{ "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 }