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diff --git a/Modern_Physics/chapter13_1.ipynb b/Modern_Physics/chapter13_1.ipynb
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+{

+ "metadata": {

+  "name": "",

+  "signature": "sha256:bee0f8df3068997ca43cf44df4f129c530dc6e34523ca501c99a2183643ee772"

+ },

+ "nbformat": 3,

+ "nbformat_minor": 0,

+ "worksheets": [

+  {

+   "cells": [

+    {

+     "cell_type": "heading",

+     "level": 1,

+     "metadata": {},

+     "source": [

+      "13: Nuclear Reactions"

+     ]

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example number 13.1, Page number 259"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#import modules\n",

+      "import math\n",

+      "from __future__ import division\n",

+      "\n",

+      "#Variable declaration\n",

+      "m1=7.0183;      #mass of 3Li7(amu)\n",

+      "m2=4.0040;      #mass of 2He4(amu)\n",

+      "m3=1.0082;      #mass of 1H1(amu)\n",

+      "N=6.026*10**26;      #Avgraodo no.(per kg atom)\n",

+      "#rxn = 3Li7 + 1H1 = 2He4 + 2He4 \n",

+      "\n",

+      "#Calculation\n",

+      "delta_m=m1+m3-(2*m2);      #deltam(amu)\n",

+      "E=delta_m*931;          #energy per disintegration(MeV)\n",

+      "n=0.1*N/7;      #no of atoms in 100 gm of lithium\n",

+      "TE=n*E;         #Total energy available(MeV)  \n",

+      "\n",

+      "#Result\n",

+      "print \"energy available per disintegration is\",round(E,2),\"MeV\"\n",

+      "print \"Total energy available is\",round(TE/1e+25,2),\"*10**25 MeV\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "energy available per disintegration is 17.22 MeV\n",

+        "Total energy available is 14.83 *10**25 MeV\n"

+       ]

+      }

+     ],

+     "prompt_number": 5

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example number 13.2, Page number 259"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#import modules\n",

+      "import math\n",

+      "from __future__ import division\n",

+      "\n",

+      "#Variable declaration\n",

+      "m1=6.015126;    #mass of 3Li7(a.m.u)\n",

+      "m2=4.002604;    #mass oh 2He4(a.m.u)\n",

+      "m3=1.00865;     #mass of 0n1(a.m.u)\n",

+      "m4=3.016049;    #mass of 1H3(a.m.u)\n",

+      "#rxn = 3Li7 + 0n1 = 2He4 + 1H3 + Q\n",

+      "\n",

+      "#Calculation\n",

+      "dm=m1+m3-(m2+m4);\n",

+      "Q=dm*931;         #energy released(MeV)\n",

+      "\n",

+      "#Result\n",

+      "print \"energy released is\",round(Q,4),\"MeV\"\n",

+      "print \"answer given in the book varies due to rounding off errors\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "energy released is 4.7695 MeV\n",

+        "answer given in the book varies due to rounding off errors\n"

+       ]

+      }

+     ],

+     "prompt_number": 8

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example number 13.3, Page number 260"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#import modules\n",

+      "import math\n",

+      "from __future__ import division\n",

+      "\n",

+      "#Variable declaration\n",

+      "m1=14.007515;     #mass of 7N14(a.m.u)\n",

+      "m2=4.003837;      #mass of 2He4(a.m.u)\n",

+      "m3=17.004533;     #mass of 8O17(a.m.u)\n",

+      "m4=1.008142;      #mass of 1H1(a.m.u)\n",

+      "#rxn = 7N14 + 2He14 = 8O17 + 1H1\n",

+      "\n",

+      "#Calculation\n",

+      "dm=m3+m4-(m1+m2);\n",

+      "Q=dm*931;           #Q value of the reaction(MeV)\n",

+      "\n",

+      "#Result\n",

+      "print \"Q value of the reaction is\",round(Q,3),\"MeV\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "Q value of the reaction is 1.232 MeV\n"

+       ]

+      }

+     ],

+     "prompt_number": 11

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example number 13.4, Page number 260"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#import modules\n",

+      "import math\n",

+      "from __future__ import division\n",

+      "\n",

+      "#Variable declaration\n",

+      "m1=14.007520;     #mass of 7N14(a.m.u)\n",

+      "m2=1.008986;      #mass oh 0n1(a.m.u)\n",

+      "#m3=mass of 6C14 in a.m.u\n",

+      "m4=1.008145;      #mass of 1H1(a.m.u)\n",

+      "#rxn = 7N14 + 0n1 = 6C14 + 1H1 + 0.55 MeV\n",

+      "\n",

+      "#Calculation\n",

+      "Q=0.55;           #energy(MeV)\n",

+      "dm=Q/931;              \n",

+      "m3=dm+m1+m2-m4;       #mass of 6C14(a.m.u)\n",

+      "\n",

+      "#Result\n",

+      "print \"mass of 6C14 is\",round(m3,5),\"a.m.u\"\n",

+      "print \"answer given in the book varies due to rounding off errors\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "mass of 6C14 is 14.00895 a.m.u\n",

+        "answer given in the book varies due to rounding off errors\n"

+       ]

+      }

+     ],

+     "prompt_number": 14

+    },

+    {

+     "cell_type": "heading",

+     "level": 2,

+     "metadata": {},

+     "source": [

+      "Example number 13.6, Page number 261"

+     ]

+    },

+    {

+     "cell_type": "code",

+     "collapsed": false,

+     "input": [

+      "#import modules\n",

+      "import math\n",

+      "from __future__ import division\n",

+      "\n",

+      "#Variable declaration\n",

+      "m0=11.01280;     #mass 5B11(a.m.u)\n",

+      "m1=4.00387;      #mass of alpha particle(a.m.u)\n",

+      "m2=14.00752;     #mass of 7N14(a.m.u)\n",

+      "#m3=mass of neutron \n",

+      "E1=5.250;       #energy of alpha particle(MeV)\n",

+      "E2=2.139;       #energy of 7N14(MeV)\n",

+      "E3=3.260;       #energy of 0n1(MeV)\n",

+      "\n",

+      "#Calculation\n",

+      "m3=(m0*931)+((m1*931)+E1)-((m2*931)+E2)-E3;          #mass of neutron(a.m.u)\n",

+      "\n",

+      "#Result\n",

+      "print \"mass of neutron is\",round(m3/931,3),\"a.m.u\""

+     ],

+     "language": "python",

+     "metadata": {},

+     "outputs": [

+      {

+       "output_type": "stream",

+       "stream": "stdout",

+       "text": [

+        "mass of neutron is 1.009 a.m.u\n"

+       ]

+      }

+     ],

+     "prompt_number": 16

+    }

+   ],

+   "metadata": {}

+  }

+ ]

+}
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