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-{

- "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": {}

-  }

- ]

-}
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