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diff --git a/backup/Modern_Physics_version_backup/chapter14.ipynb b/backup/Modern_Physics_version_backup/chapter14.ipynb deleted file mode 100755 index b6ba7ca8..00000000 --- a/backup/Modern_Physics_version_backup/chapter14.ipynb +++ /dev/null @@ -1,201 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:f3dfb0a49e3f8552798e92e62c041d1f0b46371fc7dc05dfc9c1bf5833f43216"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "14: Nuclear Fission And Fusion"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 14.1, Page number 269"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#import modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "E1=7.8; #avg. B.E per nucleon(MeV)\n",
- "E2=8.6; #for fissin fragments(MeV)\n",
- "\n",
- "#Calculation\n",
- "FER=(234*E2)-(236*E1); #Fission energy released(MeV)\n",
- "\n",
- "#Result\n",
- "print \"Fission energy released is\",FER,\"MeV\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Fission energy released is 171.6 MeV\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 14.2, Page number 273"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#import modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "m1=235.044; #mass of 92U235(a.m.u)\n",
- "m2=97.905; #mass of 42Mo98(a.m.u)\n",
- "m3=135.917; #mass of 54Xe136(a.m.u)\n",
- "#rxn = 0n1 + 92U235 = 42Mo98 + 54Xe136 + 4 -1e0 + 2 0n1\n",
- "\n",
- "#Calculation\n",
- "LHSm=1.009+m1;\n",
- "RHSm=m2+m3+(4*0.00055)+(2*1.009);\n",
- "delta_m=LHSm-RHSm; #mass defect(a.m.u)\n",
- "E=delta_m*931; #energy released(MeV)\n",
- "\n",
- "#Result\n",
- "print \"mass defect is\",round(delta_m,3),\"a.m.u\"\n",
- "print \"energy released is\",int(E),\"MeV\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "mass defect is 0.211 a.m.u\n",
- "energy released is 196 MeV\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 14.3, Page number 274"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#import modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "m1=1.00813; #mass of 1H1(a.m.u)\n",
- "m2=4.00386; #mass of 2He4(a.m.u)\n",
- "SC=1.35; #solar constant(kW/m^2)\n",
- "d=1.5*10**11; #distance b/w earth and sum(m)\n",
- "e=1.6*10**-19; #the charge on electron(C)\n",
- "Na=6.02*10**26; #Avgraodo no.(per kg mole)\n",
- "#rxn = 4 1H1 = 2He4 + 2 1e0\n",
- "\n",
- "#Calculation\n",
- "dm=(4*m1)-m2;\n",
- "E=dm*931; #energy produced(MeV)\n",
- "EP=E/4; #energy produced per atom(MeV)\n",
- "EP=EP*10**6*e; #conversion in J\n",
- "EPkg=EP*Na; #energy produced by 1 kg of hydrogen\n",
- "SC=SC*1000; #conversion(J/s-m^2)\n",
- "SA=4*math.pi*d**2; #surface area of sphere\n",
- "ER=SC*SA; #energy recieved per second\n",
- "m=ER/EPkg; #mass of hydrogen consumed(tonnes/second)\n",
- "\n",
- "#Result\n",
- "print \"mass of hydrogen consumed is\",round(m/10**11,3),\"*10**8 tonnes/second\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "mass of hydrogen consumed is 5.941 *10**8 tonnes/second\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 14.4, Page number 275"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#import modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "m1=2.01478; #mass of 1H2(a.m.u)\n",
- "m2=4.00388; #mass of 2He4(a.m.u)\n",
- "#rxn 1H2 + 1H2 = 2He4 + Q\n",
- "\n",
- "#Calculation\n",
- "Q=2*m1-m2; #energy liberated(MeV) \n",
- "Q=Q*931; #conversion in MeV\n",
- "\n",
- "#Result\n",
- "print \"energy liberated is\",round(Q,1),\"MeV\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "energy liberated is 23.9 MeV\n"
- ]
- }
- ],
- "prompt_number": 11
- }
- ],
- "metadata": {}
- }
- ]
-}
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