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diff --git a/backup/Modern_Physics_version_backup/Chapter7.ipynb b/backup/Modern_Physics_version_backup/Chapter7.ipynb new file mode 100755 index 00000000..b8966938 --- /dev/null +++ b/backup/Modern_Physics_version_backup/Chapter7.ipynb @@ -0,0 +1,251 @@ +{ + "metadata": { + "name": "Chapter7" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 7:The Hydrogen Atom in Wave Mechanics" + ] + }, + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Example 7.2 Page 213" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#initiation of variable\n", + "from math import exp\n", + "import math\n", + "from scipy import integrate\n", + "# calculating radial probability P= (4/ao^3)*integral(r^2 * e^(-2r/ao)) between the limits 0 and ao for r\n", + "\n", + "#calculation\n", + "def integrand(x):\n", + " return ((x**2)*exp(-x))/2.0\n", + "Pr=integrate.quad(integrand,0,2,args=());#simplifying where as x=2*r/a0; hence the limits change between 0 to 2\n", + "\n", + "#result\n", + "print \"Hence the probability of finding the electron nearer to nucleus is\",round(Pr[0],3);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Hence the probability of finding the electron nearer to nucleus is 0.323\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.3 Page 213" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#initiation of variable\n", + "from math import exp\n", + "import math\n", + "from scipy import integrate\n", + "# employing the formula for probability distribution similarly as done in Exa-7.2 \n", + "#calculation\n", + "def integrand(x):\n", + " return (1.0/8)*((4.0*x**2)-(4.0*x**3)+(x**4))*exp(-x)\n", + "Pr1= integrate.quad(integrand,0,1,args=()) #x=r/ao; similrly limits between 0 and 1.\n", + "\n", + "#result\n", + "print\"The probability for l=0 electron is\",round(Pr1[0],5)\n", + "\n", + "#part2\n", + "def integrand(x):\n", + " return (1.0/24)*(x**4)*(exp(-x))\n", + "Pr2=integrate.quad(integrand,0,1); #x=r/ao; similarly limits between 0 and 1.\n", + "\n", + "#result\n", + "print\"The probability for l=1 electron is\",round(Pr2[0],5)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The probability for l=0 electron is 0.03432\n", + "The probability for l=1 electron is 0.00366\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.4 Page 215" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#initiation of variable\n", + "from math import exp, sqrt\n", + "import math\n", + "from scipy import integrate\n", + "l=1.0; #given value of l\n", + "\n", + "#calculation\n", + "am1=sqrt(l*(l+1)); #angular momentum==sqrt(l(l+1)) h\n", + "l=2.0 #given l\n", + "am2=sqrt(l*(l+1));\n", + "\n", + "#result\n", + "print\"The angular momenta are found out to be\", round(am1,3),\" h and\",round(am2,3),\" h respectively for l=1 and l=2.\";\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The angular momenta are found out to be 1.414 h and 2.449 h respectively for l=1 and l=2.\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.5 Page 216" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#initiation of variable\n", + "from math import sqrt\n", + "print \"The possible values for m are [+2,-2] and hence any of the 5 components [-2h,2h] are possible for the L vector.\";\n", + "print \"Length of the vector as found out previously is %.2f*h.\",round(sqrt(6),4);#angular momentum==sqrt(l(l+1)) h" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The possible values for m are [+2,-2] and hence any of the 5 components [-2h,2h] are possible for the L vector.\n", + "Length of the vector as found out previously is %.2f*h. 2.4495\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.6 Page 223" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#initiation of variable\n", + "uz=9.27*10**-24; t=1.4*10**3; x=3.5*10**-2; #various constants and given values\n", + "m=1.8*10**-25;v=750; # mass and velocity of the particle\n", + "\n", + "#calculation\n", + "d=(uz*t*(x**2))/(m*(v**2)); #net separtion \n", + "\n", + "#result\n", + "print\"The distance of separation in mm is\",round(d*10**3,3);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The distance of separation in mm is 0.157\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.7 Page 227" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#initiation of variable\n", + "n1=1.0;n2=2.0;hc=1240.0; #hc=1240 eV.nm\n", + "E=(-13.6)*((1/n2**2)-(1/n1**2)); #Energy calculation\n", + "\n", + "#calculation\n", + "w=hc/E; #wavelength\n", + "u=9.27*10**-24; B=2; #constants\n", + "delE= u*B/(1.6*10**-19); #change in energy\n", + "delw=((w**2/hc))*delE; #change in wavelength\n", + "\n", + "#result\n", + "print\"The change in wavelength in nm. is\",round(delw,4);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The change in wavelength in nm. is 0.0014\n" + ] + } + ], + "prompt_number": 12 + } + ], + "metadata": {} + } + ] +}
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