{ "metadata": { "name": "", "signature": "sha256:43e40533bff9f31c3775eeac8a38c708406dc4765f3bc2853584ba2600307abe" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 1: The Nucleus" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 1.3.1, Page 31" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Variable declaration\n", "h = 6.626e-034; # Planck's constant, Js\n", "e = 1.602e-019; # Charge on an electron, C\n", "red_h = h/(2*math.pi*e*1e+06); # Reduced Planck's constant, MeV\n", "lamda = 5.0e-015; # de_Broglie wavelength of neutron, m\n", "\n", "#Calculations\n", "p = red_h/lamda; # Momentum of the neutron, MeV-s/m\n", "\n", "#Result\n", "print \"The momentum of the neutron from de-Broglie relation : %5.3e MeV-s/m\"%p" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The momentum of the neutron from de-Broglie relation : 1.317e-07 MeV-s/m\n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 1.3.2, Page 32" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable declaration\n", "E = [[0,0,0],[0,0,0],[0,0,0]]; # Declare a cell array of empty matrices for nuclides information\n", "E[0][0] = 'C'; # Assign element 'C' to (1,1) cell\n", "E[1][0] = 'N'; # Assign element 'N' to (2,1) cell\n", "E[2][0] = 'O'; # Assign element 'o' to (3,1) cell\n", "E[0][1] = 6; # Assign atomic No. 6 to (1,2) cell\n", "E[1][1] = 7; # Assign atomic No. 7 to (2,2) cell\n", "E[2][1] = 8; # Assign atomic No. 8 to (3,2) cell\n", "E[0][2] = [12,13,14,16]; # Assign mass numbers for 'C' to (1,3) cell\n", "E[1][2] = [14,15,16,17]; # Assign mass numbers for 'N' to (2,3) cell\n", "E[2][2] = [14,15,16,17]; # Assign mass numbers for 'O' to (3,3) cell\n", "\n", "#Calculations&Results\n", "# Isotopes\n", "print \"\\nIsotopes:\"\n", "print \"\\n=========\"\n", "for i in range(0,3): # Search for the three elements one-by-one\n", " print \"\\n(Z = %d)\"%(E[i][1])\n", " for j in range(0,4):\n", " print \"\\t%s(%d)\"%(E[i][0],E[i][2][j]),\n", " \n", "# Isotones\n", "print \"\\n\\nIsotones:\";\n", "print \"\\n========\"\n", "for N in range(6,10): # Search for the neutron numbers from 6 to 9\n", " print \"\\n(N = %d)\\n\"%N;\n", " for i in range(0,3):\n", " for j in range(0,4):\n", " if E[i][2][j]- E[i][1] == N: # N = A-Z\n", " print \"\\t%s(%d)\"%(E[i][0],E[i][2][j]),\n", " \n", "# Isobars\n", "print \"\\n\\nIsobars:\"\n", "print \"\\n=======\"\n", "for A in range(14,18): # Search for the mass numbers from 14 to 17\n", " print \"\\n(A = %d)\\n\"%A \n", " for i in range(1,3):\n", " for j in range(0,3):\n", " if E[i][2][j] == A:\n", " print \"\\t%s(%d)\"%(E[i-1][0],E[i][2][j]),\n", " \n", " " ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", "Isotopes:\n", "\n", "=========\n", "\n", "(Z = 6)\n", "\tC(12) \tC(13) \tC(14) \tC(16) \n", "(Z = 7)\n", "\tN(14) \tN(15) \tN(16) \tN(17) \n", "(Z = 8)\n", "\tO(14) \tO(15) \tO(16) \tO(17) \n", "\n", "Isotones:\n", "\n", "========\n", "\n", "(N = 6)\n", "\n", "\tC(12) \tO(14) \n", "(N = 7)\n", "\n", "\tC(13) \tN(14) \tO(15) \n", "(N = 8)\n", "\n", "\tC(14) \tN(15) \tO(16) \n", "(N = 9)\n", "\n", "\tN(16) \tO(17) \n", "\n", "Isobars:\n", "\n", "=======\n", "\n", "(A = 14)\n", "\n", "\tC(14) \tN(14) \n", "(A = 15)\n", "\n", "\tC(15) \tN(15) \n", "(A = 16)\n", "\n", "\tC(16) \tN(16) \n", "(A = 17)\n", "\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 1.4.1, Page 33" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "#Variable declaration\n", "m = 9.1e-031; # Mass of the electron, Kg\n", "C = 3e+08; # Velocity of the light,m/s\n", "\n", "#Calculations\n", "E = m*C**2/1.6e-013; # Energy of the electron at rest, MeV\n", "\n", "#Result\n", "print \"Energy of the electron at rest : %5.3f MeV\"%E" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Energy of the electron at rest : 0.512 MeV\n" ] } ], "prompt_number": 5 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 1.4.2, Page 33" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable declaration\n", "r = 3.46e-015; # Radius of the nucleus, m\n", "r0 = 