path: root/Applied_Physics/Chapter5.ipynb

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  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 5:Principles of Quantum Mechanics"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.1 , Page no:102"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "from __future__ import division\n",
      "\n",
      "#given\n",
      "lambda1=1; #in Angstrom (wavelength)\n",
      "m=1.67E-27; #in Kg (mass of neutron)\n",
      "h=6.625E-34; #in J-s (Planck's constant)\n",
      "e=1.6E-19; #in C (charge of electron)\n",
      "\n",
      "#calculate\n",
      "lambda2=lambda1*1E-10; #since lambda is in Angstrom\n",
      "#Since lambda=h/(m*v)\n",
      "#Therefore we have\n",
      "v=h/(m*lambda2); #calculation of velocity\n",
      "K=m*v**2/2; #calculation of kinetic energy\n",
      "K1=K/e; #changing unit fro J to eV\n",
      "\n",
      "#result\n",
      "print\"The velocity is v=\",v,\"m/s\";\n",
      "print\"The kinetic energy is K=\",K,\"J\";\n",
      "print\"\\t\\t       =\",K1,\"eV\";"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The velocity is v= 3967.06586826 m/s\n",
        "The kinetic energy is K= 1.31409056886e-20 J\n",
        "\t\t       = 0.0821306605539 eV\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.2 , Page no:103"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "from __future__ import division\n",
      "\n",
      "#given\n",
      "K=50; #in eV (Kinetic energy)\n",
      "m0=9.1E-31; #in Kg (mass of electron)\n",
      "h=6.625E-34; #in J-s (Planck's constant)\n",
      "e=1.6E-19; #in C (charge of electron)\n",
      "\n",
      "#calculate\n",
      "K=K*e; #changing unit from eV to J\n",
      "#Since K=m*v^2/2\n",
      "#Therefore v=sqrt(2*K/m)\n",
      "#Since lambda=h/(m*v)\n",
      "#Therefore we have\n",
      "lambda1=h/math.sqrt(2*m0*K); #calculation of wavelength\n",
      "lambda2=lambda1*1E10; #changing unit from m to Angstrom\n",
      "\n",
      "#result\n",
      "print\"The wavelength is =\",lambda1,\"m\";\n",
      "print\"\\t\\t  =\",round(lambda2,3),\"Angstrom\";"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The wavelength is = 1.73622176082e-10 m\n",
        "\t\t  = 1.736 Angstrom\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.3 , Page no:104"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "from __future__ import division\n",
      "\n",
      "#given\n",
      "E=2000; #in eV (Kinetic energy)\n",
      "m=9.1E-31; #in Kg (mass of electron)\n",
      "h=6.625E-34; #in J-s (Planck's constant)\n",
      "e=1.6E-19; #in C (charge of electron)\n",
      "\n",
      "#calculate\n",
      "E=E*e; #changing unit from eV to J\n",
      "#Since E=m*v^2/2\n",
      "#Therefore v=sqrt(2*E/m)\n",
      "#Since lambda=h/(m*v)\n",
      "#Therefore we have\n",
      "lambda1=h/math.sqrt(2*m*E); #calculation of wavelength\n",
      "lambda2=lambda1*1E9; #changing unit from m to nanometer\n",
      "\n",
      "#result\n",
      "print\"The wavelength is =\",lambda1,\"m\";\n",
      "print\"\\t\\t  =\",round(lambda2,4),\"nm\";"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The wavelength is = 2.74520764367e-11 m\n",
        "\t\t  = 0.0275 nm\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.4 , Page no:104"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "from __future__ import division\n",
      "\n",
      "#given\n",
      "m_e=9.1E-31; #in Kg (mass of electron)\n",
      "m_n=1.676E-27; #in Kg (mass of neutron)\n",
      "h=6.625E-34; #in J-s (Planck's constant)\n",
      "c=3E8; #in m/s (velocity of light)\n",
      "\n",
      "#calculate\n",
      "E_e=m_e*c**2; #rest mass energy of electron\n",
      "E_n=2*E_e; #given (kinetic energy of neutron)\n",
      "#Since K=m*v^2/2\n",
      "#Therefore v=sqrt(2*K/m)\n",
