{ "metadata": { "name": "", "signature": "sha256:0e9faaea32136a2f476b53b6ab2d5d2eb5330fb68ebd80aaad9bbe7ed1328c93" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "12: Radioactivity" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example number 12.1, Page number 351" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#importing modules\n", "import math\n", "from __future__ import division\n", "\n", "#Variable declaration\n", "M7Li3=7.018232; #mass of 7li3(amu)\n", "Malpha=4.003874; #mass of alpha particle(amu)\n", "Mpr=1.008145; #mass of proton(amu)\n", "Ey=9.15; #K.E energy of product nucleus\n", "\n", "#Calculation\n", "#xMy -> x-mass no., M-element, y-atomic no.\n", "#reaction:- 7li3 + 1H1-> 4He2 + 4He2\n", "deltaM=M7Li3+Mpr-2*Malpha; #mass defect(amu)\n", "Q=deltaM*931; #mass defect(MeV)\n", "Ex=2*Ey-Q; #K.E of incident particle(MeV)\n", "\n", "#Result\n", "print \"kinetic energy of incident proton is\",round(Ex,4),\"MeV\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "kinetic energy of incident proton is 0.9564 MeV\n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example number 12.2, Page number 351" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#importing modules\n", "import math\n", "from __future__ import division\n", "\n", "#Variable declaration\n", "M235U=235; #atomic mass of 235U\n", "m=10**-3; #mass of fissions(gm)\n", "N=6.023*10**23; #avagadro number\n", "Eperfi=200*10**6; #energy per fission(eV)\n", "T=10**-6; #time(s)\n", "\n", "#Calculation\n", "E=Eperfi*1.6*10**-19; #energy per fission(J)\n", "A=M235U; \n", "P=((m*N)/A)*(E/T); #power explosion(Watt)\n", "\n", "#Result\n", "print \"power of explosion is\",P,\"Watt\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "power of explosion is 8.20153191489e+13 Watt\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example number 12.4, Page number 352" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#importing modules\n", "import math\n", "from __future__ import division\n", "\n", "#Variable declaration\n", "n=0.4; #efficiency\n", "N=6.06*10**26; #avagadro number\n", "Eperfi=200*10**6; #energy per fission(eV)\n", "P=100*10**6; #electric power(W)\n", "A=235;\n", "\n", "#Calculation\n", "E=Eperfi*1.6*10**-19; #energy per fission(J)\n", "T=24*60*60; #time(sec)\n", "N235=P*T/(E*n); #number of atoms in 235 kg of U235\n", "m=(A*N235)/N; #mass of 235U consumed/day(kg)\n", "\n", "#Result\n", "print \"mass of 235U consumed/day is\",int(m*10**3),\"g\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "mass of 235U consumed/day is 261 g\n" ] } ], "prompt_number": 13 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example number 12.5, Page number 352" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#importing modules\n", "import math\n", "from __future__ import division\n", "\n", "#Variable declaration\n", "M2H1=2.01474; #mass of M2H1(amu)\n", "M3H1=3.01700; #mass of M3H1(amu)\n", "M1n0=1.008986; #mass of M1n0(amu)\n", "M4He2=4.003880; #mass of M4He2(amu)\n", "\n", "#Calculation\n", "#thermonuclear reaction in hydrogen bomb explosion \n", "#2H1 + 3H1 -> 4He2 + 1n0\n", "Mreac=M2H1+M3H1; #mass of reactants(amu)\n", "Mprod=M4He2+M1n0; #mass of products(amu)\n", "Q=Mreac-Mprod; #amount of energy released per reaction(J)\n", "Q=Q*931; #amount of energy released per reaction(MeV)\n", "\n", "#Result\n", "print \"amount of energy released per reaction is\",round(Q,3),\"MeV\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "amount of energy released per reaction is 17.572 MeV\n" ] } ], "prompt_number": 15 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example number 12.6, Page number 353" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#importing modules\n", "import math\n", "from __future__ import division\n", "\n", "#Variable declaration\n", "M7Li3=7.01818; #mass of Li atom(amu)\n", "M1H1=1.0081; #mass of H atom(amu)\n", "M1n0=1.009; #mass of neutron(amu)\n", "\n", "#Calculation\n", "BEpernu=(1/7)*((3*M1H1)+(4*M1n0)-M7Li3); #binding energy per nucleon(J)\n", "BEpernu=BEpernu*931; #binding energy per nucleon(MeV)\n", "\n", "#Result\n", "print \"binding energy per nucleon is\",BEpernu,\"MeV\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "binding energy per nucleon is 5.60196 MeV\n" ] } ], "prompt_number": 16 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example number 12.7, Page number 353" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#importing modules\n", "import math\n", "from __future__ import division\n", "\n", "#Variable declaration\n", "m=10*10**3; #mass of U235(gm)\n", "N=6.02*10**23; #avagadro number\n", "Eperfi=200*10**6; #energy per fission(eV)\n", "A=235;\n", "\n", "#Calculation\n", "E=Eperfi*1.6*10**-19; #energy(J)\n", "T=24*60*60; #time(s)\n", "P=((m*N)/A)*(E/T); #power output(Watt)\n", "\n", "#Result\n", "print \"power output is\",round(P/10**9,3),\"*10**9 Watt\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "power output is 9.488 *10**9 Watt\n" ] } ], "prompt_number": 19 } ], "metadata": {} } ] }