From d36fc3b8f88cc3108ffff6151e376b619b9abb01 Mon Sep 17 00:00:00 2001 From: kinitrupti Date: Fri, 12 May 2017 18:40:35 +0530 Subject: Revised list of TBCs --- Quantum_mechanics_by_M.C.Jain/chapter3.ipynb | 519 --------------------------- 1 file changed, 519 deletions(-) delete mode 100755 Quantum_mechanics_by_M.C.Jain/chapter3.ipynb (limited to 'Quantum_mechanics_by_M.C.Jain/chapter3.ipynb') 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 -- cgit