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{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# 9: Nuclear Reactions"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example number 1, Page number 272"
]
},
{
"cell_type": "code",
"execution_count": 8,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"number of nuclei produced per second is 1.66 *10**5\n",
"answer given in the book is wrong\n"
]
}
],
"source": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration \n",
"rho=19.3*10**3; #density(kg/m**3)\n",
"N=6.02*10**23; #avagadro number\n",
"M=197; #molecular weight\n",
"a=2*10**12; #neutrons/m**2 sec\n",
"A=5*10**-4; #area(m**2)\n",
"sigma=94*10**-28; #reaction cross section(m**2)\n",
"t=0.3*10**-3; #thickness(m)\n",
"\n",
"#Calculations\n",
"n=rho*N/M; #number of nuclei per unit volume(per m**3)\n",
"N0=a*A; #number of neutrons hitting the target\n",
"N0_N=N0*(1-math.exp(-n*sigma*t)); #number of nuclei produced per second\n",
"\n",
"#Result\n",
"print \"number of nuclei produced per second is\",round(N0_N/10**5,2),\"*10**5\"\n",
"print \"answer given in the book is wrong\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example number 2, Page number 275"
]
},
{
"cell_type": "code",
"execution_count": 20,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"energy of neutron is 19.768 MeV\n",
"energy of Be is 5.007 MeV\n",
"angle of recoil of Be atom is 44.855 or 45 degrees\n"
]
}
],
"source": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration \n",
"M1=2.01472; #molecular mass of H(amu)\n",
"M0=7.01784; #molecular mass of Li(amu)\n",
"M2=8.00776; #molecular mass of Be(amu)\n",
"M3=1.00893; #molecular mass of n(amu)\n",
"Ek1=10; #energy(MeV)\n",
"\n",
"#Calculations\n",
"M1M0=M1+M0; #mass of interacting particles(amu)\n",
"M2M3=M2+M3; #mass of product particles(amu)\n",
"Q=(M1M0-M2M3)*931; #decrease in mass(MeV)\n",
"Ek3=(Q+(Ek1*(1-(M1/M2))))/(1+(M3/M2)); #energy of neutron(MeV)\n",
"Ek2=Q+Ek1-Ek3; #energy of Be(MeV)\n",
"phi=math.atan(math.sqrt(Ek3*M3/(Ek1*M1))); #angle of recoil of Be atom(rad)\n",
"phi=phi*180/math.pi; #angle of recoil of Be atom(degrees)\n",
"\n",
"#Result\n",
"print \"energy of neutron is\",round(Ek3,3),\"MeV\" \n",
"print \"energy of Be is\",round(Ek2,3),\"MeV\" \n",
"print \"angle of recoil of Be atom is\",round(phi,3),\"or\",int(round(phi)),\"degrees\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example number 3, Page number 277"
]
},
{
"cell_type": "code",
"execution_count": 22,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"energy of emitted protons is 5.3 MeV\n"
]
}
],
"source": [
"#importing modules\n",
"import math\n",
"from __future__ import division\n",
"\n",
"#Variable declaration \n",
"Q=-3.9; #Q value of reaction(MeV)\n",
"M1=1.0087; #molecular mass of incident neutron(amu)\n",
"M2=18.99; #molecular mass of O nucleus(amu)\n",
"M3=1.0078; #molecular mass of proton(amu)\n",
"Ek1=10; #energy of incident neutron(MeV)\n",
"\n",
"#Calculations\n",
"x=1-(M1/M2);\n",
"y=1+(M3/M2);\n",
"Ek3=(Q+Ek1*x)/y; #energy of emitted protons(MeV)\n",
"\n",
"#Result\n",
"print \"energy of emitted protons is\",round(Ek3,1),\"MeV\""
]
}
],
"metadata": {
"kernelspec": {
"display_name": "Python 2",
"language": "python",
"name": "python2"
},
"language_info": {
"codemirror_mode": {
"name": "ipython",
"version": 2
},
"file_extension": ".py",
"mimetype": "text/x-python",
"name": "python",
"nbconvert_exporter": "python",
"pygments_lexer": "ipython2",
"version": "2.7.11"
}
},
"nbformat": 4,
"nbformat_minor": 0
}
|
