{ "metadata": { "name": "", "signature": "sha256:6270762d6231ec0577735987cb74ce2b57535e026b96bfac7596018acfb359e4" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 12: The Atomic Nuleus" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 12.1, Page 432" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Variable declaration\n", "h = 6.62e-034; # Planck's constant, Js\n", "h_bar = h/(2*math.pi); # Reduced Planck's constant, Js\n", "m = 1.67e-027; # Rest mass of proton, kg\n", "e = 1.602e-019; # Energy equivalent of 1 eV, J\n", "c = 3.00e+008; # Speed of light, m/s\n", "delta_x = 8.0e-015; # Size of the nucleus, m\n", "\n", "#Calculations\n", "delta_p = h_bar/(2*delta_x*e); # Uncertainty in momentum of proton from Heisenberg Uncertainty Principle, eV-s/m\n", "p_min = delta_p; # Minimum momentum of the proton inside the nucleus, eV-s/m\n", "K = (p_min*c)**2*e/(2*m*c**2*1e+006); # The minimum kinetic energy of the proton in a medium sized nucleus, MeV\n", "\n", "#Result\n", "print \"The minimum kinetic energy of the proton in a medium sized nucleus = %4.2f MeV\"%K" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The minimum kinetic energy of the proton in a medium sized nucleus = 0.08 MeV\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 12.2, Page 436" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Variable declaration\n", "c = 3.00e+008; # Speed of light, m/s\n", "e = 1.602e-019; # Energy equivalent of 1 eV, J\n", "m_e = 0.511; # Rest mass energy of electron, MeV\n", "m_p = 938.3; # Rest mass energy of proton, MeV\n", "h = 6.62e-034; # Planck's constant, Js\n", "A = 40; # Mass number of Ca-40\n", "r0 = 1.2; # Nuclear radius constant, fm\n", "\n", "#Calculations&Results\n", "R = r0*A**(1./3); # Radius of Ca-40 nucleus, fm\n", "print \"The radius of Ca-40 nucleus = %3.1f fm\"%R\n", "lamda = 2.0; # de-Broglie wavelength to distinguish a distance at least half the radius, fm\n", "# Electron energy\n", "E = math.ceil(math.sqrt(m_e**2+(h*c/(lamda*e*1e+006*1e-015))**2)); # Total energy of the probing electron, MeV\n", "K = E - m_e; # Kinetic energy of probing electron, MeV\n", "print \"The kinetic energy of probing electron = %3d MeV\"%math.ceil(K)\n", "# Proton energy\n", "E = math.ceil(math.sqrt(m_p**2+(h*c/(lamda*e*1e+006*1e-015))**2)); # Total energy of the probing electron, MeV\n", "K = E - m_p; # Kinetic energy of probing electron, MeV\n", "print \"The kinetic energy of probing proton = %3d MeV\"%math.ceil(K)\n", "\n", "#answers differ due to rounding errors" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The radius of Ca-40 nucleus = 4.1 fm\n", "The kinetic energy of probing electron = 620 MeV\n", "The kinetic energy of probing proton = 187 MeV\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 12.3, Page 437" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable declaration\n", "A_U238 = 238; # Mass number of U-238\n", "A_He4 = 4; # Mass number of He-4\n", "r0 = 1.2; # Nuclear radius constant, nm\n", "\n", "#Calculations&Results\n", "R_U238 = r0*A_U238**(1./3); # Radius of U-238 nucleus, fm\n", "R_He4 = r0*A_He4**(1./3); # Radius of He-4 nucleus, fm\n", "print \"The radii of U-238 and He-4 nuclei are %3.1f fm and %3.1f fm repectively\"%(R_U238, R_He4)\n", "ratio = R_U238/R_He4; # Ratio of the two radii\n", "print \"The ratio of radius to U-238 to that of He-4 = %3.1f\"%ratio" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The radii of U-238 and He-4 nuclei are 7.4 fm and 1.9 fm repectively\n", "The ratio of radius to U-238 to that of He-4 = 3.9\n" ] } ], "prompt_number": 4 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 12.4, Page 438" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable declaration\n", "h = 6.62e-034; # Planck's constant, Js\n", "c = 3.00e+008; # Speed of light, m/s\n", "e = 1.602e-019; # Energy equivalent of 1 eV, J\n", "B = 2.0; # Applied magnetic field, T\n", "mu_N = 3.15e-008; # Nucleon magnetic moment, eV/T\n", "\n", "#Calculations\n", "mu_p = 2.79*mu_N; # Proton