{ "metadata": { "name": "", "signature": "sha256:8da78b3914613cbb3785b48315fd900bd210e4e84b50eb2e4aa86b6821a0f0e1" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 3 Atoms and the Bohr model" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.1 Page no 39" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Given\n", "E=-3.4 #ev\n", "h=6.63*10**-34 #Js\n", "\n", "#Calculation\n", "import math\n", "n=math.sqrt(-13.6/E)\n", "M=(n*h)/(2.0*math.pi)\n", "\n", "#Result\n", "print\"Angular momentum of electron is given by \",round(M,36),\" Js\" " ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Angular momentum of electron is given by 2.11e-34 Js\n" ] } ], "prompt_number": 7 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.2 Page no 40" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Given\n", "E=13.6 #ev\n", "n1=4\n", "n2=2\n", "\n", "#Calculation\n", "energy=E*((1/2.0**2)-(1/4.0**2))\n", "\n", "#Result\n", "print\"Energy of photon emitted in the transition is \",energy,\"ev\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Energy of photon emitted in the transition is 2.55 ev\n" ] } ], "prompt_number": 7 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.3 Page no 40" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Given\n", "n1=3\n", "n2=2\n", "E1=-1.5 #ev\n", "E2=-3.4 #ev\n", "h=6.63*10**-34 #Js\n", "c=3*10**8 #m/s\n", "e=1.6*10**-19\n", "\n", "#Calculation\n", "v=(h*c)/((E1-E2)*e)\n", "\n", "#Result\n", "print\"Wavelength is \",round(v,10),\"m\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Wavelength is 6.543e-07 m\n" ] } ], "prompt_number": 131 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.4 Page no 40" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Given\n", "v=1200 #A\n", "R=1.097*10**7 #m-1\n", "n1=2.0\n", "n2=3.0\n", "\n", "#Calculation\n", "v1=(R*(1-(1/n1**2)))\n", "v2=(R*(1-(1/n2**2)))\n", "V=v1/v2\n", "V1=V*v\n", "\n", "#Result\n", "print\"Wavelength of the second line is \", V1,\"A\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Wavelength of the second line is 1012.5 A\n" ] } ], "prompt_number": 136 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.5 Page no 41" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Given\n", "R=1.097*10**7 #m-1\n", "n=2\n", "\n", "#Calculation\n", "v=n**2/(3.0*R)\n", "v1=1/R # for n=infinite\n", "\n", "#Result\n", "print\"longest wavelength is \",round(v*10**10,0),\"A\"\n", "print\"shortest wavelength is \",round(v1*10**10,1),\"A\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "longest wavelength is 1215.0 A\n", "shortest wavelength is 911.6 A\n" ] } ], "prompt_number": 138 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.6 Page no 41" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Given\n", "E=47.2 # 3ev\n", "n1=2\n", "n2 =3\n", "\n", "#Calculation\n", "import math\n", "Z=math.sqrt(E/(13.6*((1/2.0**2)-(1/3.0**2))))\n", "\n", "#Result\n", "print\"Atomic number of the atom is \",round(Z,0)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Atomic number of the atom is 5.0\n" ] } ], "prompt_number": 13 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.7 Page no 42" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Given\n", "Z=1.0\n", "n=1.0 #for the ground state of hydrogen\n", "Z1=4 #for Be++\n", "n1=2.0\n", "\n", "#Calculation\n", "import math\n", "n1=math.sqrt((n**2/Z)*Z1)\n", "r=(Z1**2/n1**2)/(Z**2/n**2) #Ratio of two energies\n", "\n", "#Result\n", "print\"nBe++= \", n1\n", "print\"comparison is \",r" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "nBe++= 2.0\n", "comparison is 4.0\n" ] } ], "prompt_number": 143 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.8 Page no 42" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Given\n", "Z=3.0\n", "n=3 #for Li++\n", "Z1=1.0\n", "n1=1 #for hydrogen\n", "\n", "#Calculation\n", "r=(n**2/Z)/(n1**2/Z1)\n", "\n", "#Result\n", "print\"orbital ratio of two states \",r" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "orbital ratio of two states 3.0\n" ] } ], "prompt_number": 144 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.9 Page no 42" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Given\n", "v=970.6 #A\n", "h=6.63*10**-34 #Js \n", "c=3*10**8 #m/s\n", "\n", "#Calculation\n", "import math\n", "E=((h*c)/(v*e))*10**10\n", "En=-13.6+E\n", "n=math.sqrt(-13.6/En)\n", "E3=-13.6/(3.0**2)\n", "vmax=(h*c)/((-E3+En)*(1.6*10**-19))\n", "\n", "#Result\n", "print\"Longest wavelength is \",round(vmax*10**10),\"A\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Longest wavelength is 17292.0 A\n" ] } ], "prompt_number": 159 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.10 Page no 43" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Given\n", "Z=2\n", "E=13.6 #ev\n", "E0=10.04 #ev\n", "\n", "#Calculation\n", "Ei=Z**2*E\n", "E1=-Ei\n", "E3=E1/(3.0**2)\n", "Ee=E0+E3\n", "\n", "#Result\n", "print\"Required stopping potential is \", round(Ee,0),\"V\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Required stopping potential is 4.0 V\n" ] } ], "prompt_number": 50 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.11 Page no 44" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Given\n", "Ei=4*2.2*10**-18 #Joule\n", "h=6.6*10**-34 #Js\n", "c=3*10**8 #m/s\n", "\n", "#Calculation\n", "E1=-Ei\n", "E2=E1/(2.0**2)\n", "v=(h*c)/(Ei+E2)\n", "\n", "#Result\n", "print\"Wavelength is \", round(v*10**10,0),\"A\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Wavelength is 300.0 A\n" ] } ], "prompt_number": 173 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.12 Page no 44" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Given\n", "n1=3\n", "n2 =1\n", "E=13.6 #ev\n", "\n", "#Calculation\n", "E1=E/(3.0**2) #Binding energy of the atom in n=3 state\n", "energy=E-E1 #Energy required for the atomic electron to jump from n=1 to n=3 state\n", "\n", "#Result\n", "print\"The electron beam must, therefore be accelerated through a potential difference of \",round(energy,2),\"V\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The electron beam must, therefore be accelerated through a potential difference of 12.09 V\n" ] } ], "prompt_number": 58 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3.13 Page no 46" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Given\n", "Rh=1.09678*10**7 #m-1\n", "Rhe=1.09722*10**7 #m-1\n", "\n", "#Calculation\n", "Mr=(Rhe-Rh)/(Rh-(Rhe/4.0)) #ratio of electron mass\n", "\n", "#Result\n", "print\"Ratio of the electron mas to the proton mass \",round(Mr*10**4,2),\"*10**-4\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Ratio of the electron mas to the proton mass 5.35 *10**-4\n" ] } ], "prompt_number": 6 } ], "metadata": {} } ] }