path: root/sample_notebooks/NirenNegandhi/ch2_2.ipynb

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  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 2: Nuclear Models"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 2.2.1, Page 66"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "# For Ca(20,40), actual binding energy is ...... \n",
      "m_p = 1.007825; # Mass of proton, amu\n",
      "m_n = 1.008665; # Mass of neutron, amu\n",
      "Z = 20; # Number of protons\n",
      "N = 20;  # Number of neutrons\n",
      "M_n = 39.962591; # Mass of the nucleus, amu\n",
      "B_actual = (M_n-Z*m_p-N*m_n)*931.49; # Actual binding energy, MeV\n",
      "# For Ca(20,40), Binding energ as per semiemperical mas formula......\n",
      "Z = 20; # Number of protons\n",
      "a_v = 15.5; # Volume constant, MeV\n",
      "a_s = 16.8; # Surface constant, MeV\n",
      "a_a = 23.0; # Asymmetric constant, MeV\n",
      "a_c = 0.7; # Coulomb constant, MeV\n",
      "a_p = 34.0; # Paring constant, MeV\n",
      "A = 40; # Mass number\n",
      "\n",
      "#Calculations\n",
      "B_semi = (a_v*A-(a_s*A**(2./3))-(a_c*Z*(Z-1)/A**(1./3))-(a_a*(A-2*Z)**2/A)-(a_p*A**(-3./4))); # Binding energy as per semiemperical mass formula\n",
      "# Percentage discrepancy between actual and semiemperical mass formula values are.......\n",
      "Per_des = -(B_semi+B_actual)/B_actual*100; # Percentage discrepancy \n",
      "\n",
      "#Result\n",
      "print \"Actual binding energy = %6.2f MeV\\nBinding energy as per semiemperical mass formula = %6.2f MeV\\nPercentage discrepancy = %.2f percent\"%(B_actual, B_semi, Per_des)\n",
      "\n",
      "#answers vary due to rounding-off errors"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Actual binding energy = -342.05 MeV\n",
        "Binding energy as per semiemperical mass formula = 343.59 MeV\n",
        "Percentage discrepancy = 0.45 percent\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 2.2.2, Page 67"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration \n",
      " # Calculation of coulomb energy for mirror nuclei : N-7 and O-8\n",
      " # For N-7 nucleus\n",
      "a_c = 0.7; # Coulomb energy constant, MeV\n",
      "Z_N = 7;  #  Atpmic no. \n",
      "A = 15; # Atomic mass\n",
      "E_C_N = a_c*Z_N*(Z_N-1)/(A**(1./3)); # Coulomb energy for N-7, MeV\n",
      "# For O-8 nucleus\n",
      "a_c = 0.7; # Coulomb energy constant, MeV\n",
      "Z_O = 8;  #  Atpmic no. \n",
      "A = 15.; # Atomic mass\n",
      "\n",
      "#Calculations\n",
      "E_C_O = a_c*Z_O*(Z_O-1)/(A**(1./3)); # Coulomb energy for O-8, MeV\n",
      "C_E_d = E_C_O-E_C_N; # Coulomb energy difference, MeV\n",
      "m_p = 1.007276*931.49; # Mass of proton, MeV\n",
      "m_n = 1.008665*931.49; # Mass of neutron, MeV\n",
      "M_d = m_n-m_p; # Mass difference of nucleons, MeV \n",
      "D_C_M = round(C_E_d-M_d); # Difference in coulomb energy and nucleon mass difference, MeV\n",
      "M_O = 15.003070*931.49; # Mass of O-8, MeV\n",
      "M_N = 15.000108*931.49; # Mass of N-7, MeV\n",
      "D_A = (M_O-M_N); # Actual mass difference, MeV\n",
      "\n",
      "#Result\n",
      "print \"Difference in Coulomb energy = %5.3f MeV\\nNucleon mass difference  = %6.4f MeV\\nDifference in Coulomb energy and nucleon mass difference = %5.3f MeV\\nActual mass difference = %.3f MeV\"%(C_E_d, M_d ,D_C_M, D_A)\n",
      "if D_A == D_C_M:\n",
      "    print \"Result is verified\"\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Difference in Coulomb energy = 3.974 MeV\n",
        "Nucleon mass difference  = 1.2938 MeV\n",
        "Difference in Coulomb energy and nucleon mass difference = 3.000 MeV\n",
        "Actual mass difference = 2.759 MeV\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 2.2.3, Page 68"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration&Calculations\n",
      "# For Kr-80,  \n",
