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{
"metadata": {
"name": "",
"signature": "sha256:9326f276d5dc99ce97d41c9ca0d5924dbd68f522091536657f41d8cfe038dc31"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"19: Nuclear reactions"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 19.1, Page number 368"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"m2H=2.014102; #atomic mass of 2H(u)\n",
"mn=1.008665; #mass of n(u)\n",
"m63Cu=62.929599; #mass of 63Cu(u)\n",
"m64Zn=63.929144; #mass of m64Zn(u)\n",
"E=931.5; #energy(MeV)\n",
"Kx=10; #energy of deutron(MeV)\n",
"Ky=15; #energy of neutron(MeV)\n",
"\n",
"#Calculation\n",
"Q=E*(m2H+m63Cu-mn-m64Zn); #Q-value(MeV)\n",
"KY=Q+Kx-Ky; #kinetic energy(MeV)\n",
"\n",
"#Result\n",
"print \"Q-value is\",round(Q,3),\"MeV\"\n",
"print \"kinetic energy is\",round(KY,3),\"MeV\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Q-value is 5.488 MeV\n",
"kinetic energy is 0.488 MeV\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 19.2, Page number 368"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"m19F=18.998404; #atomic mass of 19F(u)\n",
"mH=1.007825; #mass of H(u)\n",
"m19O=19.003577; #mass of 19O(u)\n",
"mn=1.008665; #mass of n(u)\n",
"E=931.5; #energy(MeV)\n",
"\n",
"#Calculation\n",
"Q=E*(m19F+mn-mH-m19O); #Q-value(MeV)\n",
"Kxmin=-Q*(1+(mn/m19F)); #threshold energy(MeV)\n",
"\n",
"#Result\n",
"print \"Q-value is\",round(Q,4),\"MeV\"\n",
"print \"threshold energy is\",round(Kxmin,2),\"MeV\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Q-value is -4.0362 MeV\n",
"threshold energy is 4.25 MeV\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 19.3, Page number 373"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"mn=1.008665; #mass of n(u)\n",
"mu=235.043924; #mass of 235U(u)\n",
"mBa=140.91440; #mass of 141Ba(u)\n",
"mKr=91.92630; #mass of Kr(u)\n",
"E=931.5; #energy(MeV)\n",
"\n",
"#Calculation\n",
"mr=mn+mu; #mass of reactants(u)\n",
"mp=mBa+mKr+(3*mn); #mass of products(u)\n",
"md=mr-mp; #mass difference(u)\n",
"E=md*E; #energy released(MeV)\n",
"\n",
"#Result\n",
"print \"energy released is\",round(E,1),\"MeV\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"energy released is 173.2 MeV\n"
]
}
],
"prompt_number": 19
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 19.4, Page number 373"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"E=200*10**6; #energy released(eV)\n",
"e=1.6*10**-19; #conversion factor from J to eV\n",
"P=300*10**6; #power(W)\n",
"t=1; #time(s)\n",
"\n",
"#Calculation\n",
"n=P*t/(E*e); #number of fissions per second\n",
"\n",
"#Result\n",
"print \"number of fissions per second is\",n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"number of fissions per second is 9.375e+18\n"
]
}
],
"prompt_number": 20
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 19.5, Page number 378"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"m2H1=2*1.66*10**-27; #mass of proton(kg)\n",
"E=931.5; #energy(MeV)\n",
"m1=2.014102;\n",
"m2=3.01609;\n",
"mH=1.007825; #mass of H(u)\n",
"\n",
"#Calculation\n",
"E=E*((2*m1)-m2-mH); #energy released(MeV)\n",
"n=0.001/m2H1; #number of nuclei\n",
"Eg=n*E/2; #energy released per gm(MeV)\n",
"\n",
"#Result\n",
"print \"energy released per gm is\",round(Eg/10**23,2),\"*10**23 MeV\"\n",
"print \"answer given in the book is wrong\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"energy released per gm is 6.02 *10**23 MeV\n",
"answer given in the book is wrong\n"
]
}
],
"prompt_number": 31
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 19.6, Page number 379"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"k=8.99*10**9; #value of k(Nm**2/C**2)\n",
"rd=1.5*10**-15; #radius of deuterium nucleus(m)\n",
"rt=1.7*10**-15; #radius of tritium nucleus(m)\n",
"e=1.6*10**-19; #conversion factor from J to eV\n",
"KE=0.225; #kinetic energy for 1 particle(MeV)\n",
"k=1.38*10**-23; #boltzmann constant(J/K)\n",
"\n",
"#Calculation\n",
"K_E=k*e**2/(e*(rd+rt)); #kinetic energy of 2 particles(MeV)\n",
"T=2*KE*e*10**6/(3*k); #temperature(K)\n",
"\n",
"#Result\n",
"print \"temperature is\",round(T/10**9),\"*10**9 K\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"temperature is 2.0 *10**9 K\n"
]
}
],
"prompt_number": 35
}
],
"metadata": {}
}
]
}
|