      "#Since lambda=h/(m*v)\n",
      "#Therefore we have\n",
      "lambda1=h/math.sqrt(2*m_n*E_n); #calculation of wavelength\n",
      "lambda2=lambda1*1E10; #changing unit from m to Angstrom\n",
      "\n",
      "#result\n",
      "print\"The wavelength is =\",lambda1,\"m\";\n",
      "print\"\\t\\t  =\",lambda2,\"Angstrom\";"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The wavelength is = 2.82733278331e-14 m\n",
        "\t\t  = 0.000282733278331 Angstrom\n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.5 , Page no:104"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "from __future__ import division\n",
      "\n",
      "#given\n",
      "V=1600; #in V (Potential)\n",
      "\n",
      "#calculate\n",
      "lambda1=12.27/math.sqrt(V); #calculation of wavelength in Angstrom\n",
      "\n",
      "#result\n",
      "print\"The wavelength is =\",round(lambda1,3),\"Angstrom\";\n",
      "print \"Note : The answer in the book is wrong \""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The wavelength is = 0.307 Angstrom\n",
        "Note : The answer in the book is wrong \n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.6 , Page no:105"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "from __future__ import division\n",
      "\n",
      "#given\n",
      "E=1000; #in eV (Kinetic energy of photon)\n",
      "K=1000; #in eV (Kinetic energy of electron)\n",
      "m0=9.1E-31; #in Kg (mass of electron)\n",
      "h=6.6E-34; #in J-s (Planck's constant)\n",
      "c=3E8; #in m/s (velocity of light)\n",
      "e=1.6E-19; #in C (charge on electron)\n",
      "\n",
      "#calculate\n",
      "E=E*e; #changing unit from eV to J\n",
      "lambda_p=h*c/E; #For photon E=hc/lambda\n",
      "lambda_p1=lambda_p*1E10; #changing unit from m to Angstrom\n",
      "#Since K=m*v^2/2\n",
      "#Therefore v=sqrt(2*K/m)\n",
      "#Since lambda=h/(m*v)\n",
      "#Therefore we have\n",
      "K=K*e; #changing unit from eV to J\n",
      "lambda_e=h/math.sqrt(2*m0*K); #calculation of wavelength\n",
      "lambda_e1=lambda_e*1E10; #changing unit from m to Angstrom\n",
      "\n",
      "#result\n",
      "print\"For photon,the wavelength is =\",lambda_p,\"m\";\n",
      "print\"\\t\\t\\t     =\",round(lambda_p1,1),\"Angstrom\";\n",
      "print\"For electron,the wavelength is =\",lambda_e,\"m\";\n",
      "print\"\\t\\t\\t       =\",round(lambda_e1,2),\"Angstrom\";"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "For photon,the wavelength is = 1.2375e-09 m\n",
        "\t\t\t     = 12.4 Angstrom\n",
        "For electron,the wavelength is = 3.86765965524e-11 m\n",
        "\t\t\t       = 0.39 Angstrom\n"
       ]
      }
     ],
     "prompt_number": 6
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.7 , Page no:105"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "from __future__ import division\n",
      "\n",
      "#given\n",
      "lambda1=1.66*10**-10; #in m (wavelength)\n",
      "m=9.1*10**-31; #in Kg (mass of electron)\n",
      "h=6.626*10**-34; #in J-s (Planck's constant)\n",
      "e=1.6E-19; #in C (charge on electron)\n",
      "\n",
      "#calculate\n",
      "#Since lambda=h/(m*v)\n",
      "#Therefore we have\n",
      "v=h/(m*lambda1); #calculation of velocity\n",
      "K=m*v**2/2; #calculation of kinetic energy\n",
      "K1=K/e; #changing unit from J to eV\n",
      "\n",
      "#result\n",
      "print\"The velocity of electron is v=\",v,\"m/s\";\n",
      "print\"The kinetic energy is K=\",K,\"J\";\n",
      "print\"\\t\\t       =\",round(K1,3),\"eV\";\n",
      "print \"Note: The answer in the book for kinetic energy is wrong \""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The velocity of electron is v= 4386336.55501 m/s\n",
        "The kinetic energy is K= 8.75417651009e-18 J\n",
        "\t\t       = 54.714 eV\n",
