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{
"metadata": {
"name": "",
"signature": "sha256:0f6dea1f19194326599a9bca2989e912ed17e32f1ffb8d9305e16c13f8cacf2c"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"18: Radioactive decay"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 18.1, Page number 347"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"N0=1; #assume\n",
"\n",
"#Calculation\n",
"f=(N0/2)/N0; #fraction after t1/2\n",
"f1=(N0/4)/N0; #fraction after 2 half lives\n",
"f2=(N0/(2**5))/N0; #fraction after 5 half lives\n",
"f3=(N0/(2**10))/N0; #fraction after 10 half lives\n",
"\n",
"#Result\n",
"print \"fraction after 2 half lives is\",f1\n",
"print \"fraction after 5 half lives is\",f2\n",
"print \"fraction after 10 half lives is\",f3"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"fraction after 2 half lives is 0.25\n",
"fraction after 5 half lives is 0.03125\n",
"fraction after 10 half lives is 0.0009765625\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 18.2, Page number 348"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"thalf=2.7*24*60*60; #half life(s)\n",
"m=1*10**-6; #mass(gm)\n",
"Na=6.02*10**23; #avagadro number(atoms/mol)\n",
"M=198; #molar mass(g/mol)\n",
"t=8*24*60*60;\n",
"\n",
"#Calculation\n",
"lamda=0.693/thalf; #decay constant(per sec)\n",
"N=m*Na/M; #number of nuclei(atoms)\n",
"A0=lamda*N; #activity(disintegrations per sec)\n",
"A=A0*math.exp(-lamda*t); #activity for 8 days(decays per sec)\n",
"\n",
"#Result\n",
"print \"decay constant is\",round(lamda*10**6,2),\"*10**-6 per sec\"\n",
"print \"activity is\",round(A0/10**9,2),\"*10**9 disintegrations per sec\"\n",
"print \"activity for 8 days is\",round(A/10**9,2),\"*10**9 decays per sec\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"decay constant is 2.97 *10**-6 per sec\n",
"activity is 9.03 *10**9 disintegrations per sec\n",
"activity for 8 days is 1.16 *10**9 decays per sec\n"
]
}
],
"prompt_number": 10
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 18.3, Page number 348"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"thalf=5570*365*24*60*60; #half life(s)\n",
"dNbydt=3.7*10**10*2*10**-3; #number of decays per sec\n",
"m=14;\n",
"Na=6.02*10**23; #avagadro number(atoms/mol)\n",
"\n",
"#Calculation\n",
"lamda=0.693/thalf; #decay constant(per sec)\n",
"N=dNbydt/lamda; #number of atoms\n",
"mN=m*N/Na; #mass of 2mCi(g)\n",
"\n",
"#Result\n",
"print \"mass of 2mCi is\",round(mN*10**4,2),\"*10**-4 g\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"mass of 2mCi is 4.36 *10**-4 g\n"
]
}
],
"prompt_number": 23
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 18.5, Page number 353"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"thalf=1.25*10**9; #half life(yr)\n",
"r=10.2; #ratio of number of atoms\n",
"\n",
"#Calculation\n",
"a=1+r;\n",
"lamda=0.693/thalf; #decay constant(per yr)\n",
"t=math.log(a)/lamda; #time(yr)\n",
"\n",
"#Result\n",
"print \"the rock is\",round(t/10**9,2),\"*10**9 yrs old\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"the rock is 4.36 *10**9 yrs old\n"
]
}
],
"prompt_number": 12
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 18.6, Page number 356"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"mU=232.037131; #atomic mass of U(u)\n",
"mHe=4.002603; #atomic mass of He(u)\n",
"E=931.5; #energy(MeV)\n",
"KE=5.32; #kinetic energy of alpha particle(MeV)\n",
"\n",
"#Calculation\n",
"mTh=mU-mHe-(KE/E); #atomic mass of Th(u)\n",
"\n",
"#Result\n",
"print \"atomic mass of Th is\",round(mTh,5),\"u\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"atomic mass of Th is 228.02882 u\n"
]
}
],
"prompt_number": 14
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 18.7, Page number 359"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"E=931.5; #energy(MeV)\n",
"mX=11.011433; #mass of 11C(u)\n",
"mXdash=11.009305; #mass of 11B(u)\n",
"me=0.511;\n",
"\n",
"#Calculation\n",
"Q=(E*(mX-mXdash))-(2*me); #Q value for decay(MeV)\n",
"\n",
"#Result\n",
"print \"maximum energy is\",round(Q,2),\"MeV.minimum energy is zero\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"maximum energy is 0.96 MeV.minimum energy is zero\n"
]
}
],
"prompt_number": 16
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 18.8, Page number 359"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"mK=39.963999; #mass of K(u)\n",
"mAr=39.962384; #mass of Ar(u)\n",
"E=931.5; #energy(MeV)\n",
"\n",
"#Calculation\n",
"Q=(mK-mAr)*E; #kinetic energy of neutrino(MeV)\n",
"\n",
"#Result\n",
"print \"kinetic energy of neutrino is\",round(Q,3),\"MeV\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"kinetic energy of neutrino is 1.504 MeV\n"
]
}
],
"prompt_number": 18
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example number 18.9, Page number 360"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration\n",
"mN=12.018613; #mass of N(u)\n",
"mC=12; #mass of C(u)\n",
"me=0.000549; #mass of me(u)\n",
"E=931.5; #energy(MeV)\n",
"Egamma=4.43; #energy of emitted gamma ray(MeV)\n",
"\n",
"#Calculation\n",
"Q=(mN-mC-(2*me))*E; #Q value(MeV)\n",
"Emax=Q-Egamma; #maximum kinetic energy(MeV)\n",
"\n",
"#Result\n",
"print \"maximum kinetic energy is\",round(Emax,2),\"MeV\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"maximum kinetic energy is 11.89 MeV\n"
]
}
],
"prompt_number": 20
}
],
"metadata": {}
}
]
}
|
