting metre to angstrom\n", - "delta_lamda=math.ceil(delta_lamda*10**4)/10**4; #rounding off to 4 decimals\n", - "lamda=(h*c)/E;\n", - "lamdaA=lamda*10**10; #converting metre to angstrom\n", - "lamda_dash=delta_lamda+lamdaA;\n", - "lamda_dash=math.ceil(lamda_dash*10**3)/10**3; #rounding off to 3 decimals\n", - "\n", - "#Result\n", - "print(\"compton shift in angstrom is\",delta_lamda);\n", - "print(\"energy of incident photon in m\",lamda);\n", - "print(\"wavelength of scattered photons in angstrom is\",lamda_dash);\n", - "\n", - "#answer for wavelength of scattered photon given in the book is wrong" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "('compton shift in angstrom is', 0.0122)\n", - "('energy of incident photon in m', 1.242375e-12)\n", - "('wavelength of scattered photons in angstrom is', 0.025)\n" - ] - } - ], - "prompt_number": 13 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example number 4.4, Page number 135" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - " \n", - "import math\n", - "\n", - "#Variable declaration\n", - "h=6.626*10**-34; #plancks constant in Js\n", - "c=3*10**8; #velocity of light in m/s\n", - "lamda=5893; #wavelength in angstrom\n", - "P=60; #output power in Watt\n", - "\n", - "#Calculation\n", - "lamda=lamda*10**-10; #wavelength in metre\n", - "E=(h*c)/lamda;\n", - "EeV=E/(1.602176565*10**-19); #converting J to eV\n", - "EeV=math.ceil(EeV*10**4)/10**4; #rounding off to 4 decimals\n", - "N=P/E;\n", - "\n", - "#Result\n", - "print(\"energy of photon in J is\",E);\n", - "print(\"energy of photon in eV is\",EeV);\n", - "print(\"number of photons emitted per se cond is\",N);\n", - "\n", - "#answer for energy in eV given in the book is wrong" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "('energy of photon in J is', 3.373154590191753e-19)\n", - "('energy of photon in eV is', 2.1054)\n", - "('number of photons emitted per se cond is', 1.7787503773015396e+20)\n" - ] - } - ], - "prompt_number": 15 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example number 4.5, Page number 136" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - " \n", - "import math\n", - "\n", - "#Variable declaration\n", - "h=6.626*10**-34; #plancks constant in Js\n", - "c=3*10**8; #velocity of light in m/s\n", - "lamda=10; #wavelength in angstrom\n", - "\n", - "#Calculation\n", - "lamda=lamda*10**-10; #wavelength in metre\n", - "E=(h*c)/lamda;\n", - "EeV=E/(1.602176565*10**-19); #converting J to eV\n", - "EeV=EeV*10**-3; #converting eV to keV\n", - "EeV=math.ceil(EeV*10**3)/10**3; #rounding off to 3 decimals\n", - "P=h/lamda;\n", - "M=h/(lamda*c);\n", - "\n", - "#Result\n", - "print(\"energy of photon in J is\",E);\n", - "print(\"energy of photon in keV is\",EeV);\n", - "print(\"momentum in kg m/sec is\",P);\n", - "print(\"mass of photon in kg is\",M);\n", - "\n", - "#answer for energy of photon in keV given in the book is wrong by 1 decimal" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "('energy of photon in J is', 1.9878e-16)\n", - "('energy of photon in keV is', 1.241)\n", - "('momentum in kg m/sec is', 6.626e-25)\n", - "('mass of photon in kg is', 2.2086666666666664e-33)\n" - ] - } - ], - "prompt_number": 18 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example number 4.6, Page number 136" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - " \n", - "import math\n", - "\n", - "#Variable declaration\n", - "h=6.626*10**-34; #plancks constant in Js\n", - "m=9.1*10**-31; #mass of the electron in kg\n", - "e=1.602*10**-19;\n", - "V=1.25; #potential difference in kV\n", - "\n", - "#Calculation\n", - "V=V*10**3; #converting kV to V\n", - "lamda=h/math.sqrt(2*m*e*V);\n", - "lamda=lamda*10**10; #converting metre to angstrom\n", - "lamda=math.ceil(lamda*10**4)/10**4; #rounding off to 4 decimals\n", - "\n", - "#Result\n", - "print(\"de Broglie wavelength in angstrom is\",lamda);" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "('de Broglie wavelength in angstrom is', 0.3471)\n" - ] - } - ], - "prompt_number": 21 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example number 4.7, Page number 136" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - " \n", - "import math\n", - "\n", - "#Variable declaration\n", - "E=45; #energy of electron in eV\n", - "E=E*1.6*10**-19; #energy in J\n", - "h=6.626*10**-34; #plancks constant in Js\n", - "m=9.1*10**-31; #mass of the electron in kg\n", - "\n", - "#Calculation\n", - "lamda=h/math.sqrt(2*m*E);\n", - "lamda=lamda*10**10; #converting metres to angstrom\n", - "lamda=math.ceil(lamda*10**4)/10**4; #rounding off to 4 decimals\n", - "\n", - "#Result\n", - "print(\"de Broglie wavelength in angstrom is\",lamda);" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "('de Broglie wavelength in angstrom is', 1.8305)\n" - ] - } - ], - "prompt_number": 24 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example number 4.8, Page number 137" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - " \n", - "import math\n", - "\n", - "#Variable declaration\n", - "v=10**7; #velocity of electron in m/sec\n", - "h=6.626*10**-34; #plancks constant in Js\n", - "m=9.1*10**-31; #mass of the electron in kg\n", - "\n", - "#Calculation\n", - "lamda=h/(m*v);\n", - "lamda=lamda*10**10; #converting metres to angstrom\n", - "lamda=math.ceil(lamda*10**4)/10**4; #rounding off to 4 decimals\n", - "\n", - "#Result\n", - "print(\"de Broglie wavelength in angstrom is\",lamda);" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "('de Broglie wavelength in angstrom is', 0.7282)\n" - ] - } - ], - "prompt_number": 25 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example number 4.9, Page number 137" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - " \n", - "import math\n", - "\n", - "#Variable declaration\n", - "V=1000; #potential difference in V\n", - "h=6.626*10**-34; #plancks constant in Js\n", - "m=1.67*10**-27; #mass of proton in kg\n", - "e=1.6*10**-19; #charge of electron in J\n", - "\n", - "#Calculation\n", - "lamda=h/math.sqrt(2*m*e*V);\n", - "\n", - "#Result\n", - "print(\"de Broglie wavelength of alpha particle in metre is\",lamda);" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "('de Broglie wavelength of alpha particle in metre is', 9.063964727801313e-13)\n" - ] - } - ], - "prompt_number": 26 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example number 4.10, Page number 138" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - " \n", - "import math\n", - "\n", - "#Variable declaration\n", - "L=25; #width of potential in armstrong\n", - "delta_x=0.05; #interval in armstrong\n", - "n=1; #particle is in its least energy\n", - "x=L/2; #particle is at the centre\n", - "pi=180; #angle in degrees\n", - "\n", - "#Calculation\n", - "pi=pi*0.0174532925; #angle in radians\n", - "L=L*10**-10; #width in m\n", - "delta_x=delta_x*10**-10; #interval in m\n", - "#probability P = integration of (A**2)*(math.sin(n*pi*x/L))**2*delta_x\n", - "#but A=math.sqrt(2/L)\n", - "#since the particle is in a small interval integration need not be applied\n", - "#therefore P=2*(L**(-1))*(math.sin(n*pi*x/L))**2*delta_x\n", - "P=2*(L**(-1))*((math.sin(n*pi*x/L))**2)*delta_x;\n", - "P=math.ceil(P*10**3)/10**3; #rounding off to 3 decimals\n", - "\n", - "#Result\n", - "print(\"probability of finding the particle is\",P);" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "('probability of finding the particle is', 0.004)\n" - ] - } - ], - "prompt_number": 27 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example number 4.11, Page number 138" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - " \n", - "import math\n", - "\n", - "#Variable declaration\n", - "n=1;\n", - "h=6.626*10**-34; #plancks constant in Js\n", - "m=9.1*10**-31; #mass of the electron in kg\n", - "L=1; #width of potential well in angstrom\n", - "\n", - "#Calculation\n", - "L=L*10**-10; #converting angstrom into metre\n", - "E=((n**2)*h**2)/(8*m*L**2);\n", - "EeV=E/(1.6*10**-19); #converting J to eV\n", - "EeV=math.ceil(EeV*10**3)/10**3; #rounding off to 3 decimals\n", - "\n", - "#Result\n", - "print(\"lowest energy of electron in J is\",E);\n", - "print(\"lowest energy of electron in eV is\",EeV);" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "('lowest energy of electron in J is', 