        "Note: The answer in the book for kinetic energy is wrong \n"
       ]
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.8 , Page no:106"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "from __future__ import division\n",
      "\n",
      "#given\n",
      "T=400; #in K (temperature)\n",
      "m=6.7E-27; #in Kg (mass of He-atom)\n",
      "h=6.6E-34; #in J-s (Planck's constant)\n",
      "k=1.376E-23; #in J/degree (Boltzmann constant)\n",
      "\n",
      "#calculate\n",
      "#Since lambda=h/(m*v)\n",
      "#E=mv^2/2;\n",
      "#Therefore lambda=h/sqrt(2*m*E)\n",
      "#E=kT\n",
      "#Therefore lambda=h/sqrt(2*m*k*T)\n",
      "lambda1=h/math.sqrt(2*m*k*T)\n",
      "lambda2=lambda1*1E10; #changing unit from m to Angstrom\n",
      "\n",
      "#result\n",
      "print\"The de-Broglie wavelength of He-atom is =\",lambda1,\"m\";\n",
      "print\"\\t\\t\\t\\t        =\",round(lambda2,4),\"Angstrom\";    "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The de-Broglie wavelength of He-atom is = 7.6851495611e-11 m\n",
        "\t\t\t\t        = 0.7685 Angstrom\n"
       ]
      }
     ],
     "prompt_number": 8
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.9 , Page no:106"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "from __future__ import division\n",
      "\n",
      "#given\n",
      "m_e=9.1E-31; #in Kg (mass of electron)\n",
      "m_p=1.6E-27; #in Kg (mass of proton)\n",
      "h=6.626E-34; #in J-s (Planck's constant)\n",
      "c=3E8; #in m/s (velocity of light)\n",
      "\n",
      "#calculate\n",
      "E=m_e*c**2; #in J (rest energy of electron)\n",
      "#Since lambda=h/(m*v)\n",
      "#E=mv^2/2;\n",
      "#herefore lambda=h/sqrt(2*m*E)\n",
      "#Also E=m_e*c^2;\n",
      "#therefore lambda=h/sqrt(2*m_p*m_e*c^2)\n",
      "lambda1=h/math.sqrt(2*m_p*m_e*c**2); #calculation of wavelength\n",
      "lambda2=lambda1*1*10**10; #changing unit from m to Angstrom\n",
      "\n",
      "#result\n",
      "print\"The de-Broglie wavelength of proton is =\",lambda1,\"m\";\n",
      "print\"\\t\\t\\t\\t       =\",lambda2,\"Angstrom\";"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The de-Broglie wavelength of proton is = 4.0929316435e-14 m\n",
        "\t\t\t\t       = 0.00040929316435 Angstrom\n"
       ]
      }
     ],
     "prompt_number": 9
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.10 , Page no:106"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "from __future__ import division\n",
      "\n",
      "#given\n",
      "V=10000; #in V (Potential)\n",
      "\n",
      "#calculate\n",
      "lambda1=12.27/math.sqrt(V); #calculation of wavelength in Angstrom\n",
      "\n",
      "#result\n",
      "print\"The wavelength is =\",round(lambda1,3),\"Angstrom\";"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The wavelength is = 0.123 Angstrom\n"
       ]
      }
     ],
     "prompt_number": 10
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.11 , Page no:107"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "from __future__ import division\n",
      "\n",
      "#given\n",
      "V=100; #in V (Potential)\n",
      "n=1; #order of diffraction\n",
      "d=2.15; #in Angstrom (lattice spacing)\n",
      "\n",
      "#calculate\n",
      "lambda1=12.27/math.sqrt(V); #calculation of wavelength in Angstrom\n",
      "#Since 2*d*sind(theta)=n*lambda\n",
      "#therefore we have\n",
      "theta=math.asin(n*lambda1/(2*d)); #calculation of glancing angle\n",
      "theta1=theta*180/3.14;\n",
      "\n",
      "#result\n",
      "print\"The wavelength is =\",round(lambda1,3),\"Angstrom\";\n",
      "print\"The glancing angle is =\",round(theta1,1),\"degree\";\n",
      "print \"Note: In question V=100 eV but the solution is using V=100V \"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The wavelength is = 1.227 Angstrom\n",
