{
"metadata": {
"name": "",
"signature": "sha256:0e9faaea32136a2f476b53b6ab2d5d2eb5330fb68ebd80aaad9bbe7ed1328c93"
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"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": {}
}
]
}