{ "metadata": { "name": "", "signature": "sha256:bb0d2851c786dd28ae3ad8cbd1e6a7fc5db7cc1384cc2031f2bb6380ed7515e2" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 14: Nuclear Physics Apllications" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 14.1, page no. 505" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "\n", "#Variable declaration\n", "\n", "MLi = 7.016003 #atomic mass of Lithium\n", "MH = 1.007825 #atomic mass of Hydrogen\n", "MHe = 4.002603 #atomic mass of Helium\n", "c2 = 931.50 #Square of speed of light (MeV/u)\n", "\n", "#Calculation\n", "\n", "Q = (MLi + MH - 2*MHe) * c2\n", "\n", "#Results\n", "\n", "print \"(a) The Q value is\",round(Q,1),\"MeV.\"\n", "\n", "\n", "\n", "#Variable declaration\n", "\n", "Kincident = 0.6 #kinetic energy of incident particle (MeV)\n", "\n", "#Calculation\n", "\n", "Kproducts = Q + Kincident\n", "\n", "#Results\n", "\n", "print \"(b) The kinetic energy of the products is\",round(Kproducts,1),\"MeV.\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(a) The Q value is 17.3 MeV.\n", "(b) The kinetic energy of the products is 17.9 MeV.\n" ] } ], "prompt_number": 6 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 14.2, page no. 509" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "\n", "#Variable declaration\n", "\n", "A = 27 #Atomic number of Aluminum\n", "d = 2.7 #density of aluminum\n", "Av = 6.02 * 10 ** 23 #Avogadro number nuclei/mol\n", "sigma = 2.0 * 10 **-31 #capture cross section (m^2)\n", "x = 0.3 * 10 ** -3 #thickness of the foil(m)\n", "R0 = 5.0 * 10 ** 12 #rate of incident particles(neutrons/cm^2.s)\n", "\n", "#Calculation\n", "\n", "n = Av * d / A * 10**6\n", "R = R0 * sigma * x * n\n", "\n", "#Result\n", "\n", "print \"The number of neutrons captured by 1.0cm^2/s is \",round(R/10**7,1),\"X 10^7 neutrons/cm^2.s\"\n", "\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The number of neutrons captured by 1.0cm^2/s is 1.8 X 10^7 neutrons/cm^2.s\n" ] } ], "prompt_number": 9 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 14.3, page no. 512" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "#Variable declaration\n", "\n", "A = 235 #atomic number of uranium\n", "m = 10**3 #mass of 235U\n", "Av = 6.02*10**23#avogadro number (nuclei/mol)\n", "Q = 208 #disintegration energy per event (MeV)\n", "\n", "#Calculation\n", "\n", "N = Av * m / A\n", "E = N * Q\n", "\n", "#Result\n", "\n", "print \"The disintegration energy is\",round(E/10**26,2),\"X 10^26 MeV.\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The disintegration energy is 5.33 X 10^26 MeV.\n" ] } ], "prompt_number": 12 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 14.4, page no. 513" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "\n", "#Variable declaration\n", "\n", "A = 235 #atomic mass of Uranium\n", "Av = 6.02 * 10**23 # Avagadro's number\n", "Q = 208 # disintegration energy per event (MeV)\n", "\n", "#Calculation\n", "\n", "N = Av / A *10**3 #No of nuclei in 1kg of 235U\n", "E = N*Q #disintegration energy (MeV)\n", "\n", "#result\n", "\n", "print \"The disintegration energy is\",round(E/10**26,2),\" X 10^26 MeV.\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The disintegration energy is 5.33 X 10^26 MeV.\n" ] } ], "prompt_number": 14 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 14.5, page no. 513" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "#Variable declaration\n", "\n", "rBa = 6.2 * 10 **-15 #nuclear radius of Ba(m)\n", "rKr = 5.4 * 10**-15 #nuclear radius of Kr(m)\n", "Z1 = 56 #atomic number of Ba\n", "Z2 = 36 #atomic number of Kr\n", "k = 1.440 * 10 **-9 #Coulomb constant (eV.nm)\n", "\n", "#Calculation\n", "\n", "r = rBa + rKr\n", "U = k * Z1 * Z2 /( round(r/10**-15) * 10**-15)\n", "\n", "#Result\n", "\n", "print \"The potential energy of the two nuclei is\",round(U/10**6),\"MeV.\" " ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The potential energy of the two nuclei is 242.0 MeV.\n" ] } ], "prompt_number": 17 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 14.6, page no. 519" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "\n", "#variable declaration\n", "\n", "e = 1.6 * 10 ** -19 #electron charge (C)\n", "r = 1.0 * 10**-14 #separation (m)\n", "k = 8.99 * 10**9 #Coulomb constant(N.m^2/C^2)\n", "\n", "#Calculation\n", "\n", "U = k * e**2/ r\n", "\n", "#result\n", "\n", "print \"(a)The height of the potential barrier is\",round(U/e/10**6,2),\"MeV.\"\n", "\n", "\n", "#Variable Declaration\n", "\n", "kB = 1.38 * 10**-23 #Boltzmann's constant (J/K)\n", "\n", "#Calculation\n", "\n", "T = (U/2.0)*(2.0/3.0)/(kB+.07*10**-23)\n", "\n", "#Result\n", "\n", "print\"(b)The effective temperature is\",round(T/10**8,1),\"X 10^8 K.\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(a)The height of the potential barrier is 0.14 MeV.\n", "(b)The effective temperature is 5.3 X 10^8 K.\n" ] } ], "prompt_number": 22 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 14.7, page no. 530" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "import math\n", "\n", "#variable declaration\n", "\n", "mu = 55 #mu for x-rays \n", "\n", "#Calculation\n", "\n", "x = math.log(2)/mu\n", "\n", "#Result\n", "\n", "print \"The half value thickness for lead is\",round(x/10**-2,2),\"X 10^-2 cm.\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The half value thickness for lead is 1.26 X 10^-2 cm.\n" ] } ], "prompt_number": 24 } ], "metadata": {} } ] }