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diff --git a/backup/Modern_Physics_version_backup/Chapter6.ipynb b/backup/Modern_Physics_version_backup/Chapter6.ipynb new file mode 100755 index 00000000..53684185 --- /dev/null +++ b/backup/Modern_Physics_version_backup/Chapter6.ipynb @@ -0,0 +1,318 @@ +{ + "metadata": { + "name": "Chapter6", + "signature": "sha256:36f31f6870acf2c11b00274dcf34bd9e9879abf6f82026373900139ccc4b5799" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 6:The Rutherford Bohr Model" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.1 Page 178" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#initiation of variable\n", + "from math import sqrt, pi\n", + "R=0.1;Z=79.0; x=1.44; #x=e^2/4*pi*epsi0\n", + "zkR2=2*Z*x/R # from zkR2= (2*Z*e^2)*R^2/(4*pi*epsi0)*R^3\n", + "mv2=10.0*10**6; #MeV=>eV\n", + "\n", + "#calculation\n", + "theta=sqrt(3.0/4)*zkR2/mv2; #deflection angle\n", + "theta=theta*(180/pi); #converting to degrees\n", + "\n", + "#result\n", + "print\"Hence the average deflection angle per collision in degrees is\",round(theta,3 );" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Hence the average deflection angle per collision in degrees.is 0.011\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.2 Page 181" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#initiation of variable\n", + "from math import sin, cos, tan, sqrt, pi\n", + "Na=6.023*10**23;p=19.3;M=197.0;\n", + "n=Na*p/M; #The number of nuclei per atom\n", + "t=2*10**-6;Z=79;K=8*10**6;x=1.44; theta=90.0*pi/180; #x=e^2/4*pi*epsi0\n", + "b1=t*Z*x/tan(theta/2)/(2*K) #impact parameter b\n", + "f1=n*pi*b1**2*t #scattering angle greater than 90\n", + "\n", + "#result\n", + "print\"The fraction of alpha particles scattered at angles greater than 90 degrees is %.1e\" %f1;\n", + "\n", + "#part b\n", + "theta=45.0*pi/180;\n", + "b2=t*Z*x/tan(theta/2)/(2*K);\n", + "f2=n*pi*b2**2*t; #scattering angle greater than 45\n", + "fb=f2-f1 #scattering angle between 45 to 90\n", + "\n", + "#result\n", + "print\"The fraction of particles with scattering angle from 45 to 90 is %.1e\" %fb;" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The fraction of alpha particles scattered at angles greater than 90 degrees is 7.5e-05\n", + "The fraction of particles with scattering angle from 45 to 90 is 3.6e-04\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.3 Page 185" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#initiation of variable\n", + "from math import sin, cos, tan, sqrt, pi\n", + "Z=79.0;x=1.44;K=8.0*10**6;z=2; #where x=e^2/4*pi*epsi0;z=2 for alpha particles\n", + "\n", + "#calculation\n", + "d=z*x*Z/K; #distance\n", + "\n", + "#result\n", + "print \"The distance of closest approach in nm. is\",d*10**-9" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The distance of closest approasch in nm. is 2.844e-14\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.4 Page 188" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#initiation of variable\n", + "sl=820.1;n0=3.0; #given values\n", + "n=4;w=sl*(n**2/(n**2-n0**2)); \n", + "#result\n", + "print \"The 3 longest possible wavelengths in nm respectively are a.\",round(w,3),; \n", + "\n", + "#partb\n", + "n=5.0;w=sl*(n**2/(n**2-n0**2)); \n", + "\n", + "#result\n", + "print \"b. (in nm)\",round(w,3),;\n", + "\n", + "#partc\n", + "n=6.0;w=sl*(n**2/(n**2-n0**2));\n", + "\n", + "#result\n", + "print \"c. (in nm )\",round(w,3);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The 3 longest possible wavelengths in nm respectively are a. 1874.514 b. (in nm) 1281.406 c. (in nm ) 1093.467\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.5 Page 189" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#initiation of variable\n", + "sl=364.5;n=3.0; #given variables and various constants are declared in the subsequent steps wherever necessary \n", + "w1=sl*(n**2/(n**2-4)); #longest wavelength of balmer \n", + "c=3.0*10**8;\n", + "f1=c/(w1*10**-9); #corresponding freq.\n", + "n0=1.0;n=2.0; \n", + "\n", + "#calculation\n", + "w2=91.13*(n**2/(n**2-n0**2)); #first longest of lymann \n", + "f2=c/(w2*10**-9); #correspoding freq\n", + "n0=1.0;n=3.0\n", + "w3=91.13*(n**2/(n**2-n0**2)); #second longest of lymann\n", + "f3=3.0*10**8/(w3*10**-9) #corresponding freq.\n", + "\n", + "#result\n", + "print \"The freq. corresponding to the longest wavelength of balmer is %.1e\" %f1,\" & First longest wavelength of Lymann is %.1e\" %f2;\n", + "print\"The sum of which s equal to %.1e\" %(f1+f2);\n", + "print\"The freq. corresponding to 2nd longest wavelength was found out to be %.1e\" %f3,\"Hence Ritz combination principle is satisfied.\";" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The freq. corresponding to the longest wavelength of balmer is 4.6e+14 & First longest wavelength of Lymann is 2.5e+15\n", + "The sum of which s equal to 2.9e+15\n", + "The freq. corresponding to 2nd longest wavelength was found out to be 2.92622261239e+15 Hence Ritz combination principle is satisfied.\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.6 Page 192" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#initiation of variable\n", + "Rinfi=1.097*10**7; #known value \n", + "n1=3.0;n2=2.0; #first 2 given states\n", + "\n", + "#calculation\n", + "w=(n1**2*n2**2)/((n1**2-n2**2)*Rinfi);\n", + "\n", + "#result\n", + "print\"Wavelength of transition from n1=3 to n2=2 in nm is\",round(w*10**9,3);\n", + "\n", + "#partb\n", + "n1=4.0;n2=2.0; #second 2 given states \n", + "w=(n1**2*n2**2)/((n1**2-n2**2)*Rinfi);\n", + "\n", + "#result\n", + "print\"Wavelength of transition from n1=3 to n2=2 in nm is\",round(w*10**9,3);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Wavelength of trnasition from n1=3 to n2=2 in nm is 656.335\n", + "Wavelength of trnasition from n1=3 to n2=2 in nm is 486.174\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.7 Page 194" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#initiation of variable\n", + "n1=3.0;n2=2.0;Z=4.0;hc=1240.0;\n", + "delE=(-13.6)*(Z**2)*((1/(n1**2))-((1/n2**2)));\n", + "\n", + "#calculation\n", + "w=(hc)/delE; #for transition 1\n", + "\n", + "#result\n", + "print \"The wavelngth of radiation for transition(2->3) in nm is\", round(w,3);\n", + "\n", + "#for transition 2\n", + "n1=4.0;n2=2.0; # n values for transition 2\n", + "delE=(-13.6)*(Z**2)*((1/n1**2)-(1/n2**2));\n", + "w=(hc)/delE;\n", + "\n", + "#result\n", + "print \"The wavelngth of radiation emitted for transition(2->4) in nm is\", round(w,3);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The wavelngth of radiation for transition(2->3) in nm is 41.029\n", + "The wavelngth of radiation emitted for transition(2->4) in nm is 30.392\n" + ] + } + ], + "prompt_number": 11 + } + ], + "metadata": {} + } + ] +}
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