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  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "chapter 10 : Structure of Nuclei"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex10.1.1 : Pg:209 "
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Energy and mass equivalence of wavelength\n",
      "e = 1.6e-019     # Energy equivalent of 1 eV, J/eV\n",
      "me = 9.1e-031     # Mass of en electron, kg\n",
      "L = 4.5e-013     # Wavelength of gamma ray, m\n",
      "h = 6.626e-034     # Planck's constant, Js\n",
      "c = 3e+08     # Speed of light, m/s\n",
      "U = h*c/L     # Energy equivalence of wavelength, J\n",
      "m = U/c**2     # Mass equivalent of wavelength, kg\n",
      "print \"The energy equivalence of wavelength %3.1e m = %4.2f MeV\"% (L, U/(e*1e+06)) \n",
      "print \"The mass equivalence of wavelength %3.1e m = %4.2f me\"% (L, m/me) \n",
      "# Result \n",
      "# The energy equivalence of wavelength 4.5e-013 m = 2.76 MeV\n",
      "# The mass equivalence of wavelength 4.5e-013 m = 5.39 me "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The energy equivalence of wavelength 4.5e-13 m = 2.76 MeV\n",
        "The mass equivalence of wavelength 4.5e-13 m = 5.39 me\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex10.1.2 : Pg:210 "
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Binding energy per nucleon for oxygen isotopes\n",
      "mp = 1.007276     # Mass of proton, amu\n",
      "mn = 1.008665     # Mass of neutron, amu\n",
      "amu = 931     # Energy equivalent of 1 amu, MeV\n",
      "# For Isotope O-16\n",
      "M_O16 = 15.990523     # Mass of O-16 isotope, amu\n",
      "Z = 8     # Number of protons\n",
      "N = 8     # Number of neutrons\n",
      "BE = (8*(mp+mn)-M_O16)*amu     # Binding energy of O-16 isotope, MeV\n",
      "BE_bar16 = BE/(Z+N)     # Binding energy per nucleon of O-16 isotope, MeV\n",
      "# For Isotope O-18\n",
      "M_O18 = 17.994768     # Mass of O-18 isotope, amu\n",
      "Z = 8     # Number of protons\n",
      "N = 10     # Number of neutrons\n",
      "BE = (8*mp+10*mn-M_O18)*amu     # Binding energy of O-18 isotope, MeV\n",
      "BE_bar18 = BE/(Z+N)     # Binding energy per nucleon of O-18 isotope, MeV\n",
      "print \"The binding energy per nucleon of O-16 isotope = %5.3f MeV\"% BE_bar16 \n",
      "print \"The binding energy per nucleon of O-18 isotope = %5.3f MeV\"% BE_bar18 \n",
      "# Result \n",
      "# The binding energy per nucleon of O-16 isotope = 7.972 MeV\n",
      "# The binding energy per nucleon of O-18 isotope = 7.763 MeV "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The binding energy per nucleon of O-16 isotope = 7.972 MeV\n",
        "The binding energy per nucleon of O-18 isotope = 7.763 MeV\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex10.2.1 : Pg:214 "
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import log, exp\n",
      "from sympy import symbols, solve\n",
      "#Range of alpha-emitters of uranium\n",
      "L1 = 4.8e-018     # Decay constant of first alpha-emitter, per sec\n",
      "L2 = 4.225e+03     # Decay constant of second alpha-emitter, per sec\n",
      "L3 = 3.786e-03     # Decay constant of third alpha-emitter, per sec\n",
      "R1 = 4.19     # Range of first alpha-emitter, cm\n",
      "R2 = 7.86     # Range of second alpha-emitter, cm\n",
      "# From Geiger Nuttal law, log R = A log L + B\n",
      "# Putting R1, L1 and R2, L2, subtracting and solving for A\n",
      "A = log(R2/R1)/log(L2/L1)     # Slope of straight line between R and L\n",
      "B = symbols('B')     # Intercept of straight line between R and L\n",
      "B = solve(log(R2)-A*log(L2)-B, B)[0]     # Other constant of Geiger-Nuttal law\n",
      "R3 = exp(A*log(L3)+B)     # Range of third alpha-emitter of uranium, cm\n",
      "print \"The range of third alpha-emitter of uranium = %5.3f cm\"% R3"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The range of third alpha-emitter of uranium = 6.554 cm\n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex10.3.1 : Pg:219 "
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Binding energy per nucleon of helium\n",
      "amu = 931     # Energy equivalent of amu, MeV\n",
      "mp = 1.007895     # Mass of proton, amu\n",
      "mn = 1.008665     # Mass of neutron, amu\n",
      "M_He = 4.00260     # Atomic weight of helium, amu\n",
      "dm = 2*(mp+mn)-M_He     # Mass difference, amu\n",
      "BE = dm*amu     # Binding energy of helium, MeV\n",
      "BE_bar = BE/4     # Binding energy per nucleon, MeV\n",
      "print \"The binding energy per nucleon of helium = %6.4f MeV\"% BE_bar"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The binding energy per nucleon of helium = 7.1035 MeV\n"
       ]
      }
     ],
     "prompt_number": 6
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex10.3.2 : Pg:220"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Energy released in the fusion of deuterium \n",
      "e = 1.6e-019     # Energy equivalent of 1 eV, J/eV\n",
      "Q = 43     # Energy released in fusion of six deuterium atoms, MeV\n",
      "N = 6.023e+026     # Avogadro's number, No. of atoms per kg\n",
      "n = N/2     # Number of atoms contained in 1 kg of deuterium\n",
      "U = Q/6*n*e*1e+06     # Energy released due to fusion of 1 kg of deuterium, J\n",
      "print \"The energy released due to fusion of 1 kg of deuterium = %5.3e J\"% U"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The energy released due to fusion of 1 kg of deuterium = 3.373e+14 J\n"