        "The glancing angle is = 16.6 degree\n",
        "Note: In question V=100 eV but the solution is using V=100V \n"
       ]
      }
     ],
     "prompt_number": 11
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.12 , Page no:107"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "from __future__ import division\n",
      "\n",
      "#given\n",
      "V=344; #in V (Potential)\n",
      "n=1; #order of diffraction\n",
      "theta=60; #in degree (glancing angle)\n",
      "\n",
      "#calculate\n",
      "lambda1=12.27/math.sqrt(V); #calculation of wavelength in Angstrom\n",
      "#Since 2*d*sind(theta)=n*lambda\n",
      "#therefore we have\n",
      "d=n*lambda1/(2*math.sin(math.radians(theta))); #calculation of spacing constant\n",
      "\n",
      "#result\n",
      "print\"The wavelength is =\",round(lambda1,3),\"Angstrom\";\n",
      "print\"The spacing of the crystal is =\",round(d,4),\"Angstrom\";"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The wavelength is = 0.662 Angstrom\n",
        "The spacing of the crystal is = 0.3819 Angstrom\n"
       ]
      }
     ],
     "prompt_number": 12
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.13 , Page no:107"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "from __future__ import division\n",
      "\n",
      "#given\n",
      "r=0.53*10**-10; #in m (radius of first Bohr orbit)\n",
      "h=6.6*10**-34; #in J-s (Planck's constant)\n",
      "m=9.1*10**-31; #in Kg (mass of electron)\n",
      "n=1; #First Bohr orbit\n",
      "pi=3.14; #value of pi used in the solution\n",
      "\n",
      "#calculate\n",
      "#Since 2*pi*r=n*lambda  and  lambda=h/(m*v)\n",
      "#Threfore we have v=h*n/(2*pi*r*m)\n",
      "v=h*n/(2*pi*r*m); #calculation of velocity\n",
      "\n",
      "#result\n",
      "print\"The velocity of electron is =\",v,\"m/s\";"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The velocity of electron is = 2179049.16859 m/s\n"
       ]
      }
     ],
     "prompt_number": 13
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.14 , Page no:108"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "from __future__ import division\n",
      "\n",
      "#given\n",
      "dx=0.2; #in Angstrom (uncertainty in the position)\n",
      "h=6.6E-34; #in J-s (Planck's constant)\n",
      "m0=9.1E-31; #in Kg (mass of electron)\n",
      "pi=3.14; #value of pi used in the solution\n",
      "\n",
      "#calculate\n",
      "dx=dx*1E-10; #since dx is in Angstrom\n",
      "#Since dx*dp=h/4*pi  (uncertainty relation)\n",
      "dp=h/(4*pi*dx); #calculation of uncertainty in the momentum\n",
      "#since dp=m*dv\n",
      "dv=dp/m0; #calculation of uncertainty in the velocity\n",
      "\n",
      "#result\n",
      "print\"The uncertainty in the momentum is dp=\",dp,\"Kg-m/\";\n",
      "print\"The uncertainty in the velocity is dv=\",dv,\"m/s\";"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The uncertainty in the momentum is dp= 2.62738853503e-24 Kg-m/\n",
        "The uncertainty in the velocity is dv= 2887240.14839 m/s\n"
       ]
      }
     ],
     "prompt_number": 14
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.15 , Page no:108"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "from __future__ import division\n",
      "\n",
      "#given\n",
      "m_e=9.1E-31; #in Kg (mass of electron)\n",
      "m_p=1.67E-27; #in Kg (mass of proton)\n",
      "dx_p=1; #in nanometer (uncertainty in position of electron)\n",
      "dx_n=1; #in nanometer (uncertainty in position of proton)\n",
      "\n",
      "#calculate\n",
      "#since dp=h/(4*pi*dx)\n",
      "#since h/(4*pi) is constant and dx is same for electron and proton \n",
      "#therefor both electron and proton have same uncertainty in the momentum\n",
