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diff --git a/Quantum_mechanics_by_M.C.Jain/chapter3_1.ipynb b/Quantum_mechanics_by_M.C.Jain/chapter3_1.ipynb new file mode 100644 index 00000000..7ef8f5c9 --- /dev/null +++ b/Quantum_mechanics_by_M.C.Jain/chapter3_1.ipynb @@ -0,0 +1,519 @@ +{
+ "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 |