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Diffstat (limited to 'backup/Modern_Physics_version_backup/chapter13.ipynb')
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diff --git a/backup/Modern_Physics_version_backup/chapter13.ipynb b/backup/Modern_Physics_version_backup/chapter13.ipynb deleted file mode 100755 index fddec6a2..00000000 --- a/backup/Modern_Physics_version_backup/chapter13.ipynb +++ /dev/null @@ -1,248 +0,0 @@ -{
- "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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