1.2e-015; # Distance of closest approach of the nucleus, m\n", "\n", "#Calculations\n", "A = round((r/r0)**3); # Mass number of the nucleus\n", "if A == 23:\n", " element = \"Na\";\n", "elif A == 24:\n", " element = \"Mg\";\n", "elif A == 27:\n", " element = \"Al\";\n", "elif A == 28:\n", " element = \"Si\";\n", "\n", "#Result\n", "print \"The mass number of the nucleus is %d and the nucleus is of %s\"%(A, element)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The mass number of the nucleus is 24 and the nucleus is of Mg\n" ] } ], "prompt_number": 7 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 1.4.3, Page 34" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Variable declaration\n", "m = 40*(1.66e-027); # Mass of the nucleus, kg \n", "r0 = 1.2e-015; # Distance of the closest approach, m\n", "A = 40; # Atomic mass of the nucleus\n", "\n", "#Calculations\n", "r = r0*A**(1./3); #Radius of the nucleus, m\n", "V = 4./3*(math.pi*r**3); # Volume of the nucleus, m^3\n", "density = m/V; # Density of the nucleus, kg/m^3\n", "\n", "#Result\n", "print \"Radius of the nucleus: %3.1e m\\nVolume of the nucleus: %5.3e m^3\\nDensity of the nucleus: %3.1e kg/m^3\"%(r,V,density)\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Radius of the nucleus: 4.1e-15 m\n", "Volume of the nucleus: 2.895e-43 m^3\n", "Density of the nucleus: 2.3e+17 kg/m^3\n" ] } ], "prompt_number": 9 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 1.4.4, Page 34" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Variable declaration\n", "m = 1.66e-027; # Mass of a nucleon, kg\n", "A = 235; # Atomic mass of U-235 nucleus\n", "M = A*m; #Mass of the U-235 nucleus, kg\n", "r0 = 1.2e-015; # Distance of closest approach, m\n", "\n", "#Calculations\n", "r = r0*(A)**(1./3); # Radius of the U-235 nucleus\n", "V = 4./3*(math.pi*r**3); # Volume of the U-235 nucleus,m^3\n", "d = M/V; # Density of the U-235 nucleus,kg/m^3\n", "\n", "#Result\n", "print \"The density of U-235 nucleus : %4.2e kg per metre cube\"%d\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The density of U-235 nucleus : 2.29e+17 kg per metre cube\n" ] } ], "prompt_number": 12 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 1.4.5, Page 35" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Variable declaration\n", "m_O = 2.7e-026; # Mass of O nucleus, kg\n", "r_O = 3e-015; # Radius of O nucleus, m\n", "V_O = 4/3*(math.pi*(r_O)**3); # Volume of O nucleus, metre cube\n", "d_O = m_O/V_O; # Density of O nucleus, kg/metre cube\n", "m_Pb = 3.4e-025; # Mass of Pb nucleus, kg\n", "r_Pb = 7.0e-015; # Radius of Pb nucleus, m\n", "\n", "#Calculations\n", "V_Pb = 4./3*(math.pi*(r_Pb)**3); # Volume of Pb nucleus, metre cube\n", "d_Pb = m_Pb/V_Pb; #Density of Pb nucleus,kg/metre cube\n", "\n", "#Results\n", "print \"The density of oxygen nucleus : %4.2e in kg/metre cube\"%d_O\n", "print \"The density of Pb nucleus : %4.2e in kg/metre cube\"%d_Pb\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The density of oxygen nucleus : 3.18e+17 in kg/metre cube\n", "The density of Pb nucleus : 2.37e+17 in kg/metre cube\n" ] } ], "prompt_number": 13 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 1.4.6, Page 35" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Variable declaration\n", "E = 5.48*1.6e-013; # Energy of alpha particle, J\n", "e = 1.6e-019; # Charge of an electron, C\n", "Z = 79; # Mas number of Au nucleus, \n", "epsilon_0 = 8.85e-012; # Permittivity of free space, \n", "\n", "#Calculations\n", "D = (2*Z*e**2)/(4*math.pi*epsilon_0*E); # Distance of closest approach, m\n", "\n", "#Result\n", "print \"The distance of closest appproach of alpha particle : %4.2e m\"%D\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The distance of closest appproach of alpha particle : 4.15e-14 m\n" ] } ], "prompt_number": 14 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 1.4.7, Page 36" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable declaration\n", "A = 208; # Mass number of