{ "metadata": { "name": "", "signature": "sha256:9326f276d5dc99ce97d41c9ca0d5924dbd68f522091536657f41d8cfe038dc31" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "19: Nuclear reactions" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example number 19.1, Page number 368" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#importing modules\n", "import math\n", "from __future__ import division\n", "\n", "#Variable declaration\n", "m2H=2.014102; #atomic mass of 2H(u)\n", "mn=1.008665; #mass of n(u)\n", "m63Cu=62.929599; #mass of 63Cu(u)\n", "m64Zn=63.929144; #mass of m64Zn(u)\n", "E=931.5; #energy(MeV)\n", "Kx=10; #energy of deutron(MeV)\n", "Ky=15; #energy of neutron(MeV)\n", "\n", "#Calculation\n", "Q=E*(m2H+m63Cu-mn-m64Zn); #Q-value(MeV)\n", "KY=Q+Kx-Ky; #kinetic energy(MeV)\n", "\n", "#Result\n", "print \"Q-value is\",round(Q,3),\"MeV\"\n", "print \"kinetic energy is\",round(KY,3),\"MeV\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Q-value is 5.488 MeV\n", "kinetic energy is 0.488 MeV\n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example number 19.2, Page number 368" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#importing modules\n", "import math\n", "from __future__ import division\n", "\n", "#Variable declaration\n", "m19F=18.998404; #atomic mass of 19F(u)\n", "mH=1.007825; #mass of H(u)\n", "m19O=19.003577; #mass of 19O(u)\n", "mn=1.008665; #mass of n(u)\n", "E=931.5; #energy(MeV)\n", "\n", "#Calculation\n", "Q=E*(m19F+mn-mH-m19O); #Q-value(MeV)\n", "Kxmin=-Q*(1+(mn/m19F)); #threshold energy(MeV)\n", "\n", "#Result\n", "print \"Q-value is\",round(Q,4),\"MeV\"\n", "print \"threshold energy is\",round(Kxmin,2),\"MeV\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Q-value is -4.0362 MeV\n", "threshold energy is 4.25 MeV\n" ] } ], "prompt_number": 5 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example number 19.3, Page number 373" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#importing modules\n", "import math\n", "from __future__ import division\n", "\n", "#Variable declaration\n", "mn=1.008665; #mass of n(u)\n", "mu=235.043924; #mass of 235U(u)\n", "mBa=140.91440; #mass of 141Ba(u)\n", "mKr=91.92630; #mass of Kr(u)\n", "E=931.5; #energy(MeV)\n", "\n", "#Calculation\n", "mr=mn+mu; #mass of reactants(u)\n", "mp=mBa+mKr+(3*mn); #mass of products(u)\n", "md=mr-mp; #mass difference(u)\n", "E=md*E; #energy released(MeV)\n", "\n", "#Result\n", "print \"energy released is\",round(E,1),\"MeV\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "energy released is 173.2 MeV\n" ] } ], "prompt_number": 19 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example number 19.4, Page number 373" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#importing modules\n", "import math\n", "from __future__ import division\n", "\n", "#Variable declaration\n", "E=200*10**6; #energy released(eV)\n", "e=1.6*10**-19; #conversion factor from J to eV\n", "P=300*10**6; #power(W)\n", "t=1; #time(s)\n", "\n", "#Calculation\n", "n=P*t/(E*e); #number of fissions per second\n", "\n", "#Result\n", "print \"number of fissions per second is\",n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "number of fissions per second is 9.375e+18\n" ] } ], "prompt_number": 20 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example number 19.5, Page number 378" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#importing modules\n", "import math\n", "from __future__ import division\n", "\n", "#Variable declaration\n", "m2H1=2*1.66*10**-27; #mass of proton(kg)\n", "E=931.5; #energy(MeV)\n", "m1=2.014102;\n", "m2=3.01609;\n", "mH=1.007825; #mass of H(u)\n", "\n", "#Calculation\n", "E=E*((2*m1)-m2-mH); #energy released(MeV)\n", "n=0.001/m2H1; #number of nuclei\n", "Eg=n*E/2; #energy released per gm(MeV)\n", "\n", "#Result\n", "print \"energy released per gm is\",round(Eg/10**23,2),\"*10**23 MeV\"\n", "print \"answer given in the book is wrong\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "energy released per gm is 6.02 *10**23 MeV\n", "answer given in the book is wrong\n" ] } ], "prompt_number": 31 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example number 19.6, Page number 379" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#importing modules\n", "import math\n", "from __future__ import division\n", "\n", "#Variable declaration\n", "k=8.99*10**9; #value of k(Nm**2/C**2)\n", "rd=1.5*10**-15; #radius of deuterium nucleus(m)\n", "rt=1.7*10**-15; #radius of tritium nucleus(m)\n", "e=1.6*10**-19; #conversion factor from J to eV\n", "KE=0.225; #kinetic energy for 1 particle(MeV)\n", "k=1.38*10**-23; #boltzmann constant(J/K)\n", "\n", "#Calculation\n", "K_E=k*e**2/(e*(rd+rt)); #kinetic energy of 2 particles(MeV)\n", "T=2*KE*e*10**6/(3*k); #temperature(K)\n", "\n", "#Result\n", "print \"temperature is\",round(T/10**9),\"*10**9 K\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "temperature is 2.0 *10**9 K\n" ] } ], "prompt_number": 35 } ], "metadata": {} } ] }