From f270f72badd9c61d48f290c3396004802841b9df Mon Sep 17 00:00:00 2001 From: kinitrupti Date: Fri, 12 May 2017 18:53:46 +0530 Subject: Removed duplicates --- Engineering_Physics_by_P._V._Naik/Chapter11.ipynb | 620 ++++++++++++++++++++++ 1 file changed, 620 insertions(+) create mode 100755 Engineering_Physics_by_P._V._Naik/Chapter11.ipynb (limited to 'Engineering_Physics_by_P._V._Naik/Chapter11.ipynb') diff --git a/Engineering_Physics_by_P._V._Naik/Chapter11.ipynb b/Engineering_Physics_by_P._V._Naik/Chapter11.ipynb new file mode 100755 index 00000000..39762859 --- /dev/null +++ b/Engineering_Physics_by_P._V._Naik/Chapter11.ipynb @@ -0,0 +1,620 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:1d5f4970753c62d94a2bd202867cc0f79046f1baac4b1a42721a5ae6844ad5f4" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "11: Nuclear Radiations and Detectors" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 11.1, Page number 227" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#importing modules\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "r0=1.2; #radius(fm)\n", + "A=7; #mass number \n", + "\n", + "#Calculation \n", + "r=r0*A**(1/3);\t #radius of Li(fm) \n", + "\n", + "#Result\n", + "print \"The radius of Li is\",round(r,4),\"fm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The radius of Li is 2.2955 fm\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 11.2, Page number 227" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#importing modules\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "M=235.043945; #atomic mass of uranium(u)\n", + "Z=92; #atomic number of uranium\n", + "mp=1.007825; #mass of proton(kg)\n", + "N=143; #no.of neutrons\n", + "mn=1.008665; #mass of neutron(kg)\n", + "A=235; #number of nucleons\n", + "\n", + "#Calculation \n", + "B=(((Z*mp)+(N*mn)-(M))/A)*931.5; #Binding energy(MeV)\n", + "\n", + "#Result\n", + "print \"The binding energy per nucleon is\",round(B,3),\"MeV\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The binding energy per nucleon is 7.591 MeV\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 11.3, Page number 227" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#importing modules\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "#After removing one neutron from Ca(A=43;Z=20) it becomes Ca(A=42;Z=20)\n", + "M=41.958622; #mass of Ca(A=42;Z=20)(kg)\n", + "mn=1.008665; #mass of neutron(kg)\n", + "E=42.95877; #mass of Ca(A=43;Z=20)(kg)\n", + "\n", + "#Calculation \n", + "C=M+mn;\n", + "D=C-E;\n", + "B=D*931.5; #Binding energy of neutron(MeV)\n", + "\n", + "#Result\n", + "print \"The binding energy of neutron is\",round(B,4),\"MeV\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The binding energy of neutron is 7.9336 MeV\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 11.4, Page number 227" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#importing modules\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "mBe=9.012182; #Atomic mass of beryllium(u)\n", + "mHe=4.002603; #Atomic mass of helium\n", + "mn=1.008665; #mass of neutron(kg)\n", + "mC=12.000000; #Atomic mass of carbon\n", + "\n", + "#Calculation \n", + "Q=(mBe+mHe-mn-mC)*931.5 #energy balance of the reaction(MeV)\n", + "\n", + "#Result\n", + "print \"The Q-value is\",round(Q,1),\"MeV\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The Q-value is 5.7 MeV\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 11.5, Page number 227" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#importing modules\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "mLi=7.016004; #mass of Lithium(A=7)(u)\n", + "mH=1.007825; #mass of Hydrogen(A=1)(u)\n", + "mHe=4.002603; #mass of helium(A=4)(u)\n", + "p=0.5; #energy of proton(MeV)\n", + "\n", + "#Calculation \n", + "Q=(mLi+mH-2*(mHe))*931.5 #energy balance of the reaction(MeV)\n", + "#The energy of 2 alpha particles is equal to the Q-value + energy of proton.\n", + "Ealpha=(Q+p)/2; #energy of each alpha particle(MeV)\n", + "\n", + "#Result\n", + "print \"The Q-value of the reaction is\",round(Q,2),\"MeV\"\n", + "print \"The energy of each alpha particle is\",round(Ealpha,3),\"MeV\"\n", + "print \"answer for energy in the book varies due to rounding off errors\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The Q-value of the reaction is 17.35 MeV\n", + "The energy of each alpha particle is 8.924 MeV\n", + "answer for energy in the book varies due to rounding off errors\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 11.6, Page number 228" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#importing modules\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "wt=1000; #weight(gm)\n", + "A=235; #mass number of uranium\n", + "N=(6.02*10**23/A)*wt; #no.of nuclei in 1kg of uranium\n", + "Q=208; #energy-balance of the reaction\n", + "\n", + "#Calculation \n", + "E=N*Q; #Energy released(MeV)\n", + "#1MeV=4.45*10^-20kWh\n", + "E=E*4.45*10**-20;\n", + "\n", + "#Result\n", + "print \"The energy released is\",round(E/10**7,3),\"*10**7 kWh\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The energy released is 2.371 *10**7 kWh\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 11.7, Page number 228" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#importing modules\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "wt=5000; #weight(gm)\n", + "A=235; #mass number of uranium\n", + "Ef=208; #Energy released per fission(MeV)\n", + "\n", + "#Calculation \n", + "N=(6.02*10**23/A)*wt; #number of nuclei in 5 Kg\n", + "E=N*Ef; #Energy(MeV)\n", + "E=E*1.6*10**-13; #Energy(J)\n", + "T=24*60*60; #time\n", + "P=E/T; #power(MW)\n", + "\n", + "#Result\n", + "print \"The power output of a nuclear reactor is\",round(P/10**6),\"MW\"\n", + "print \"answer given in the book is wrong\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The power output of a nuclear reactor is 4934.0 MW\n", + "answer given in the book is wrong\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 11.8, Page number 228" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#importing modules\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "A=235; #mass number of uranium\n", + "p=1000; #amount of electric power produced(MW)\n", + "e=0.32; #energy conversion efficiency of the plant\n", + "f=200; #fission energy per event(MeV)\n", + "\n", + "#Calculation \n", + "I=p/e; #Input nuclear energy(MW)\n", + "TE=I*(10**6)*3600*24*365; #total energy(J)\n", + "EF=f*(10**6)*1.6*10**-19; #Energy released per fission event(J)\n", + "N=TE/EF; #Number of nuclei required\n", + "M=N*A/(6.02*10**23); #corresponding mass(g)\n", + "\n", + "#Result\n", + "print \"The amount of uranium required is\",round(M*10**-3,1),\"kg\"\n", + "print \"answer in the book varies due to rounding off errors\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The amount of uranium required is 1202.2 kg\n", + "answer in the book varies due to rounding off errors\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 11.9, Page number 229" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#importing modules\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "q=1.6*10**-19; #charge of the particle(c)\n", + "B=1; #magnetic field(T)\n", + "m=1.67*10**-27; #mass of proton(kg)\n", + "r=0.5; #radius(m)\n", + "\n", + "#Calculation \n", + "omega=(q*B)/m; #angular frequency(radian/s)\n", + "v=(omega/(2*math.pi))*10**-8; #frequency(MHz)\n", + "s=omega*r; #speed of proton(m/s)\n", + "K=(m*(s**2))*(1/2)*6.27*10**12; #kinetic energy of protons emerging from cyclotron(MeV)\n", + "\n", + "#Result\n", + "print \"The frequency of oscillator to accelerate protons is\",round(omega/10**8,2),\"*10**8 radian/s\"\n", + "print \"The speed of proton is\",round(s/10**7,1),\"*10**7 m/s\"\n", + "print \"The kinetic energy of protons emerging from the cyclotron is\",int(K),\"MeV\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The frequency of oscillator to accelerate protons is 0.96 *10**8 radian/s\n", + "The speed of proton is 4.8 *10**7 m/s\n", + "The kinetic energy of protons emerging from the cyclotron is 12 MeV\n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 11.10, Page number 229" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#importing modules\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "rho=1.83*10**17; #average density of carbon nucleus(kg/m^3)\n", + "m=12; #mass(u)\n", + "e=1.66*10**-27;\n", + "\n", + "#Calculation \n", + "r=(m*e/((4/3)*math.pi*rho))**(1/3)*10**15; #radius(fm)\n", + "\n", + "#Result\n", + "print \"The radius is\",round(r,2),\"fm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The radius is 2.96 fm\n" + ] + } + ], + "prompt_number": 30 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 11.11, Page number 229" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#importing modules\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "q=1.6*10**-19; #charge of the particle(c)\n", + "B=5; #magnetic field(T)\n", + "m=9.1*10**-31; #mass of electron(kg)\n", + "\n", + "#Calculation \n", + "v=(q*B)/(2*math.pi*m); #cyclotron frequency(Hz)\n", + "\n", + "#Result\n", + "print \"cyclotron frequency of electron is\",round(v/10**11,1),\"*10**11 Hz\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "cyclotron frequency of electron is 1.4 *10**11 Hz\n" + ] + } + ], + "prompt_number": 32 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 11.12, Page number 229" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#importing modules\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "k=1.5; #maximum kinetic energy(MeV)\n", + "m=1.67*10**-27; #mass of proton(kg)\n", + "q=1.6*10**-19; #charge of particle(c)\n", + "r=0.35; #radius(m)\n", + "\n", + "#Calculation \n", + "B=math.sqrt(k*10**6*q*2*m)/(q*r); #magnetic field(T)\n", + "\n", + "#Result\n", + "print \"The mgnetic field is\",round(B,1),\"T\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The mgnetic field is 0.5 T\n" + ] + } + ], + "prompt_number": 34 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 11.13, Page number 229" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#importing modules\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "m=1.67*10**-27; #mass of proton(kg)\n", + "q=1.6*10**-19; #charge of particle(q)\n", + "v=25; #cyclotron frequency(MHz)\n", + "\n", + "#Calculation \n", + "B=(v*10**6*2*math.pi*m)/q; #magnetic field(T)\n", + "\n", + "#Result\n", + "print \"The required magnetic field is\",round(B,4),\"T\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The required magnetic field is 1.6395 T\n" + ] + } + ], + "prompt_number": 37 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example number 11.14, Page number 229" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#importing modules\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "v=20; #cyclotron frequency(MHz)\n", + "B=1.3; #magnetic field(T)\n", + "\n", + "#Calculation \n", + "d=2*math.pi*v*10**6/B; #charge to mass ratio of proton(C/kg)\n", + "\n", + "#Result\n", + "print \"charge to mass ratio of proton is\",round(d/10**6,2),\"*10**6 C/kg\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "charge to mass ratio of proton is 96.66 *10**6 C/kg\n" + ] + } + ], + "prompt_number": 39 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file -- cgit