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diff --git a/Fundamental_Of_Physics_by_D_Haliday/40-More_About_Matter_waves.ipynb b/Fundamental_Of_Physics_by_D_Haliday/40-More_About_Matter_waves.ipynb new file mode 100644 index 0000000..87f2b64 --- /dev/null +++ b/Fundamental_Of_Physics_by_D_Haliday/40-More_About_Matter_waves.ipynb @@ -0,0 +1,211 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 40: More About Matter waves" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 40.1: Sample_Problem_1.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('quantum.sci', -1)\n", +"\n", +"//Given that\n", +"L = 100*10^-12 //in m\n", +"\n", +"//Sample Problem 40-1a\n", +"printf('**Sample Prblem 40-1a**\n')\n", +"n = 1 //for min energy\n", +"Emin = Ediff(n, 0, Me, L)\n", +"printf('The least possible energy is %1.2fev\n', Emin/conv)\n", +"\n", +"//Sample Problem 40-1b\n", +"printf('\n**Sample Prblem 40-1b**\n')\n", +"n2 = 3\n", +"n1 = 1\n", +"deltaE13 = Ediff(n2, n1, Me, L)\n", +"printf('The energy to be transferred is %1.2fev\n', deltaE13/conv)\n", +"\n", +"//Sample Problem 40-1c\n", +"printf('\n**Sample Prblem 40-1c**\n')\n", +"lambda = wavelength(deltaE13)\n", +"printf('The wavelength of photon is %1.2fnm\n', lambda*10^9)\n", +"\n", +"//Sample Problem 40-1d\n", +"printf('\n**Sample Prblem 40-1d**\n')\n", +"deltaE12 = Ediff(2, 1, Me, L)\n", +"lambda1 = wavelength(deltaE12)\n", +"deltaE23 = Ediff(3, 2, Me, L)\n", +"lambda2 = wavelength(deltaE23)\n", +"printf('The possible wavelength of photon is :\n')\n", +"printf('\t %1.2fnm\n', lambda*10^9)\n", +"printf('\t %1.2fnm\n', lambda1*10^9)\n", +"printf('\t %1.2fnm', lambda2*10^9)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 40.3: Sample_Problem_3.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"L = 100*10^-12 //in m\n", +"\n", +"//Sample Problem 40-3a\n", +"printf('**Sample Prblem 40-3a**\n')\n", +"P = integrate('2/L*(sin(%pi/L*x))^2', 'x', 0, L/3)\n", +"printf('The probability is equal to %1.2f\n', P)\n", +"\n", +"//Sample Problem 40-3b\n", +"printf('\n**Sample Prblem 40-3b**\n')\n", +"P = integrate('2/L*(sin(%pi/L*x))^2', 'x', L/3, 2*L/3)\n", +"printf('The probability is equal to %1.2f\n', P)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 40.4: Sample_Problem_4.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('quantum.sci', -1)\n", +"\n", +"//Given that\n", +"L = 100*10^-12 //in m\n", +"Uo = 450*conv //in ev\n", +"l = 2*10^-9 //in m\n", +"\n", +"//Sample Problem 40-4a\n", +"printf('**Sample Prblem 40-4a**\n')\n", +"E1 = Ediff(1, 0, Me, L)\n", +"E = Uo - E1\n", +"lambda = wavelength(E)\n", +"printf('The wavelength of the free electron is %1.2fnm\n', lambda*10^9)\n", +"\n", +"//Sample Problem 40-4b\n", +"printf('\n**Sample Prblem 40-4b**\n')\n", +"E2 = Energy(l)\n", +"K = E2 - E\n", +"printf('The electron energy is %1.2fev', K/conv)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 40.6: Sample_Problem_6.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"exec('quantum.sci', -1)\n", +"\n", +"//Sample Problem 40-6a\n", +"printf('**Sample Prblem 40-6a**\n')\n", +"//lyman series : least energetic photon => 2 to 1 transition\n", +"deltaE21 = -(13.6) * (1/2^2 - 1/1^2)*conv\n", +"lambda = wavelength(deltaE21)\n", +"printf('The wavelength of the least energetic photon in lyman series is %1.2fnm\n', lambda*10^9)\n", +"\n", +"//Sample Problem 40-6b\n", +"printf('\n**Sample Prblem 40-6b**\n')\n", +"//lyman series limit => 1 to infinity transition\n", +"deltaE = -(13.6) * (0 - 1/1^2)*conv\n", +"lambda = wavelength(deltaE)\n", +"printf('The wavelength of the lyman series limit is %1.2fnm\n', lambda*10^9)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 40.8: Sample_Problem_8.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"//Given that\n", +"p = 0.90\n", +"\n", +"//Sample Problem 40-8\n", +"printf('**Sample Prblem 40-8**\n')\n", +"funcprot(0)\n", +"function [P] = f(x)\n", +" P = 1 - (%e^(-2*x))*(1 + 2*x + 2*x^2) - p\n", +"endfunction\n", +"p = fsolve(1, f)\n", +"printf('The possible value of radius is %1.2f*a', p)" + ] + } +], +"metadata": { + "kernelspec": { + "display_name": "Scilab", + "language": "scilab", + "name": "scilab" + }, + "language_info": { + "file_extension": ".sce", + "help_links": [ + { + "text": "MetaKernel Magics", + "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md" + } + ], + "mimetype": "text/x-octave", + "name": "scilab", + "version": "0.7.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} |