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  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 13: Nuclear Interactiions and Applications"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 13.1, Page 479"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "N_A = 6.02e+023;    # Avogadro's number\n",
      "e = 1.6e-019;    # Charge on an electron, C\n",
      "q = 2*e;    # Charge on the alpha particle, C\n",
      "rho = 1.9;    # Density of carbon target, atoms/cc\n",
      "N_M = 1;    # Number of atoms per molecule\n",
      "M_g = 12;    # Gram atomic mass of C12 isotope, g/mol\n",
      "sigma = 25e-031;    # Total cross section for the reaction, Sq.m\n",
      "t = 1e-006;    # Thickness of carbon target, m\n",
      "I_beam = 1e-006;    # Beam current of akpha particle, ampere\n",
      "time = 3600;    # Time for which the alpha particle beam is incident on the target, s\n",
      "\n",
      "#Calculations\n",
      "n = rho*N_A*N_M/M_g;    # Number of nuclei per unit volume, per cc\n",
      "P = n*t*sigma*1e+006;     # Probability of scattering of alpha particles\n",
      "N_I = I_beam*time/q;    # Number of incident alpha particles\n",
      "N_n = N_I*P;    # Number of neutrons produced in the reaction\n",
      "\n",
      "#Result\n",
      "print \"The number of neutrons produced in the reaction = %3.1e neutrons\"%N_n\n",
      "#answer differs due to rounding errors"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The number of neutrons produced in the reaction = 2.7e+09 neutrons\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 13.2, Page 480"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "sigma_n = 3;    # Differential cross setion of production of the neutron, mb/sr\n",
      "sigma_p = 0.2;    # Differential cross setion of production of the proton, mb/sr\n",
      "\n",
      "#Calculations\n",
      "# As P_n = sigma_n and P_p = sigma_p, so\n",
      "P_ratio = sigma_n/sigma_p;    # The likelihood of a neutron production than a proton\n",
      "\n",
      "#Result\n",
      "print \"The likelihood of the neutron production than the proton = %2d\"%P_ratio"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The likelihood of the neutron production than the proton = 15\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 13.3, Page 481"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "u = 931.5;    # Energy equivalent of 1 amu, MeV\n",
      "M_He = 4.002603;    # Mass of He-4 nucleus, u\n",
      "M_N = 14.003074;    # Mass of N-14 nucleus, u\n",
      "M_H = 1.007825;    # Mass of hydrogen nucleus, u\n",
      "M_O = 16.999132;    # Mass of O-16 nucleus, u\n",
      "K_alpha = 7.7;    # The kinetic energy of alpha particle, MeV\n",
      "\n",
      "#Calculations\n",
      "Q = (M_He + M_N - M_H - M_O)*u;    # The ground state Q-value of the nuclear reaction, MeV\n",
      "# As Q = K_p + K_O - K_alpha, solving for K_p + K_O\n",
      "K = Q + K_alpha;    # The sum of kinetic energy of the products, MeV\n",
      "\n",
      "#Results\n",
      "print \"The ground state Q-value of the endoergic nuclear reaction = %5.3f MeV\"%Q\n",
      "print \"The sum of kinetic energy of the products = %3.1f MeV\"%K"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The ground state Q-value of the endoergic nuclear reaction = -1.192 MeV\n",
        "The sum of kinetic energy of the products = 6.5 MeV\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 13.4, Page 485"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "u = 931.5;    # Energy equivalent of 1 amu, MeV\n",
      "K_lab = 14.6;    # Kinetic energy of incident aplha particle, MeV\n",
      "Mx = 4;    # Mass number of projectile nucleus\n",
      "MX = 12;    # Mass number of target nucleus\n",
      "M_He = 4.002603;    # Mass of He nucleus, u\n",
      "M_C = 12.0 # Mass of carbon nucleus, u\n",
      "M_O = 15.994915;    # Mass of oxygen nucleus, u\n",
      "\n",
      "#Calculations\n",
      "K_cm = MX/(Mx + MX)*K_lab;    # Kinetic energy available in the centre of mass, MeV\n",
      "E_O = (M_He + M_C - M_O)*u;    # The ground state excitation energy of O-16, MeV\n",
      "E_final_O = K_cm + E_O;    # The final excitation energy of O-16, MeV\n",
      "\n",
      "#Result\n",
      "print \"The final excitation energy of O-16 = %4.2f MeV\"%E_final_O"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The final excitation energy of O-16 = 7.16 MeV\n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 13.5, Page 487"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "\n",
      "#Variable declaration\n",
      "u = 931.5;    # Energy equivalent of 1 amu, MeV\n",
      "M_U235 = 235.0439;    # Mass of U-235 nucleus, u\n",
      "m_n = 1.0087;    # Mass of a neutron, u\n",
      "M_Zr99 = 98.9165;    # Mass of Zr-99 nucleus, u\n",
      "M_Te134 = 133.9115;    # Mass of Te-134 nucleus, u\n",
      "\n",
      "#Calculations\n",
      "Q = (M_U235 + m_n - M_Zr99 - M_Te134 - 3*m_n)*u;    # Q-value of the reaction, MeV\n",
      "\n",
      "#Result\n",