      "m_p = 1.007825; # Mass of proton, amu\n",
      "m_n = 1.008665; # Mass of neutron, amu\n",
      "Z = 36; # Number of protons\n",
      "N_80 = 44;  # Number of neutrons\n",
      "M_n_80 = 79.91628; # Mass of Kr nucleus\n",
      "BE_Kr_80 = (Z*m_p+N_80*m_n-M_n_80)*931.49; #  Binding energy for Kr-80, MeV\n",
      "# For Kr-81,\n",
      "N_81 = 45; # Number of neutrons\n",
      "M_n_81 = 80.91661; # Mass of Kr-81 nucleus\n",
      "BE_Kr_81 = (Z*m_p+N_81*m_n-M_n_81)*931.49; # Binding energy for Kr-81 nucleus\n",
      "# For Kr-82\n",
      "N_82 = 46;  # Number of neutrons\n",
      "M_n_82 = 81.913482; # Mass of Kr nucleus\n",
      "BE_Kr_82 = (Z*m_p+N_82*m_n-M_n_82)*931.49;  # Binding energy for Kr-82,MeV\n",
      "# For Kr-83 \n",
      "N_83 = 47; # Number of protons\n",
      "M_n_83 = 82.914134; # Mass of Kr-83 nucleus\n",
      "BE_Kr_83 = (Z*m_p+N_83*m_n-M_n_83)*931.49; # Binding energy for Kr-83, MeV\n",
      "E_sep_81 = BE_Kr_81-BE_Kr_80; # Energy seperation of neutron for Kr-81, MeV\n",
      "E_sep_82 = BE_Kr_82-BE_Kr_81; # Energy seperation of neutron for Kr-82, MeV\n",
      "E_sep_83 = BE_Kr_83-BE_Kr_82; # Energy seperation of neutron for Kr-83, MeV\n",
      "\n",
      "#Result\n",
      "print \"Energy seperation of neutron for Kr-81 = %4.2f MeV\\nEnergy seperation of neutron for Kr-82 = %4.2f MeV\\nEnergy seperation of  neutron for Kr-83 = %5.2f MeV\"%(E_sep_81, E_sep_82, E_sep_83)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Energy seperation of neutron for Kr-81 = 7.76 MeV\n",
        "Energy seperation of neutron for Kr-82 = 10.99 MeV\n",
        "Energy seperation of  neutron for Kr-83 =  7.46 MeV\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 2.2.4, Page 68"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import sympy\n",
      "from sympy.solvers import solve\n",
      "from sympy import Symbol\n",
      "\n",
      "#Variable declaration\n",
      "a_v = 15.5; # Volume energy coefficient, MeV\n",
      "a_s = 16.8; # Surface energy coefficient MeV\n",
      "a_c = 0.7; # Coulomb energy coefficient, MeV\n",
      "a_a = 23.0; # Asymmetric energy coefficient, MeV\n",
      "a_p = 34.0; # Pairing energy coefficient, MeV\n",
      "A = 75; #  Given atomic mass \n",
      "\n",
      "#Calculations\n",
      "z = Symbol('z')\n",
      "B =solve((((-a_c*(2*z-1))/A**1./3)+((4*a_a*(A-2*z))/A)),z) # Binding energy as per liquid drop model\n",
      "\n",
      "#Result\n",
      "print \"Most stable isotope of A = 75 corresponds to Z = %.d\"%B[0]\n",
      "\n",
      "#answer varies due to usage of sympy module"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Most stable isotope of A = 75 corresponds to Z = 37\n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 2.2.5, Page 69"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import sympy\n",
      "from sympy.solvers import solve\n",
      "from sympy import Symbol\n",
      "\n",
      "#Variable declaration\n",
      "a_v = 15.5; # Volume energy, MeV \n",
      "a_s = 16.8; # Surface energy, MeV \n",
      "a_c = 0.7; # Coulomb energy, MeV\n",
      "a_a = 23.0; # Asymmetric energy, MeV\n",
      "a_p = 34.0; # Pairing energy, MeV\n",
      "\n",
      "z = Symbol('z')\n",
      "A = 27\n",
      "#Calculations\n",
      "# For A = 27;\n",
      "A = 27\n",
      "Z_27 = ((4*a_a)+(a_c/A**1./3))/(((2*a_c)/A**1./3)+(8*a_a/A))\n",
      "\n",
      "# For A = 118 \n",
      "A = 118\n",
      "Z_118 = ((4*a_a)+(a_c/A**1./3))/(((2*a_c)/A**1./3)+(8*a_a/A))\n",
      "\n",
      "# For A = 238\n",
      "A = 238\n",
      "Z_238 = ((4*a_a)+(a_c/A**1./3))/(((2*a_c)/A**1./3)+(8*a_a/A))\n",
      "\n",
      "#Result\n",
      "print \"Most stable isotopes for A = 27, A = 118, A = 238 corresponds to z = %d, %d and %d respectively\"%(Z_27, Z_118, Z_238)\n",
      "\n",