Pb-208\n", "r0 = 1.2e-015; # Distance of closest approach, m\n", "\n", "#Calculations\n", "r = r0*((A)**(1./3)); # Radius of Pb-208, m\n", "\n", "#Result\n", "print \"The radius of Pb-208 : %4.2e m\"%r\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The radius of Pb-208 : 7.11e-15 m\n" ] } ], "prompt_number": 15 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 1.5.1, Page 36" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable declaration\n", "amu = 931.49; # Atomic mass unit, MeV\n", "M_p = 1.00758; # Mass of proton, amu\n", "M_n = 1.00897; # Mass of neutron, amu\n", "M_He = 4.0028; # Mass of He nucleus, amu\n", "Z = 2; # Atomic number\n", "N = 2; # Number of neutron\n", "\n", "#Calculations\n", "M_defect = Z*M_p+N*M_n-M_He; # Mass defect, amu\n", "BE_MeV = M_defect*amu; # Binding energy, MeV\n", "BE_J = M_defect*1.49239e-010; # Binding energy, J\n", "\n", "#Results\n", "print \"The binding energy (in MeV): %5.2f\"%BE_MeV\n", "print \"The binding energy (in J): %4.2e\"%BE_J\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The binding energy (in MeV): 28.22\n", "The binding energy (in J): 4.52e-12\n" ] } ], "prompt_number": 17 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 1.5.2, Page 37" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable declaration\n", "amu = 1.49239e-010; # Atomic mass unit, J\n", "M_C = 12; # Mass of C-12, amu\n", "M_a = 4.0026; # Mass of alpha particle, amu\n", "\n", "#Calculations\n", "M_3a = 3*M_a; # Mass of 3 alpha particle, amu\n", "D = M_C-M_3a; # Difference in two masses, amu\n", "E = D*amu; # Required energy,J\n", "\n", "#Result\n", "print \"The energy required to break 3 alpha particles : %4.2e J\"%E\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The energy required to break 3 alpha particles : -1.16e-12 J\n" ] } ], "prompt_number": 19 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 1.5.3, Page 37" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable declaration\n", "M_p = 1.007895; # Mass of proton, amu\n", "M_n = 1.008665; # Mass of neutron, amu\n", "M_He = 4.0026; # Mass of He-nucleus, amu\n", "Z = 2; # Number of proton\n", "N = 2; # Number of neutron\n", "\n", "#Calculations\n", "D_m = ((Z*M_p)+(N*M_n)-M_He); # Mass defect, amu\n", "amu = 931.49; # Atomic mass unit, MeV\n", "E = D_m*amu; # Required energy, MeV\n", "\n", "#Result\n", "print \"The energy required to knock out nucleons from the He nucleus = %5.2f MeV\"%E\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The energy required to knock out nucleons from the He nucleus = 28.43 MeV\n" ] } ], "prompt_number": 21 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 1.5.4, Page 38" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable declaration\n", "M_Fe = 55.934939; # Mass of Fe-56, amu\n", "M_p = 1.007825; # Mass of proton, amu\n", "M_n = 1.008665; # Mass of neutron, amu\n", "Z = 26; # Atomic number of Fe-56\n", "N = 30; # Number of neutron in Fe-56\n", "amu = 931.49; # Atomic mass unit, MeV\n", "\n", "#Calculations\n", "BE = ((Z*M_p)+(N*M_n)-M_Fe)*amu; # Binding energy of Fe-56, MeV\n", "\n", "#Result\n", "print \"The binding energy of Fe-56 : %6.4f MeV\"%BE\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The binding energy of Fe-56 : 492.2561 MeV\n" ] } ], "prompt_number": 23 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 1.5.5, Page 38" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable declaration\n", "amu = 931.49; # Atomic mass unit, MeV\n", "M_p = 1.007825; # Mass of proton, amu\n", "M_n = 1.008663; # Mass of neutron, amu\n", "A = 2; # Mass number of deutron, amu\n", "M_D = 2.014103; # Mass of deuteron nucleus, amu\n", "\n", "#Calculations\n", "M_Defect = (M_p+M_n-M_D)*amu; # Mass defect of the nucleus, MeV\n", "P_fraction = (M_D - A)/A; # Packing fraction of nucleus\n", "\n", "#Results\n", "print \"Mass defect %4.2f MeV\\n Packing fraction %7.5f\"%(M_Defect,P_fraction);\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Mass defect 2.22 MeV\n", " Packing