6.030752197802197e-18)\n", - "('lowest energy of electron in eV is', 37.693)\n" - ] - } - ], - "prompt_number": 28 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example number 4.12, Page number 139" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - " \n", - "import math\n", - "\n", - "#Variable declaration\n", - "n=1;\n", - "h=6.626*10**-34; #plancks constant in Js\n", - "m=9.1*10**-31; #mass of the electron in kg\n", - "L=1; #width of potential well in angstrom\n", - "\n", - "#Calculation\n", - "L=L*10**-10; #converting angstrom into metre\n", - "E=(2*(n**2)*h**2)/(8*m*L**2);\n", - "E=E/(1.6*10**-19); #converting J to eV\n", - "E=math.ceil(E*10**3)/10**3; #rounding off to 3 decimals\n", - "\n", - "#Result\n", - "print(\"lowest energy of system in eV is\",E);" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "('lowest energy of system in eV is', 75.385)\n" - ] - } - ], - "prompt_number": 29 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example number 4.13, Page number 139" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - " \n", - "#importing modules\n", - "import math\n", - "\n", - "#Variable declaration\n", - "h=6.626*10**-34; #plancks constant in Js\n", - "m=9.1*10**-31; #mass of the electron in kg\n", - "L=1; #width of potential well in angstrom\n", - "\n", - "#Calculation\n", - "L=L*10**-10; #converting angstrom into metre\n", - "#according to pauli's exclusion principle, 1st electron occupies n1=1 and second electron occupies n2=2\n", - "n1=1;\n", - "n2=2;\n", - "E=((2*(n1**2)*h**2)/(8*m*L**2))+(((n2**2)*h**2)/(8*m*L**2));\n", - "E=E/(1.6*10**-19); #converting J to eV\n", - "E=math.ceil(E*10**3)/10**3; #rounding off to 3 decimals\n", - "\n", - "#Result\n", - "print(\"lowest energy of system in eV is\",E);\n", - "print(\"quantum numbers are\");\n", - "print(\"n=1,l=0,mL=0,mS=+1/2\");\n", - "print(\"n=1,l=0,mL=0,mS=-1/2\");\n", - "print(\"n=2,l=0,mL=0,mS=+1/2\");" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "('lowest energy of system in eV is', 226.154)\n", - "quantum numbers are\n", - "n=1,l=0,mL=0,mS=+1/2\n", - "n=1,l=0,mL=0,mS=-1/2\n", - "n=2,l=0,mL=0,mS=+1/2\n" - ] - } - ], - "prompt_number": 30 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example number 4.14, Page number 140" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - " \n", - "#Variable declaration\n", - "n=1;\n", - "h=6.626*10**-34; #plancks constant in Js\n", - "L=100; #width of potential well in angstrom\n", - "\n", - "#Calculation\n", - "L=L*10**-10; #converting angstrom into metre\n", - "E=0.025; #lowest energy in eV\n", - "E=E*(1.6*10**-19); #converting eV to J\n", - "m=((n**2)*h**2)/(8*E*L**2);\n", - "\n", - "#Result\n", - "print(\"mass of the particle in kg is\",m);" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "('mass of the particle in kg is', 1.3719961249999998e-31)\n" - ] - } - ], - "prompt_number": 31 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example number 4.15, Page number 141" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - " \n", - "#importing modules\n", - "import math\n", - "\n", - "#Variable declaration\n", - "k=1.38*10**-23;\n", - "T=6000; #temperature in K\n", - "h=6.626*10**-34; #plancks constant in Js\n", - "c=3*10**8; #velocity of light in m/s\n", - "lamda1=450; #wavelength in nm\n", - "lamda2=460; #wavelength in nm\n", - "\n", - "#Calculation\n", - "lamda1=lamda1*10**-9; #converting nm to metre\n", - "lamda2=lamda2*10**-9; #converting nm to metre\n", - "new1=c/lamda1;\n", - "new2=c/lamda2;\n", - "new=(new1+new2)/2;\n", - "A=math.exp((h*new)/(k*T));\n", - "rho_v=(8*math.pi*h*new**3)/(A*c**3);\n", - "\n", - "#Result\n", - "print(\"energy density of the black body in J/m^3 is\",rho_v);\n", - "\n", - "#answer given in the book is wrong" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "('energy density of the black body in J/m^3 is', 9.033622836188887e-16)\n" - ] - } - ], - "prompt_number": 32 - }, - { - "cell_type": "code", - "collapsed": false, - "input": [], - "language": "python", - "metadata": {}, - "outputs": [] - } - ], - "metadata": {} - } - ] -}
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