From 476705d693c7122d34f9b049fa79b935405c9b49 Mon Sep 17 00:00:00 2001 From: prashantsinalkar Date: Tue, 14 Apr 2020 10:19:27 +0530 Subject: Initial commit --- .../5-The_Schrodinger_Equation.ipynb | 112 +++++++++++++++++++++ 1 file changed, 112 insertions(+) create mode 100644 Modern_Physics_by_K_S_Krane/5-The_Schrodinger_Equation.ipynb (limited to 'Modern_Physics_by_K_S_Krane/5-The_Schrodinger_Equation.ipynb') diff --git a/Modern_Physics_by_K_S_Krane/5-The_Schrodinger_Equation.ipynb b/Modern_Physics_by_K_S_Krane/5-The_Schrodinger_Equation.ipynb new file mode 100644 index 0000000..c499103 --- /dev/null +++ b/Modern_Physics_by_K_S_Krane/5-The_Schrodinger_Equation.ipynb @@ -0,0 +1,112 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 5: The Schrodinger Equation" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.1: Displacement_and_velocity_of_the_object.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear \n", +"clc\n", +"disp('Exa-5.1');//The solution involves very complex symbolic equation solving and approximations. Hence only answers are displayed\n", +"printf('The displacement and velocity of the ball are found out in 2 steps\n step1-before reaching the surface of water and \n step2-Inside water till it rises back to surface\n');\n", +"printf('The values are as follows: v1(t)=-g*t and y1(t)=H-((g/2)*t^2))\n');\n", +"printf('In region 2: v2(t)=(-B/m*sqrt(2*H/g))+(B/m-g)*t; y2(t)= H+ HB/mg -B/m*sqrt(2*H/g)+ (B/m-g)');" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.2: Solution_for_a_and_b.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear \n", +"clc\n", +"disp('Exa-5.2(a)');\n", +"h=1.05*10^-34;m=9.11*10^-31;L=10^-10; // all the values are taken in SI units\n", +"E1=h^2*%pi^2/(2*m*L^2); E2=4*E1; //Energies are calculated\n", +"delE=(E2-E1)/(1.6*10^-19); //Difference in energy is converted to eV\n", +"printf('Energy to be supplied is %.0f eV.\n',delE);\n", +"disp('Exa-5.2(b)');\n", +"x1=0.09*10^-10;x2=0.11*10^-10 //limits of the given region\n", +"probGnd=(2/L)*integrate('(sin(%pi*x/L)^2)','x',x1,x2);\n", +"printf('The percentage probablility of finding an electron in the ground state is %.2f.\n',probGnd*100);\n", +"disp('Exa-5.2(c)');\n", +"x1=0,x2=0.25*10^-10;\n", +"probExc=(2/L)*integrate('(sin(2*%pi*x/L)^2)','x',x1,x2);\n", +"printf('The probablility of finding an electron in the excited state is %.2f.\n',probExc);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.3: Proof_for_average_value_of_x.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear \n", +"clc\n", +"disp('Ex-5.3');\n", +"x1=0;x2=L;\n", +"xavg=(2/L)*integrate('sin(%pi*x/L)^2','x',x1,x2);\n", +"printf('The average value of x is found out to be L/2 which apparently is independent of Qunatum state.');" + ] + } +], +"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 +} -- cgit