magnetic moment, eV/T\n", "delta_E = 2*mu_p*B; # Energy difference between the up and down proton states, eV\n", "f = delta_E*e/h; # Frequency of electromagnetic radiation that flips the proton spins, Hz\n", "lamda = c/f; # Wavelength of electromagnetic radiation that flips the proton spins, m\n", "\n", "#Results\n", "print \"The energy difference between the up and down proton states = %3.1e eV\"%delta_E\n", "print \"The frequency of electromagnetic radiation that flips the proton spins = %2d MHz\"%(f/1e+006)\n", "print \"The wavelength of electromagnetic radiation that flips the proton spins = %3.1f m\"%lamda" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The energy difference between the up and down proton states = 3.5e-07 eV\n", "The frequency of electromagnetic radiation that flips the proton spins = 85 MHz\n", "The wavelength of electromagnetic radiation that flips the proton spins = 3.5 m\n" ] } ], "prompt_number": 5 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 12.5, Page 443" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable declaration\n", "c = 3.00e+008; # Speed of light, m/s\n", "e = 1.602e-019; # Energy equivalent of 1 eV, J\n", "u = 931.5; # Energy equivalent of 1 amu, MeV\n", "m_n = 1.008665; # Mass of a neutron, u\n", "M_H = 1.007825; # Mass of a proton, u\n", "M_He = 4.002603; # Mass of helium nucleus\n", "M_Be = 8.005305; # Mass of Be-8, u\n", "\n", "#Calculations&Results\n", "B_Be = (4*m_n+4*M_H - M_Be)*u;\n", "B_Be_2alpha = (2*M_He - M_Be)*u;\n", "print \"The binding energy of Be-8 = %4.1f MeV and is positive\"%B_Be\n", "if (B_Be_2alpha > 0):\n", " print \"The Be-4 is stable w.r.t. decay into two alpha particles.\";\n", "else:\n", " print \"The Be-4 is unstable w.r.t. decay into two alpha particles since the decay has binding energy of %5.3f MeV\"%B_Be_2alpha \n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The binding energy of Be-8 = 56.5 MeV and is positive\n", "The Be-4 is unstable w.r.t. decay into two alpha particles since the decay has binding energy of -0.092 MeV\n" ] } ], "prompt_number": 6 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 12.6, Page 444" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable declaration\n", "Z = 92; # Atomic number of U-238\n", "A = 238; # Mass number of U-238\n", "\n", "#Calculations\n", "E_Coul = 0.72*Z*(Z-1)*A**(-1./3); # Total Coulomb energy of U-238, MeV\n", "\n", "#Result\n", "print \"The total Coulomb energy of U-238 = %3d MeV\"%E_Coul" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The total Coulomb energy of U-238 = 972 MeV\n" ] } ], "prompt_number": 7 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 12.8, Page 447" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable declaration\n", "u = 931.5; # Energy equivalent of 1 amu, MeV\n", "m_p = 1.007825; # Mass of proton, amu\n", "m_n = 1.008665; # Mass of neutron, amu\n", "M_Ne = 19.992440; # Mass of Ne-20 nucleus, amu\n", "M_Fe = 55.934942; # Mass of Fe-56 nucleus, amu\n", "M_U = 238.050783; # Mass of U-238 nucleus, amu\n", "A_Ne = 20; # Mass number of Ne-20 nucleus\n", "A_Fe = 56; # Mass number of Fe-56 nucleus\n", "A_U = 238; # Mass number of U-238 nucleus\n", "\n", "#Calculations\n", "BE_Ne = (10*m_p+10*m_n-M_Ne)*u; # Binding energy of Ne-20 nucleus, MeV\n", "BE_Fe = (26*m_p+30*m_n-M_Fe)*u; # Binding energy of Fe-56 nucleus, MeV\n", "BE_U = (92*m_p+146*m_n-M_U)*u; # Binding energy of U-238 nucleus, MeV\n", "\n", "#Results\n", "print \"The binding energy per nucleon for Ne-20 : %4.2f MeV/nucleon\"%(BE_Ne/A_Ne)\n", "print \"The binding energy per nucleon for Fe-56 : %4.2f MeV/nucleon\"%(BE_Fe/A_Fe)\n", "print \"The binding energy per nucleon for U-238 : %4.2f MeV/nucleon\"%(BE_U/A_U)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The binding energy per nucleon for Ne-20 : 8.03 MeV/nucleon\n", "The binding energy per nucleon for Fe-56 : 8.79 MeV/nucleon\n", "The binding energy per nucleon for U-238 : 7.57 MeV/nucleon\n" ] } ], "prompt_number": 8 