      "#since dv=dp/m  and  dp is same for both\n",
      "#therefore dv_e/dv_p=m_p/m_e\n",
      "#therefore\n",
      "K=m_p/m_e; #ratio of uncertainty in the velocity of electron and proton\n",
      "\n",
      "#result\n",
      "print\"The ratio of uncertainty in the velocity of electron to that of proton is =\",round(K);"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The ratio of uncertainty in the velocity of electron to that of proton is = 1835.0\n"
       ]
      }
     ],
     "prompt_number": 15
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.16 , Page no:108"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "from __future__ import division\n",
      "\n",
      "#given\n",
      "dx=5*10**-15; #in m (radius of nucleus or uncertainty in the position)\n",
      "h=6.6*10**-34; #in J-s (Planck's constant)\n",
      "m=1.67E-27; #in Kg (mass of proton)\n",
      "pi=3.14; #value of pi used in the solution\n",
      "e=1.6E-19; #in C (charge of electron)\n",
      "\n",
      "#calculate\n",
      "#Since dx*dp=h/4*pi  (uncertainty relation)\n",
      "dp=h/(4*pi*dx); #calculation of uncertainty in the momentum\n",
      "p=dp; #minimum value of momentum to calculate mimimum kinetic energy\n",
      "K=(p**2)/(2*m); #calculation of minimum kinetic energy of proton\n",
      "K1=K/e; #changing unit from J to eV\n",
      "K2=K1/1E6; #changing unit from eV to MeV\n",
      "\n",
      "#result\n",
      "print\"The minimum uncertainty in the momentum of proton is dp =\",dp,\"Kg-m/s\";\n",
      "print\"The minimum kinetic energy of proton is K=\",K,\"J\";\n",
      "print\"\\t\\t\\t\\t \\t =\",K1,\"eV\";\n",
      "print\"\\t\\t\\t\\t\\t =\",K2,\"MeV\";"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The minimum uncertainty in the momentum of proton is dp = 1.05095541401e-20 Kg-m/s\n",
        "The minimum kinetic energy of proton is K= 3.30690803067e-14 J\n",
        "\t\t\t\t \t = 206681.751917 eV\n",
        "\t\t\t\t\t = 0.206681751917 MeV\n"
       ]
      }
     ],
     "prompt_number": 16
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.17 , Page no:109"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "from __future__ import division\n",
      "\n",
      "#given\n",
      "K=1; #in KeV (kinetic energy of electron)\n",
      "dx=1; #in Angstrom (uncertainty in the position)\n",
      "h=6.63E-34; #in J-s (Planck's constant)\n",
      "m=9.1E-31; #in Kg (mass of electron)\n",
      "pi=3.14; #value of pi used in the solution\n",
      "e=1.6E-19; #in C (charge of electron)\n",
      "\n",
      "#calculate\n",
      "dx=dx*1E-10; #since dx is in Angstrom\n",
      "#Since dx*dp=h/4*pi  (uncertainty relation)\n",
      "dp=h/(4*pi*dx); #calculation of uncertainty in the momentum\n",
      "K=K*1E3*1.6E-19; #changing unit from KeV to J\n",
      "p=math.sqrt(2*m*K); #calculation of momentum \n",
      "poc=(dp/p)*100; #calculation of percentage of uncertainty\n",
      "\n",
      "#result\n",
      "print\"The uncertainty in the momentum of electron is dp=\",dp,\"Kg-m/s\";\n",
      "print\"The momentum of electron is p=\",p,\"Kg-m/s\";\n",
      "print\"The percentage of uncertainty in the momentum  is =\",round(poc,1);"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The uncertainty in the momentum of electron is dp= 5.27866242038e-25 Kg-m/s\n",
        "The momentum of electron is p= 1.70645832062e-23 Kg-m/s\n",
        "The percentage of uncertainty in the momentum  is = 3.1\n"
       ]
      }
     ],
     "prompt_number": 17
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.18 , Page no:109"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "from __future__ import division\n",
      "\n",
      "#given\n",
      "v=6.6E4; #m/s (speed of electron)\n",
      "poc=0.01; #percentage of uncertainty\n",
      "h=6.63E-34; #in J-s (Planck's constant)\n",
