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diff --git a/backup/Principles_of_Physics_by_F.J.Bueche_version_backup/Chapter26.ipynb b/backup/Principles_of_Physics_by_F.J.Bueche_version_backup/Chapter26.ipynb new file mode 100755 index 00000000..ab561997 --- /dev/null +++ b/backup/Principles_of_Physics_by_F.J.Bueche_version_backup/Chapter26.ipynb @@ -0,0 +1,216 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 26: Energy levels and Spectra" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex26.1:pg-1208" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The ionization energy of the hydrogen atom is E= 13.6 eV\n" + ] + } + ], + "source": [ + " #Example 26_1\n", + " \n", + " \n", + "#To find the ionization energy of the hydrogen atom\n", + "e=13.6 #units in eV\n", + "print \"The ionization energy of the hydrogen atom is E=\",round(e,1),\" eV\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex26.2:pg-1209" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The wavelength of fourth line in Paschen series is= 1002.0 nm\n" + ] + } + ], + "source": [ + " #Example 26_2\n", + " \n", + " \n", + " #To find the wavelength of fourth line in Paschen series\n", + "n1=3 #Units in constant\n", + "n2=7 #Units in constant\n", + "r=1.0974*10**7 #units in meter**-1\n", + "lamda=round((1/r)*((n1**2*n2**2)/(n2**2-n1**2))*10**9) #Units in nm\n", + "print \"The wavelength of fourth line in Paschen series is=\",round(lamda),\" nm\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex26.3:pg-1209" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The energy associated with line 1 is E1= -54.4 eV\n", + "The energy associated with line 2 is E2= -13.6 eV\n", + "The energy associated with line 3 is E3= -6.04 eV\n", + "\n", + "The first line of balmer series is lamda= 163.0 nm and belongs to the ultraviolet region\n" + ] + } + ], + "source": [ + " #Example 26_3\n", + " \n", + " \n", + " #To draw the energy level diagram and the find the first line of balmer type series\n", + "n=1\n", + "e1=-54.4/n**2 #units in ev\n", + "n=2\n", + "e2=-54.4/n**2 #units in ev\n", + "n=3\n", + "e3=-54.4/n**2 #units in ev\n", + "print \"The energy associated with line 1 is E1=\",round(e1,1),\" eV\\nThe energy associated with line 2 is E2=\",round(e2,1),\" eV\\nThe energy associated with line 3 is E3=\",round(e3,2),\" eV\\n\"\n", + "e1=1 #units in eV\n", + "e2=7.6 #Units in eV\n", + "lamda1=1240 #units in nm\n", + "lamda=(e1/e2)*lamda1 #Units in nm\n", + "print \"The first line of balmer series is lamda=\",round(lamda),\" nm and belongs to the ultraviolet region\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex26.4:pg-1209" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The longest wavelength of light capable of ionizing hydrogen atom is lamda= 91.2 nm\n" + ] + } + ], + "source": [ + " #Example 26_4\n", + " \n", + " \n", + " #To find the longest wavelength of light capable of ionizing hydrogen atom\n", + " #First method\n", + "R=1.097*10**7 #Units in meter**-1\n", + "lamda=(1/R)*10**9 #Units in meters\n", + " #Second method\n", + "E=13.6 #units in eV\n", + "e1=1 #units in eV\n", + "lamda3=1240.0 #Units in eV\n", + "lamda2=(e1/E)*(lamda3) #Units in nm\n", + "print \"The longest wavelength of light capable of ionizing hydrogen atom is lamda=\",round(lamda2,1),\" nm\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex26.5:pg-1210" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The energy difference between n=1 and n=2 level is E= 17713.0 eV\n" + ] + } + ], + "source": [ + " #Example 26_5\n", + " \n", + " \n", + " #To find the energy difference between the n is 1 and n is 2 level\n", + "e1=1 #Units in eV\n", + "lamda2=1240 #Units in eV\n", + "lamda3=0.07 #Units in eV\n", + "e2=lamda2/lamda3 #Units in eV\n", + "e=e2-e1 #Units in eV\n", + "print \"The energy difference between n=1 and n=2 level is E=\",round(e),\" eV\"\n", + " #In textbook answer is prinred wrong as E=18000 eV the correct answer is E=17713 eV \n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.11" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} |