      "#Incorrect answers in the textbook"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Most stable isotopes for A = 27, A = 118, A = 238 corresponds to z = 13, 58 and 118 respectively\n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 2.2.6, Page 69"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "# Mirror nuclei : Na-11 and Mg-12 \n",
      "m_p = 1.007276; #  Mass of proton, amu\n",
      "m_n = 1.008665; # Mass of neutron, amu\n",
      "M_Mg  = 22.994124; #  Atomic mass of Mg-12, amu\n",
      "M_Na = 22.989768; # Atomic mass of Na-11, amu\n",
      "A = 23; # Mass number\n",
      "Z_Mg = 12; # Atomic number of Mg-12\n",
      "e = 1.6e-019; # Charge of the electron, C\n",
      "K = 8.98e+09; # Coulomb force constant\n",
      "\n",
      "#Calculations\n",
      "a_c = A**(1./3)/(2*Z_Mg-1)*((M_Mg-M_Na)+(m_n-m_p))*931.47; # Coulomb coefficient, MeV \n",
      "r_0 = 3./5*K*e**2/(a_c*1.6e-013); # Nuclear radius, m\n",
      "\n",
      "#Result\n",
      "print \"Coulomb coefficient = %4.2f MeV\\nNuclear radius = %3.1e m\"%(a_c, r_0)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Coulomb coefficient = 0.66 MeV\n",
        "Nuclear radius = 1.3e-15 m\n"
       ]
      }
     ],
     "prompt_number": 6
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 2.2.7, Page 71"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "Z = 92; # Atomic number of U-236\n",
      "e = 1.6e-019; # Charge of an electron, C\n",
      "A = 236; # Mass number of U-236\n",
      "K = 8.98e+09; # Coulomb constant,\n",
      "r_o = 1.2e-015; # Distance of closest approach, m\n",
      "a_s = -16.8; # Surface constant\n",
      "\n",
      "#Calculations\n",
      "E_c = -(3*K*Z*(Z-1)*e**2)/(5*r_o*A**(1./3)*1.6e-013); # Coulomb energy, MeV\n",
      "E_s = a_s*A**(2./3); # Surface energy, MeV   \n",
      "\n",
      "#Result\n",
      "print \"Coulomb energy for U(92,236)  = %5.1f MeV \\nSurface energy for U(92,236) = %5.1f MeV  \"%(E_c, E_s)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Coulomb energy for U(92,236)  = -973.3 MeV \n",
        "Surface energy for U(92,236) = -641.6 MeV  \n"
       ]
      }
     ],
     "prompt_number": 8
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 2.3.1, Page 126"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "#Variable declaration\n",
      "t_prime =  1600; # Half life of radioactive material, years\n",
      "t = 2000; # Total time, years\n",
      "lamda = 0.6931/t_prime; # Decay constant, years^(-1)\n",
      "m0 = 1; # The mass of radioactive substance at t0, mg\n",
      "\n",
      "#Calculations\n",
      "m = m0* math.exp(-(lamda*t)); # Ratio of total number of atoms and number of atoms disintegrat, mg\n",
      "a = 1-m; # The amount of radioactive substance decayed, mg \n",
      "\n",
      "#Result\n",
      "print \"The amount of radioactive substance decayed : %6.4f mg\"%a\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The amount of radioactive substance decayed : 0.5795 mg\n"
       ]
      }
     ],
     "prompt_number": 9
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 2.3.4, Page 126"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "#Variable declaration\n",
      "t_prime =  1600; # Half life of radioactive material, years\n",
      "t = 2000; # Total time, years\n",
      "lamda = 0.6931/t_prime; # Decay constant, years^(-1)\n",
      "m0 = 1; # The mass of radioactive substance at t0, mg\n",
      "\n",
      "#Calculations\n",
      "m = m0* math.exp(-(lamda*t)); # Ratio of total number of atoms and number of atoms disintegrat, mg\n",
      "a = 1-m; # The amount of radioactive substance decayed, mg \n",
      "\n",
      "#Result\n",
      "print \"The amount of radioactive substance decayed : %6.4f mg\"%a\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The amount of radioactive substance decayed : 0.5795 mg\n"
       ]
      }
     ],
     "prompt_number": 10
    }
   ],
   "metadata": {}
  }
 ]
}