      "m=9E-31; #in Kg (mass of electron)\n",
      "pi=3.14; #value of pi used in the solution\n",
      "\n",
      "#calculate\n",
      "p=m*v; #calculation of momentum\n",
      "dp=(poc/100)*p; #calculation of uncertainty in the momentum\n",
      "#Since dx*dp=h/4*pi  (uncertainty relation)\n",
      "dx=h/(4*pi*dp); #calculation of uncertainty in the position\n",
      "\n",
      "#result\n",
      "print\"The momentum of electron is p=\",p,\"Kg-m/s\";\n",
      "print\"The uncertainty in the momentum of electron is dp=\",dp,\"Kg-m/s\";\n",
      "print\"The uncertainty in the position of electron is dx=\",dx,\"Kg-m/s\";"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The momentum of electron is p= 5.94e-26 Kg-m/s\n",
        "The uncertainty in the momentum of electron is dp= 5.94e-30 Kg-m/s\n",
        "The uncertainty in the position of electron is dx= 8.88663707135e-06 Kg-m/s\n"
       ]
      }
     ],
     "prompt_number": 18
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.19 , Page no:111"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "from __future__ import division\n",
      "\n",
      "#given\n",
      "lambda1=1; #in Angstrom (wavelength)\n",
      "pi=3.14; #value of pi used in the solution\n",
      "dlambda=1E-6; #uncertainty in wavelength\n",
      "\n",
      "#calculate\n",
      "lambda2=lambda1*1E-10; #sinc lambda is in Angstrom\n",
      "#By uncertainty principle, dx*dp>=h/(4*pi)  --(1)\n",
      "#since p=h/lambda  -----(2)\n",
      "#Or p*lambda=h \n",
      "#diffrentiting this equation\n",
      "#p*dlambda+lambda*dp=0\n",
      "#dp=-p*dlambda/lambda     ----(3)\n",
      "#from (2) and (3)  dp=-h*dlambda/lambda^2   ---(4)\n",
      "#from (1) and(4)  dx*dlambda>=lambda^2/4*pi\n",
      "#Or dx=lambda^2/(4*pi*dlambda)\n",
      "dx=lambda2**2/(4*pi*dlambda); #calculation of uncertainty in the position\n",
      "\n",
      "#result\n",
      "print\"The uncertainty in the position of X-ray photon is dx=\",dx,\"m\";\n",
      "print \"Note: wavelength accuracy is given as 1 in 1E8 but in book solution has used 1 in 1E6 \"\n",
      "print \" ANSWEER IS WRONG IN THE TEXTBOOK\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The uncertainty in the position of X-ray photon is dx= 7.96178343949e-16 m\n",
        "Note: wavelength accuracy is given as 1 in 1E8 but in book solution has used 1 in 1E6 \n",
        " ANSWEER IS WRONG IN THE TEXTBOOK\n"
       ]
      }
     ],
     "prompt_number": 19
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.20 , Page no:111"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "from __future__ import division\n",
      "\n",
      "#given\n",
      "dt=1*10**-8; #in sec (average life time)\n",
      "pi=3.14; #value of pi used in the solution\n",
      "\n",
      "#calculate\n",
      "#Since dE*dt>=h/(4*pi)  (uncertainty relation for energy)\n",
      "#and E=h*v      v is the frequency\n",
      "#therefore we have dv>=1/(4*pi*dt)\n",
      "dv=1/(4*pi*dt); #calculation of minimum uncertainty in the frequency\n",
      "\n",
      "#result\n",
      "print\"The minimum uncertainty in the frequency of the photon is dv=\",dv,\"sec^-1\";"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The minimum uncertainty in the frequency of the photon is dv= 7961783.43949 sec^-1\n"
       ]
      }
     ],
     "prompt_number": 20
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.21 , Page no:111"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "from __future__ import division\n",
      "\n",
      "#given\n",
      "dt=1E-12; #in sec (average life time)\n",
      "h=6.63E-34; #in J-s (Planck'c constant)\n",
      "pi=3.14; #value of pi used in the solution\n",
      "e=1.6*1E-19; #in C (charge of electron)\n",
      "\n",
      "#calculate\n",
