{ "metadata": { "name": "" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 4:Atomic Structure" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example no:4.1,Page no:125" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable declaration \n", "E= -13.6; #Energy required to separate electron and proton, eV\n", "e= 1.6*(10**(-19)); #charge of an electron, C\n", "E= E*e; #converting to J\n", "Po= 8.85*(10**(-12)); #Permittivity of free space, F/m\n", "\n", "#Calculation\n", "import math\n", "r= e**2/(8*(math.pi)*Po*E); #radius, m\n", "r= -r;\n", "m= 9.1*(10**(-31)); #mass of electron, kg\n", "v=e/math.sqrt(4*(math.pi)*Po*m*r); #velocity, m/s\n", "\n", "#Result\n", "print\"The orbital radius of the electron is:%.2g\"%r,\"m\"\n", "print\"The velocity of electron is:%.2g\"%v,\"m/s\"\n", " \n", " " ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The orbital radius of the electron is:5.3e-11 m\n", "The velocity of electron is:2.2e+06 m/s\n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example no:4.2,Page no:135" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable declaration \n", "n1=1.0; #initial state\n", "n2=3.0; #final state\n", "E= -13.6; #energy in ground state, eV\n", "\n", "#Calculation\n", "dE= E*((1/n2**2)-(1/n1**2)); #Change in energy, eV\n", "\n", "#Result\n", "print\"The energy change of Hydrogen atom is: \",round(dE,1),\"eV\"\n", " " ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The energy change of Hydrogen atom is: 12.1 eV\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example no:4.3,Page no:135" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable declaration\n", "#Part(a)\n", "Rn= 10.0**(-5); #radius of Rydberg atom, m\n", "Ao= 5.29*(10**(-11)); #Bohr radius, m\n", "\n", "#Calculation\n", "n= math.sqrt(Rn/Ao); #Quantum number\n", "E1= -13.6; #Ground state energy level, eV\n", "En= E1/n**2.0; #Nth state energy level, eV\n", "\n", "#Result\n", "print\"(a).The quantum number of the Rydberg atom is: \",round(n)\n", "print\"(b).The energy ofthe rydberg atom is:%.3g\"%En,\"eV\"\n", " " ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(a).The quantum number of the Rydberg atom is: 435.0\n", "(b).The energy ofthe rydberg atom is:-7.19e-05 eV\n" ] } ], "prompt_number": 4 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example no:4.4,Page no:138" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable declaration\n", "n1= 3.0; #initial state\n", "n2= 2.0; #final state\n", "R= 1.097*(10**7); #Rydberg's constant, m**(-1)\n", "\n", "#Calculation\n", "k= (1/n2**2)-(1/n1**2);\n", "l= 1/(k*R); #longest wavelength, m\n", "l= l*(10**9); #converting to nm\n", "\n", "#Result\n", "print\"The longest in Balmer series of Hydrogen, in nm, is: \",round(l),\"nm\"\n", " " ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The longest in Balmer series of Hydrogen, in nm, is: 656.0 nm\n" ] } ], "prompt_number": 5 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example no:4.5,Page no:139" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable declaration\n", "n1=1.0; #initial state\n", "n2=2.0; #final state\n", "E1= 2.18*(10**(-18)); #Rydberg's constant, J\n", "h= 6.63*(10**(-34)); #Planck's constant, J.s\n", "\n", "#Calculation\n", "f1= (E1/h)*(2.0/n1**3); #Frequency for first orbit, rev/s\n", "f2= (E1/h)*(2.0/n2**3); #Frequency for second orbit, rev/s\n", "print\"Ans (A):Frequency of revolution for orbit n=1 is,f1: %.3g\"%f1,\"rev/s\"\n", "print\"Frequency of revolution for orbit n=2 is,f2:%.2e\"%f2,\"rev/s\"\n", "print\"which is equivalent to 0.823*10**15 rev/s 'without' any scientific notation\\n\"\n", "#Part (b)\n", "n1=2.0; #initial orbit\n", "n2=1.0; #final orbit\n", "f= (E1/(h))*((1.0/(n2**2))-(1.0/n1**3)); #frequency, Hz\n", "print\"Ans(B):Frequency of emitted photon is: %.3g\"%f,\"Hz\\n\"\n", "#Part (c)\n", "n= 2.0; #orbit\n", "f= f2; #from part (a)\n", "dt= 10.0**(-8); # time duration, s\n", "N= f*dt; #Number of revolutions\n", "#Result\n", "print\"Ans(C):Number of revolutions the electron makes is:%.3g\"%N\n", " " ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Ans (A):Frequency of revolution for orbit n=1 is,f1: 6.58e+15 rev/s\n", "Frequency of revolution for orbit n=2 is,f2:8.22e+14 rev/s\n", "which is equivalent to 0.823*10**15 rev/s 'without' any scientific notation\n", "\n", "Ans(B):Frequency of emitted photon is: 2.88e+15 Hz\n", "\n", "Ans(C):Number of revolutions the electron makes is:8.22e+06\n" ] } ], "prompt_number": 6 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example no:4.7,Page no:142" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable declaration\n", "#Part (a) \n", "Me= 9.1*(10**(-31)); #mass of electron, kg\n", "m= 207*Me; #mass of muon, kg\n", "\n", "#Calculation\n", "Mp= 1836*Me; #mass of proton, kg\n", "Mreduced= (m*Mp)/(m+Mp); #reduced mass, kg\n", "Ao= 5.29*(10**(-11)); #Bohr's orbit for n=1, m\n", "r1= Ao; #expected orbit for atom, m\n", "r2= (Me/Mreduced)*r1; #reduced radius of orbit, m\n", "#Part (b)\n", "E=-13.6; # energy for elctron in n=1, eV\n", "Ereduced= (Mreduced/Me)*E; #energy for eectron in mounic atom, eV\n", "Ereduced= Ereduced/(10**3);#converting to keV\n", "\n", "#Result\n", "print\"(A)Radius of the mounic atom formed, in m, is:%.3g\"%r2,\"m\"\n", "print\"(B)Ionisation energy for the muonic atom is: \",round(Ereduced,2),\"keV\"\n", " " ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(A)Radius of the mounic atom formed, in m, is:2.84e-13 m\n", "(B)Ionisation energy for the muonic atom is: -2.53 keV\n" ] } ], "prompt_number": 7 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example no:4.8,Page no:156" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable declaration \n", "I= 7.7; #Intensity of beam, MeV\n", "Dgold= 1.93*(10**4); #density of gold foil used, kg/m**3\n", "u= 1.66*(10**(-27)); #atomic mass unit, kg\n", "Mgold= 197*u; #atomic mass of gold, per atom\n", "\n", "#Calculation\n", "n= Dgold/Mgold; #number of atoms per unit volume, atoms/m**3\n", "Zgold= 79; #atomic number of gold\n", "e= 1.6*(10**(-19)); #electronis charge, C\n", "KE= (I*e)/(10**(-6)); #converting to J\n", "angle= 45; #degree\n", "p=1/math.tan(math.radians(angle/2));\n", "Po= 8.85*(10**(-12)); #Permittivity of free space, F/m\n", "t= 3*(10**(-7)); #thickness of foil, m\n", "f= (math.pi)*n*t*(((Zgold*(e**2))/(4*(math.pi)*Po*KE))**2)*(p**2) #using Rutherford scattering formula\n", "\n", "#Result\n", "print\"f=%.g\"%f\n", "print\"Fraction of the beam scattered through 45 degree or more is: \",round(f*100,3),\"%\"\n", " " ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "f=7e-05\n", "Fraction of the beam scattered through 45 degree or more is: 0.007 %\n" ] } ], "prompt_number": 8 } ], "metadata": {} } ] }