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 12.10, Page 451" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Variable declaration\n", "N_A = 6.023e+023; # Avogadro's number\n", "T = 138*24*3600; # Half life of Po-210, s\n", "R = 2000; # Activity of Po-210, disintegrations/s\n", "M = 0.210; # Gram molecular mass of Po-210, kg\n", "\n", "#Calculations\n", "f = 1/3.7e+010; # Conversion factor to convert from decays/s to Ci\n", "A = R*f/1e-006; # Activity of Po-210, micro-Ci\n", "N = R*T/math.log(2); # Number of radioactive nuclei of Po-210, nuclei\n", "m = N*M/N_A; # Mass of Po-210 sample, kg\n", "\n", "#Results\n", "print \"The activity of Po-210 = %5.3f micro-Ci\"%A\n", "print \"The mass of Po-210 sample = %3.1e kg\"%m" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The activity of Po-210 = 0.054 micro-Ci\n", "The mass of Po-210 sample = 1.2e-14 kg\n" ] } ], "prompt_number": 9 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 12.11, Page 452" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Variable declaration\n", "T = 110; # Half life of F-18, min\n", "f_remain = 0.01; # Fraction of the F-18 sample remained\n", "\n", "#Calculations\n", "t = -math.log(0.01)/(math.log(2)*60)*T; # Time taken by the F-18 sample to decay to 1 percent of its initial value, h\n", "\n", "#Result\n", "print \"The time taken for 99 percent of the F-18 sample to decay = %4.1f h\"%t" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The time taken for 99 percent of the F-18 sample to decay = 12.2 h\n" ] } ], "prompt_number": 10 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 12.12, Page 452" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Variable declaration\n", "N_A = 6.023e+023; # Avogadro's number\n", "M = 10e+03; # Mass of the U-235, g\n", "M_U235 = 235; # Molecular mass of U-235, g\n", "t_half = 7.04e+008; # Half life of U-235, y\n", "\n", "#Calculations\n", "N = M*N_A/M_U235; # Number of U-235 atoms in 10 kg sample\n", "R = math.log(2)*N/t_half; # The alpha activity of 10 kg sample of U-235, decays/y\n", "\n", "#Result\n", "print \"The alpha activity of 10 kg sample of U-235 = %3.1e Bq\"%(R/(365.25*24*60*60))" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The alpha activity of 10 kg sample of U-235 = 8.0e+08 Bq\n" ] } ], "prompt_number": 11 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 12.13, Page 453" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable declaration\n", "u = 931.5; # Energy equivalent of 1 u, MeV\n", "M_U230 = 230.033927; # Atomic mass of U-230, u\n", "m_n = 1.008665; # Mass of a neutron, u\n", "M_H = 1.007825; # Mass of hydrogen, u\n", "M_U229 = 229.033496; # Gram atomic mass of U-230\n", "\n", "#Calculations&Results\n", "Q = (M_U230 - M_U229 - m_n)*u; # Q-value of the reaction emitting a neutron\n", "if (Q < 0):\n", " print \"The neutron decay in this reaction is not possible.\"\n", "else:\n", " print \"The neutron decay in this reaction is possible.\"\n", "\n", "Q = (M_U230 - M_U229 - M_H)*u; # Q-value of the reaction emitting a proton\n", "if (Q < 0):\n", " print \"The proton decay in this reaction is not possible.\"\n", "else:\n", " print \"The proton decay in this reaction is possible.\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The neutron decay in this reaction is not possible.\n", "The proton decay in this reaction is not possible.\n" ] } ], "prompt_number": 14 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 12.15, Page 461" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable declaration\n", "u = 931.5; # Energy equivalent of 1 u, MeV\n", "M_Fe55 = 54.938298; # Atomic mass of Fe-55, u\n", "M_Mn55 = 54.938050; # Atomic mass of Mn-55, u\n", "m_e = 0.000549; # Mass of the electron, u\n", "\n", "#Calculations&Results\n", "Q = (M_Fe55 - M_Mn55 - 2*m_e)*u; # Q-value of the reaction undergoing beta+ decay, MeV\n", "if (Q < 0):\n", " print \"The beta+ decay is not allowed for Fe-55\"\n", "else:\n", " print \"The beta+ decay is allowed for Fe-55\"\n", "\n", "Q = (M_Fe55 - M_Mn55)*u; # Q-value of the reaction undergoing electron capture, MeV\n", "if (Q < 0):\n", " print \"Fe-55 may not undergo electron capture\"\n", "else:\n", " print \"Fe-55 may undergo electron capture\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The beta+ decay is not allowed for Fe-55\n", "Fe-55 may undergo electron capture\n" ] } ], "prompt_number": 13 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 12.16, Page 462" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable declaration\n", "def check_allowance(Q, decay_type):\n", " if (Q < 0):\n", "\tprint \"The %s is not allowed for Ac-226\"%decay_type\n", " else:\n", "\tprint \"The %s is allowed for Ac-226\"%decay_type\n", "\n", "\n", "u = 931.5; # Energy equivalent of 1 u, MeV\n", "M_Ac226 = 226.026090; # Atomic mass of Ac-226, u\n", "M_Fr222 = 222.017544; # Atomic mass of Fr-222, u\n", "M_He4 = 4.002603; # Atomic mass of He-4, u\n", "M_Th226 = 226.024891; # Atomic mass of M_Th226, u\n", "M_Ra226 = 226.025403; # Atomic mass of M_Ra226, u\n", "m_e = 0.000549; # Mass of the electron, u\n", "\n", "#Calculations\n", "# Alpha Decay\n", "Q = (M_Ac226 - M_Fr222 - M_He4)*u; # Q-value of the reaction undergoing alpha decay, MeV\n", "check_allowance(Q, \"alpha decay\");\n", "# Beta- Decay\n", "Q = (M_Ac226 - M_Th226)*u; # Q-value of the reaction undergoing beta- decay, MeV\n", "check_allowance(Q, \"beta- decay\");\n", "# Beta+ Decay\n", "Q = (M_Ac226 - M_Ra226 - 2*m_e)*u; # Q-value of the reaction undergoing beta+ decay, MeV\n", "check_allowance(Q, \"beta+ decay\");\n", "# Electron Capture\n", "Q = (M_Ac226 - M_Ra226)*u; # Q-value of the reaction undergoing electron capture, MeV\n", "check_allowance(Q, \"electron capture\");\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The alpha decay is allowed for Ac-226\n", "The beta- decay is allowed for Ac-226\n", "The beta+ decay is not allowed for Ac-226\n", "The electron capture is allowed for Ac-226\n" ] } ], "prompt_number": 15 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 12.17, Page 463" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable declaration\n", "u = 931.5; # Energy equivalent of 1 u, MeV\n", "E_ex = 0.072; # Energy of the excited state, MeV\n", "\n", "#Calculations\n", "M = 226*u*1e+0; # Energy equivalent of atomic mass of Th-226, MeV\n", "x = E_ex/(2*M); # The error introduced in the gamma ray energy by approximation\n", "\n", "#Result\n", "print \"The error introduced in the gamma ray energy by approximation = %3.1e\"%x" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The error introduced in the gamma ray energy by approximation = 1.7e-07\n" ] } ], "prompt_number": 16 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 12.18, Page 467" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Variable declaration\n", "t_half = 4.47e+009; # The half life of uranium ore, years\n", "R_prime = 0.60; # The ratio of Pb206 abundance to that of U238\n", "\n", "#Calculations\n", "t = t_half/math.log(2)*math.log(R_prime + 1); # Age of the uranuim ore, years\n", "\n", "#Result\n", "print \"The age of U-238 ore = %3.1e years\"%t" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The age of U-238 ore = 3.0e+09 years\n" ] } ], "prompt_number": 17 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 12.19, Page 469" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Variable declaration\n", "t_half = 5730; # The half life of uranium ore, years\n", "R0 = 1.2e-012; # The initial ratio of C14 to C12 at the time of death\n", "R = 1.10e-012; # The final ratio of C14 to C12 t years after death\n", "\n", "#Calculations\n", "# As R = R0*math.exp(-lambda*t), solving for t\n", "t = -math.log(R/R0)*t_half/math.log(2); # Age of the bone, years\n", "\n", "#Result\n", "print \"The %3d years age of bone does not date from the Roman Empire.\"%math.ceil(t)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The 720 years age of bone does not date from the Roman Empire.\n" ] } ], "prompt_number": 18 } ], "metadata": {} } ] }