       ]
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex10.3.3 : Pg: 220"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Mass of deuterium nucleus \n",
      "amu = 1.6e-027     # Mass of a nucleon, kg\n",
      "mp = 1.007895     # Mass of proton, amu\n",
      "mn = 1.008665     # Mass of neutron, amu\n",
      "BE = 2/931     # Binding energy of two nucleons, amu\n",
      "M_D = (mp+mn-BE)*amu     # Mass of a deuterium nucleus, kg\n",
      "print \"The mass of deuterium nucleus = %5.3e kg\"% M_D"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The mass of deuterium nucleus = 3.226e-27 kg\n"
       ]
      }
     ],
     "prompt_number": 9
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex10.3.4 : Pg: 220 "
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Binding energy per nucleon of Ni-64\n",
      "amu = 931     # Mass of a nucleon, MeV\n",
      "MH = 1.007825     # Mass of hydrogen, amu\n",
      "Me = 0.000550     # Mass of electron, amu\n",
      "Mp = MH-Me     # Mass of proton, amu\n",
      "Mn = 1.008665     # Mass of neutron, amu\n",
      "m_Ni = 63.9280     # Mass of Ni-64 atom, amu\n",
      "MNi = m_Ni-28*Me     # Mass of ni-64 nucleus, amu\n",
      "m = (28*Mp+36*Mn)-MNi     # Mass difference, amu\n",
      "BE = m*amu     # Binding energy of Ni-64, MeV\n",
      "BE_bar = BE/64     # Binding energy per nucleon of Ni-64, MeV\n",
      "print \"The binding energy per nucleon of Ni-64 = %4.2f MeV\"% BE_bar "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The binding energy per nucleon of Ni-64 = 8.77 MeV\n"
       ]
      }
     ],
     "prompt_number": 10
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex10.3.5 : Pg: 221 "
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Energy released during fusion of two deuterons\n",
      "e = 1.6e-019     # Energy equivalent of 1 eV, J/eV\n",
      "x = 1.1     # Binding energy per nucleon of deuterium, MeV\n",
      "y = 7.0     # Binding energy per nucleon of helium-4, MeV\n",
      "E = (y - 2*x)*1e+06*e     # Energy released when two deutron nuclei fuse together, MeV\n",
      "print \"The binding energy per nucleon of deuterium = %4.2e J\"%E "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The binding energy per nucleon of deuterium = 7.68e-13 J\n"
       ]
      }
     ],
     "prompt_number": 12
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex10.6 : Pg: 221"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from __future__ import division\n",
      "#Binding energy and packing fraction of helium\n",
      "amu = 931.0     # Energy equivalent of amu, MeV\n",
      "mp = 1.00814     # Mass of proton, amu\n",
      "mn = 1.00898     # Mass of neutron, amu\n",
      "m_He = 4.00387     # Mass of helium, amu\n",
      "A = 4.0     # Mass number of helium\n",
      "m = 2.0*(mp+mn)-m_He     # Mass difference, amu\n",
      "dm = m_He - A     # Mass defect of He\n",
      "BE = dm*amu     # Binding energy of He, MeV\n",
      "p = dm/A     # Packing fraction of He\n",
      "print \"The binding energy of helium = %6.3f MeV\"% BE  \n",
      "print \"The packing fraction of helium = %5.3e\"% p \n",
      "#Answer is wrong in the book."
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The binding energy of helium =  3.603 MeV\n",
        "The packing fraction of helium = 9.675e-04\n"
       ]
      }
     ],
     "prompt_number": 21
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex10.7 : Pg: 222 "
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Mass of Yukawa particle\n",
      "h = 6.626e-034     # Reduced Planck's constant, Js\n",
      "e = 1.6e-019     # Charge on an electron, coulomb\n",
      "R0 = 1.2e-015     # Nuclear radius constant, m\n",
      "R = 2*R0     # Range of nuclear force, m\n",
      "v = 1e+08     # Speed of the particle, m/s\n",
      "S = R     # Distance travelled by particle within the nucleus, m\n",
      "dt = S/v     # time taken by the particle to travel across the nucleus, s\n",
      "# From Heisenberg's uncertainty principle, dE.dt = h_bar, solving for dE\n",
      "dE = h/(1e+06*e*dt)     # Energy of Yukawa paeticle, MeV\n",
      "m = dE/0.51     # Approximate mass of Yukawa particle, electronic mass unit\n",
      "print \"The mass of Yukawa particle = %3d me\"% m"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The mass of Yukawa particle = 338 me\n"
       ]
      }
     ],
     "prompt_number": 19
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Ex10.8 : Pg:222"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import pi\n",
      "#Maximum height of the potential barrier for alpha penetration \n",
      "epsilon_0 = 8.854e-12     # Absolute electrical permittivity of free space, coulomb square per newton per metre square\n",
      "Z = 92     # Atomic number of U-92 nucleus\n",
      "z = 2     # Atomic number of He nucleus\n",
      "e = 1.6e-019     # Charge on an electron, coulomb\n",
      "R = 9.3e-015     # Radius of residual nucleus, m\n",
      "U = 1/(4*pi*epsilon_0)*Z*z*e**2/(R*1.6e-013)     # Maximum height of potential barrier, MeV\n",
      "print \"The maximum height of the potential barrier for alpha penetration = %2d MeV\"% U "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The maximum height of the potential barrier for alpha penetration = 28 MeV\n"
       ]
      }
     ],
     "prompt_number": 20
    }
   ],
   "metadata": {}
  }
 ]
}