fraction 0.00705\n" ] } ], "prompt_number": 26 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 1.5.6, Page 38" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable declaration\n", "m_p = 1.007825; # Mass of proton, amu\n", "m_n = 1.008665; # Mass of neutron, amu\n", "m_He = 4.002634; # Mass of He-4 nucleus, amu\n", "amu = 931.47; # Atomic mass unit, MeV\n", "A = 4 # Mass number of He-4 nucleus\n", "\n", "#Calculations\n", "BE = (2*m_p+2*m_n-m_He)*amu; # Binding energy of He-4 nucleus, MeV\n", "Av_BE = BE/A; # Average binding energy or binding energy per nucleon, MeV\n", "\n", "#Result\n", "print \"The binding energy per nucleon : %4.2f MeV\"%Av_BE\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The binding energy per nucleon : 7.07 MeV\n" ] } ], "prompt_number": 28 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 1.6.1, Page 39" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable declaration\n", "l1 = 1; # Orbital qunatum number for p-state nucleon\n", "l2 = 2; # Orbital qunatum number for d-state nucleon\n", "\n", "#Calculations&Result\n", "# Display the value of L within the for loop\n", "print \"The possible L values will be\"\n", "for i in range(abs(l1-l2),abs(l1+l2+1)): # Coupling of l-orbitals \n", " print \"\\t %1d\"%i,\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The possible L values will be\n", "\t 1 \t 2 \t 3\n" ] } ], "prompt_number": 30 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 1.6.2, Page 40" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import numpy\n", "#Variable declaration\n", "# Get the l value from the user\n", "l = 3; # Orbital qunatum number for f-state proton\n", "s = 0.5; # Magnitude of spin quantum number\n", "\n", "#Calculations&Result\n", "# Display the value of j within the for loop\n", "\n", "print \"The j values will be between\"\n", "for i in numpy.arange((l-s),(l+s+1)): # l-s Coupling \n", " print \"\\t %3.1f\"%i,\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The j values will be between\n", "\t 2.5 \t 3.5\n" ] } ], "prompt_number": 32 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 1.11.1, Page 40" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable declaration\n", "V = 1000; # Potential difference, volts\n", "R = 0.122; # Radius of the circular path, m\n", "B = 1500e-04; # Magnetic field, tesla\n", "e = 1.602e-019; # Charge of the electron, C\n", "amu = 1.673e-027; # Atomic mass unit, kg\n", "\n", "#Calculations\n", "v = (2*V)/(R*B); # Speed of the ion, m/s\n", "M = 2*e*V/v**2; # Mass of the ion, kg\n", "A = M/amu; # Mass number\n", "\n", "#Result\n", "print \" Speed > %5.3e m/s \\n Mass > %5.3e kg \\n Mass number > %5.2f \"%(v, M, A)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " Speed > 1.093e+05 m/s \n", " Mass > 2.682e-26 kg \n", " Mass number > 16.03 \n" ] } ], "prompt_number": 34 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 1.11.2, Page 41" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "import numpy\n", "\n", "#Variable declaration\n", "amu = 1.673e-027; # Atomic mass unit, kg\n", "E = 5e+04; # Electric field, V/m\n", "B1 = 0.4; # Magnetic field, tesla\n", "v = E/B1; # Velocity of ions, m/s\n", "B = 0.8; # Magnetic field, tesla\n", "e = 1.602e-019; #charge of electron,C\n", "\n", "#Calculations\n", "m_Ar = [36,38,40] # Masses of three isoptopes of Ar, amu\n", "\n", "r_Ar = [0,0,0]; # Array of radii of three Ar ions, mm\n", "for i in range(len(r_Ar)):\n", " r_Ar[i] = (m_Ar[i]*amu*v)/(B*e)*1e+03; # Radius of Ar ion orbit, mm\n", "\n", "d1 = 2*(r_Ar[1]-r_Ar[0]); # Distance b/w first and second line, mm\n", "d2 = 2*(r_Ar[2]-r_Ar[1]); # Distance b/w second and third line, mm\n", "\n", "#Results\n", "print \"The distance between successive lines due to three different isotopes : %3.1f mm and %3.1f mm\"%(d1,d2)\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The distance between successive lines due to three different isotopes : 6.5 mm and 6.5 mm\n" ] } ], "prompt_number": 36 } ], "metadata": {} } ] }