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diff --git a/backup/Modern_Physics_version_backup/Chapter3.ipynb b/backup/Modern_Physics_version_backup/Chapter3.ipynb deleted file mode 100755 index 72876741..00000000 --- a/backup/Modern_Physics_version_backup/Chapter3.ipynb +++ /dev/null @@ -1,298 +0,0 @@ -{ - "metadata": { - "name": "Chapter3" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 3:The particle like properties of electromagnetic radiation" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.1 Page 69" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#initiation of variable\n", - "from math import sin,pi\n", - "w=0.250; theta=26.3;n=1 # n=1 for hydrogen atom and rest all are given values\n", - "\n", - "#calculation\n", - "d=n*w/(2*sin(theta*pi/180)); # bragg's law\n", - "\n", - "#result\n", - "print \"Hence the atomic spacing in nm is\",round(d,3);" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Hence the atomic spacing in nm is 0.282\n" - ] - } - ], - "prompt_number": 1 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.2 Page 73" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#initiation of variable\n", - "from math import pi,sin\n", - "I=120.0;r=0.1*10**-9;Eev=2.3 #I-intensity in W/m^2 r in m & E in electron volt\n", - "A=pi*r**2;K=1.6*10**-19; # A=area and K is conversion factor from ev to joules\n", - "\n", - "#calculation\n", - "t= Eev*K/(I*A); #time interval\n", - "\n", - "#result\n", - "print \"The value of time interval was found out to be in sec is\",round(t,3);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The value of time interval was found out to be in sec is 0.098\n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.3 Page 76" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#initiation of variable\n", - "from math import pi,sin\n", - "w=650.0*10**-9;h=6.63*10**-34;c=3*10**8; #given values and constant taken in comfortable units\n", - "\n", - "#calculation\n", - "E=h*c/w; \n", - "E1=E/(1.6*10**-19);\n", - "\n", - "#result\n", - "print \"The Energy of the electron in J \",E,\"which is equivalent to in eV is \", round(E1,3);\n", - "print \"The momentum of electron is p=E/c i.e is \", round(E1,3);\n", - "\n", - "#part b\n", - "E2=2.40; #given energy of photon.\n", - "\n", - "#calculation\n", - "w2=h*c*10**9/(E2*1.6*10**9); #converting the energy in to eV and nm \n", - "\n", - "#result\n", - "print \"The wavelength of the photon in m is\",round(w2*10**28,0)" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The Energy of the electron in J 3.06e-19 which is equivalent to in eV is 1.912\n", - "The momentum of electron is p=E/c i.e is 1.912\n", - "The wavelength of the photon in m is 518.0\n" - ] - } - ], - "prompt_number": 8 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.4 Page 78" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#initiation of variable\n", - "hc=1240.0; phi=4.52 #both the values are in eV\n", - "\n", - "#calcualtion\n", - "w1=hc/phi; \n", - "\n", - "#result\n", - "print \"The cutoff wavelength of the tungsten metal in nm is \",round(w1,3);\n", - "\n", - "#part b\n", - "w2=198.0; #given value of wavelength \n", - "\n", - "#calculation\n", - "Kmax=(hc/w2)-phi;\n", - "\n", - "#result\n", - "print 'The max value of kinetic energy in eV is',round(Kmax,3);\n", - "\n", - "#part c\n", - "Vs=Kmax;\n", - "\n", - "#result\n", - "print \"The numerical value of the max kinetic energy is same as stopping potential in volts.Hence in V is\",round(Vs,3);" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The cutoff wavelength of the tungsten metal in fnm is 274.336\n", - "The max value of kinetic energy in eV is 1.743\n", - "The numerical value of the max kinetic energy is same as stopping potential in volts.Hence in V is 1.743\n" - ] - } - ], - "prompt_number": 5 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.5 Page 80" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#initiation of variable\n", - "T1=293.0; Kw=2.898*10**-3;\n", - "\n", - "#calculation\n", - "w1=Kw/T1;\n", - "\n", - "#result\n", - "print \"The wavelength at which emits maximum radiation in um. is\",round(w1*10**6,3);\n", - "\n", - "#part b\n", - "w2=650.0*10**-9; \n", - "T2=Kw/w2;\n", - "\n", - "#result\n", - "print 'The temperature of the object must be raised to in K. is',round(T2,3);\n", - "\n", - "#part c\n", - "x=(T2/T1)**4; \n", - "\n", - "#result\n", - "print \"Thus the thermal radiation at higher temperature in times the room (lower) tempertaure. is\",round(x,3);" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The wavelength at which emits maximum radiation in um. is 9.891\n", - "The temperature of the object must be raised to in K. is 4458.462\n", - "Thus the thermal radiation at higher temperature in times the room (lower) tempertaure. is 53612.939\n" - ] - } - ], - "prompt_number": 8 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.6 Page 82" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#initiation of variable\n", - "#part a\n", - "from math import cos, sin, pi,atan\n", - "w1=0.24;wc=0.00243;theta=60.0; #given values w=wavelength(lambeda)\n", - "\n", - "#calculation\n", - "w2=w1+(wc*(1-cos(theta*pi/180))); \n", - "\n", - "#result\n", - "print \"The wavelength of x-rays after scattering in nm is\",round(w2,5);\n", - "\n", - "#part b;\n", - "hc=1240;\n", - "E2=hc/w2;E1=hc/w1; \n", - "\n", - "#result\n", - "print \"The energy of scattered x-rays in eV is\",round(E2,3);\n", - "\n", - "#part c\n", - "K= E1-E2; #The kinetic energy is the difference in the energy before and after the collision;\n", - "\n", - "print \"The kinetic energy of the x-rays in eV is\",round(K,3);\n", - "\n", - "#part d\n", - "phi2=atan(E2*sin(theta*pi/180)/(E1-E2*cos(theta*pi/180)))\n", - "\n", - "#result\n", - "print \"The direction of the scattered eletron in degrees is\",round(phi2*180/pi,3);" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The wavelength of x-rays after scattering in nm is 0.24121\n", - "The energy of scattered x-rays in eV is 5140.642\n", - "The kinetic energy of the x-rays in eV is 26.025\n", - "The direction of the scattered eletron in degrees is 59.749\n" - ] - } - ], - "prompt_number": 9 - } - ], - "metadata": {} - } - ] -}
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