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diff --git a/Engineering_Physics_by_Uma_Mukherji/Chapter12.ipynb b/Engineering_Physics_by_Uma_Mukherji/Chapter12.ipynb new file mode 100755 index 00000000..0efc0c8d --- /dev/null +++ b/Engineering_Physics_by_Uma_Mukherji/Chapter12.ipynb @@ -0,0 +1,287 @@ +{
+ "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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