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Diffstat (limited to 'Quantum_mechanics_by_M.C.Jain/chapter3.ipynb')
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diff --git a/Quantum_mechanics_by_M.C.Jain/chapter3.ipynb b/Quantum_mechanics_by_M.C.Jain/chapter3.ipynb deleted file mode 100755 index 7ef8f5c9..00000000 --- a/Quantum_mechanics_by_M.C.Jain/chapter3.ipynb +++ /dev/null @@ -1,519 +0,0 @@ -{
- "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": {}
- }
- ]
-}
\ No newline at end of file |