      "#Since dE*dt>=h/(4*pi)  (uncertainty relation for energy)\n",
      "dE=h/(4*pi*dt); #calculation of minimum uncertainty in the energy\n",
      "dE1=dE/e; #changing unit from J to eV\n",
      "\n",
      "#result\n",
      "print\"The uncertainty in the energy of the photon is dE=\",dE,\"J\";\n",
      "print\"\\t\\t\\t\\t\\t         =\",dE1,\"eV\";"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The uncertainty in the energy of the photon is dE= 5.27866242038e-23 J\n",
        "\t\t\t\t\t         = 0.000329916401274 eV\n"
       ]
      }
     ],
     "prompt_number": 21
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.22 , Page no:111"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "from __future__ import division\n",
      "\n",
      "#given\n",
      "dT=2.5E-14; #in sec (average life time)\n",
      "h=6.63E-34; #in J-s (Planck'c constant)\n",
      "pi=3.14; #value of pi used in the solution\n",
      "e=1.6*1E-19; #in C (charge of electron)\n",
      "\n",
      "#calculate\n",
      "#Since dE*dt>=h/(4*pi)  (uncertainty relation for energy)\n",
      "dE=h/(4*pi*dT); #calculation of minimum uncertainty in the energy\n",
      "dE1=dE/e; #changing unit from J to eV\n",
      "\n",
      "#result\n",
      "print\"The uncertainty in the energy of the photon is dE=\",dE,\"J\";\n",
      "print\"\\t\\t\\t\\t\\t         =\",dE1,\"eV\";"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The uncertainty in the energy of the photon is dE= 2.11146496815e-21 J\n",
        "\t\t\t\t\t         = 0.013196656051 eV\n"
       ]
      }
     ],
     "prompt_number": 22
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.23 , Page no:112"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "from __future__ import division\n",
      "\n",
      "#given\n",
      "a=2; #in Angstrom (length of the box)\n",
      "m=9.1E-31; #in Kg (mass of electron)\n",
      "h=6.626E-34; #in J-s (Planck'c constant)\n",
      "n2=2; n4=4; #two quantum states\n",
      "e=1.6*1E-19; #in C (charge of electron)\n",
      "\n",
      "#calculate\n",
      "a=a*1E-10; #since a is in Angstrom \n",
      "#Since  E_n=n^2*h^2/(8*m*a^2)   (Energy corresponding to nth quantum state)\n",
      "E2=n2**2*h**2/(8*m*a**2); #calculation of energy corresponding to the 2nd quantum state\n",
      "E21=E2/e; #changing unit from J to eV\n",
      "E4=n4**2*h**2/(8*m*a**2); #calculation of energy corresponding to the 4nd quantum state\n",
      "E41=E4/e; #changing unit from J to eV\n",
      "\n",
      "#result\n",
      "print\"The energy corresponding to the 2nd quantum state is E2=\",E2,\"J\";\n",
      "print\"\\t\\t\\t\\t\\t\\t       =\",E21,\"eV\";\n",
      "print\"The energy corresponding to the 4nd quantum state is E4=\",E4,\"J\";\n",
      "print\"\\t\\t\\t\\t\\t\\t       =\",E41,\"eV\";"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The energy corresponding to the 2nd quantum state is E2= 6.0307521978e-18 J\n",
        "\t\t\t\t\t\t       = 37.6922012363 eV\n",
        "The energy corresponding to the 4nd quantum state is E4= 2.41230087912e-17 J\n",
        "\t\t\t\t\t\t       = 150.768804945 eV\n"
       ]
      }
     ],
     "prompt_number": 23
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.24 , Page no:113"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "from __future__ import division\n",
      "\n",
      "#given\n",
      "a=1E-10; #in m (width of the well)\n",
      "m=9.1E-31; #in Kg (mass of electron)\n",
      "h=6.626E-34; #in J-s (Planck'c constant)\n",
      "n1=1; n2=2; n3=3; #ground and first two excited states\n",
      "e=1.6*1E-19; #in C (charge of electron)\n",
      "\n",
      "#calculate\n",
      "#Since  E_n=n^2*h^2/(8*m*a^2)   (Energy corresponding to nth quantum state)\n",
      "E1=n1**2*h**2/(8*m*a**2); #calculation of energy corresponding to the Ground state\n",