      "print \"The ground state Q-value of the induced fission reaction = %3d MeV\"%math.ceil(Q)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The ground state Q-value of the induced fission reaction = 185 MeV\n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 13.6, Page 488"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "u = 931.5;    # Energy equivalent of 1 amu, MeV\n",
      "m_n = 1.0087;    # Mass of a neutron, u\n",
      "M_U235 = 235.0439;    # Mass of U-235 nucleus, u\n",
      "M_U236 = 236.0456;    # Mass of U-236 nucleus, u\n",
      "M_U238 = 238.0508;    # Mass of U-238 nucleus, u\n",
      "M_U239 = 239.0543;    # Mass of U-239 nucleus, u\n",
      "\n",
      "#Calculations\n",
      "E_U236 = (m_n + M_U235 - M_U236)*u;    # Excitation energy of U-236 nucleus, MeV\n",
      "E_U239 = (m_n + M_U238 - M_U239)*u;    # Excitation energy of U-239 nucleus, MeV\n",
      "\n",
      "#Results\n",
      "print \"The excitation energy of U-236 nucleus = %3.1f MeV\"%E_U236\n",
      "print \"The excitation energy of U-239 nucleus = %3.1f MeV\"%E_U239"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The excitation energy of U-236 nucleus = 6.5 MeV\n",
        "The excitation energy of U-239 nucleus = 4.8 MeV\n"
       ]
      }
     ],
     "prompt_number": 6
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 13.7, Page 490"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "t = 30e-003;    # Time during which the number of fissions is to be calculated, s\n",
      "E = 185;    # Energy produced for each fission, MeV\n",
      "delta_t = 5e-006;    # Average time during which a neutron is captured, s\n",
      "\n",
      "#Calculations\n",
      "fs = t/delta_t;    # Number of fission cycles within 30 ms\n",
      "N = (1.01)**fs;    # Number of fissions that occur in 30 ms\n",
      "E_total = N*E;    # Total energy produced in 30 ms, MeV\n",
      "\n",
      "#Results\n",
      "print \"The total number of fissions that occur in %d ms = %3.1e\"%(t/1e-003, N)\n",
      "print \"The total energy produced = %3.1e MeV\"%E_total\n",
      "\n",
      "#Incorrect solution in textbook"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The total number of fissions that occur in 30 ms = 8.5e+25\n",
        "The total energy produced = 1.6e+28 MeV\n"
       ]
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 13.8, Page 500"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "u = 931.5;    # Energy equivalent of 1 amu, MeV\n",
      "M_He = 4.002603;    # Mass of He nucleus, u\n",
      "M_C = 12.0 # Mass of carbon nucleus, u\n",
      "M_O = 15.994915;    # Mass of oxygen nucleus, u\n",
      "\n",
      "#Calculations\n",
      "Q = (M_He + M_C - M_O)*u;    # Q-value of the reaction, MeV\n",
      "\n",
      "#Result\n",
      "print \"The energy expended in the fusion reaction inside supergiant star = %3.1f MeV\"%Q"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The energy expended in the fusion reaction inside supergiant star = 7.2 MeV\n"
       ]
      }
     ],
     "prompt_number": 8
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 13.9, Page 502"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "k = 1.38e-023;    # Boltzmann constant, J/K\n",
      "r = 3e-015;    # Distance at which the nuclear force becomes effective, m\n",
      "e = 1.6e-019;  # Charge on an electron, C\n",
      "K = 9e+009;    # Coulomb's constant, N-Sq.m/C^2\n",
      "\n",
      "#Calculations\n",
      "V = K*e**2/r;    # Coulomb potential energy, J\n",
      "# As V = 3/2*k*T, solving for T\n",
      "T = 2./3*V/k;    # The ignition temperature needed for the fusion reaction between deuterium and a tritium, K\n",
      "\n",
      "#Result\n",
      "print \"The ignition temperature needed for the fusion reaction between a deuterium and a tritium = %3.1e K\"%T"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The ignition temperature needed for the fusion reaction between a deuterium and a tritium = 3.7e+09 K\n"
       ]
      }
     ],
     "prompt_number": 9
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 13.10, Page 509"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "k = 1.38e-023;    # Boltzmann constant, J/K\n",
      "e = 1.6e-019;  # Energy equivalent of 1 eV, J\n",
      "h = 6.62e-034;    # Planck's oconstant, Js\n",
      "m = 1.67e-027;    # Mass of the neutron, kg\n",
      "lamda = 0.060e-009;    # Wavelength of the neutron beam, m\n",
      "\n",
      "#Calculations\n",
      "p = h/lamda;    # Momentum of the neutron from de-Broglie relation, kg-m/s\n",
      "K = p**2/(2*m*e);    # Kinetic energy of the neutron needed to study atomic structure, eV\n",
      "# As K = 3/2*k*T, solving for T\n",
      "T = 2./3*K*e/k;    # The temperature of the neutron needed to study atomic structure, K\n",
      "\n",
      "#Result\n",
      "print \"The energy and temperature of the neutron needed to study the atomic structure of solids = %4.2f eV and %4d K respectively\"%(K, T)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The energy and temperature of the neutron needed to study the atomic structure of solids = 0.23 eV and 1760 K respectively\n"
       ]
      }
     ],
     "prompt_number": 10
    }
   ],
   "metadata": {}
  }
 ]
}