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diff --git a/sample_notebooks/NirenNegandhi/ch2_2.ipynb b/sample_notebooks/NirenNegandhi/ch2_2.ipynb new file mode 100755 index 00000000..9f685b2b --- /dev/null +++ b/sample_notebooks/NirenNegandhi/ch2_2.ipynb @@ -0,0 +1,453 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:93b6ee47cf464df569e4895dab3158be49c4f81380cffe2b4a954430fff24e01" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "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": {} + } + ] +}
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