      "E11=E1/e; #changing unit from J to eV\n",
      "E2=n2**2*h**2/(8*m*a**2); #calculation of energy corresponding to the 1st excited state\n",
      "E21=E2/e; #changing unit from J to eV\n",
      "E3=n3**2*h**2/(8*m*a**2); #calculation of energy corresponding to the 2nd excited state\n",
      "E31=E3/e; #changing unit from J to eV\n",
      "\n",
      "#result\n",
      "print\"The energy corresponding to the ground state is E1=\",E1,\"J\";\n",
      "print\"\\t\\t\\t\\t\\t =\",E11,\"eV\";\n",
      "print\"The energy corresponding to the 1st excited state is E2=\",E2,\"\";\n",
      "print\"\\t\\t\\t\\t\\t =\",E21,\"eV\";\n",
      "print\"The energy corresponding to the 2nd excited state is E3=\",E3,\"J\";\n",
      "print\"\\t\\t\\t\\t\\t =\",E31,\"eV\";\n",
      "print \" There is slight variation in the answer due to round off error\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The energy corresponding to the ground state is E1= 6.0307521978e-18 J\n",
        "\t\t\t\t\t = 37.6922012363 eV\n",
        "The energy corresponding to the 1st excited state is E2= 2.41230087912e-17 \n",
        "\t\t\t\t\t = 150.768804945 eV\n",
        "The energy corresponding to the 2nd excited state is E3= 5.42767697802e-17 J\n",
        "\t\t\t\t\t = 339.229811126 eV\n",
        " There is slight variation in the answer due to round off error\n"
       ]
      }
     ],
     "prompt_number": 24
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.25 , Page no:113"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "from __future__ import division\n",
      "\n",
      "#given\n",
      "dx=1E-8; #in m (length of box or uncertainty in the position)\n",
      "h=6.626E-34; #in J-s (Planck'c constant)\n",
      "m=9.1E-31; #in Kg (mass of electron)\n",
      "\n",
      "#calculate\n",
      "#From uncertainty principle dx*dp=h and dp=m*dv\n",
      "#therefore we have\n",
      "dv=h/(m*dx); #calculation of minimum uncertainty in the velocity\n",
      "dv1=dv*1E-3; #changing unit from m/s to Km/s\n",
      "\n",
      "#result\n",
      "print\"The minimum uncertainty in the velocity of electron i dv=\",dv,\"m/s\";\n",
      "print\"\\t\\t\\t\\t\\t\\t        =\",round(dv1,2),\"Km/s\";\n",
      "print \"Note: There is slight variation in the answer due to round off error\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The minimum uncertainty in the velocity of electron i dv= 72813.1868132 m/s\n",
        "\t\t\t\t\t\t        = 72.81 Km/s\n",
        "Note: There is slight variation in the answer due to round off error\n"
       ]
      }
     ],
     "prompt_number": 25
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.26 , Page no:113"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "from __future__ import division\n",
      "\n",
      "#given\n",
      "a=4E-10; #in m (length of the box)\n",
      "m=9.1E-31; #in Kg (mass of electron)\n",
      "h=6.626E-34; #in J-s (Planck'c constant)\n",
      "n1=1; #ground state\n",
      "e=1.6*1E-19; #in C (charge of electron)\n",
      "\n",
      "#calculate\n",
      "#Since  E_n=n^2*h^2/(8*m*a^2)   (Energy corresponding to nth quantum state)\n",
      "E1=n1**2*h**2/(8*m*a**2); #calculation of energy corresponding to the ground state\n",
      "E11=E1/e; #changing unit from J to eV\n",
      "\n",
      "#result\n",
      "print\"The minimum energy of electron is E1=\",E1,\"J (Note: The answer in the book  is wrong due to printing error)\";\n",
      "print\"\\t\\t\\t\\t    =\",round(E11,3),\"eV\";\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The minimum energy of electron is E1= 3.76922012363e-19 J (Note: The answer in the book  is wrong due to printing error)\n",
        "\t\t\t\t    = 2.356 eV\n"
       ]
      }
     ],
     "prompt_number": 26
    }
   ],
   "metadata": {}
  }
 ]
}