From df60071cf1d1c18822d34f943ab8f412a8946b69 Mon Sep 17 00:00:00 2001 From: hardythe1 Date: Wed, 3 Jun 2015 15:27:17 +0530 Subject: add books --- _A_Textbook_Of_Engineering_Physics/Chapter15.ipynb | 661 +++++++++++++++++++++ 1 file changed, 661 insertions(+) create mode 100755 _A_Textbook_Of_Engineering_Physics/Chapter15.ipynb (limited to '_A_Textbook_Of_Engineering_Physics/Chapter15.ipynb') diff --git a/_A_Textbook_Of_Engineering_Physics/Chapter15.ipynb b/_A_Textbook_Of_Engineering_Physics/Chapter15.ipynb new file mode 100755 index 00000000..fd70c5b8 --- /dev/null +++ b/_A_Textbook_Of_Engineering_Physics/Chapter15.ipynb @@ -0,0 +1,661 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:00168ce8e6485ed62b7ab93f835949895122e8d417f4d6143363c166514fa2f2" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Chapter15-Atomic Nucleus And Nuclear Energy" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex1-pg-pg427" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "##Example 15.1\n", + "##calculation of binding energy per nucleon\n", + "\n", + "##given values\n", + "Mp=1.00814;##mass of proton in amu\n", + "Mn=1.008665;##mass of nucleon in amu\n", + "M=7.01822;##mass of Lithium nucleus in amu\n", + "amu=931.;##amu in MeV\n", + "n=7-3;##no of neutrons in lithium nucleus\n", + "\n", + "##calculation\n", + "ET=(3*Mp+4*Mn-M)*amu;##total binding energy in MeV\n", + "E=ET/7.;##7 1s the mass number\n", + "print'%s %.2f %s'%('Binding energy per nucleon in MeV is',E,'');\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Binding energy per nucleon in MeV is 5.43 \n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex2-pg427" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "##Example 15.2\n", + "##calculation of energy \n", + "\n", + "##given values\n", + "M1=15.00001;##atomic mass of N15 in amu\n", + "M2=15.0030;##atomic mass of O15 in amu\n", + "M3=15.9949;##atomic mass of O16 in amu\n", + "amu=931.4;##amu in MeV\n", + "mp=1.0072766;##restmass of proton\n", + "mn=1.0086654;##restmass of neutron\n", + "\n", + "##calculation\n", + "Q1=(M3-mp-M1)*amu;\n", + "print'%s %.2f %s'%('energy required to remove one proton from O16 is',Q1,'');\n", + "Q2=(M3-mn-M2)*amu;\n", + "print'%s %.2f %s'%('energy required to remove one neutron from O16 is',Q2,'');\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "energy required to remove one proton from O16 is -11.54 \n", + "energy required to remove one neutron from O16 is -15.62 \n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex3-pg428" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "##Example 15.3\n", + "##calculation of binding energy \n", + "\n", + "##given values\n", + "Mp=1.00758;##mass of proton in amu\n", + "Mn=1.00897;##mass of nucleon in amu\n", + "M=4.0028;##mass of Helium nucleus in amu\n", + "amu=931.4;##amu in MeV\n", + "\n", + "##calculation\n", + "E1=(2*Mp+2*Mn-M)*amu;##total binding energy\n", + "print'%s %.2f %s'%('Binding energy in MeV is',E1,'');\n", + "E2=E1*10**6*1.6*10**-19;\n", + "print'%s %.3e %s'%('binding energy in Joule is',E2,'');\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Binding energy in MeV is 28.22 \n", + "binding energy in Joule is 4.515e-12 \n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex4-pg435" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "##Example 15.4\n", + "##calculation of amount of unchanged material\n", + "\n", + "##given values\n", + "T=2;##half life in years\n", + "k=.6931/T;##decay constant\n", + "M=4.0028;##mass of Helium nucleus in amu\n", + "amu=931.4;##amu in MeV\n", + "No=1.;##initial amount in g\n", + "\n", + "##calculation\n", + "N=No*(math.e**(-k*2*T));\n", + "print'%s %.2f %s'%('amount of material remaining unchanged after four years(in gram) is',N,'');\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "amount of material remaining unchanged after four years(in gram) is 0.25 \n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex5-pg435" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "##Example 15.5\n", + "##calculation of amount of halflife\n", + "\n", + "##given values\n", + "t=5.;##time period in years\n", + "amu=931.4;##amu in MeV\n", + "No=5.;##initial amount in g\n", + "N=5.-(10.5*10**-3);##amount present after 5 years\n", + "\n", + "\n", + "##calculation\n", + "k=math.log(N/No)/t;##decay constant\n", + "T=-.693/k;\n", + "print'%s %.2f %s'%('halflife in years is',T,'');" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "halflife in years is 1648.27 \n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex6-pg435" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "##Example 15.6\n", + "##calculation of activity\n", + "\n", + "##given values\n", + "t=28.;##half life in years\n", + "m=10**-3;##mass of sample\n", + "M=90.;##atomic mass of strontium\n", + "NA=6.02*10**26;##avogadro's number\n", + "\n", + "\n", + "##calculation\n", + "n=m*NA/M;##no of nuclei in 1 mg sample\n", + "k=.693/(t*365*24.*60.*60.);##decay constant\n", + "A=k*n;\n", + "print'%s %.3e %s'%('activity of sample(in disintegrations per second) is',A,'');\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "activity of sample(in disintegrations per second) is 5.250e+12 \n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex7-pg439" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "##Example 15.7\n", + "##calculation of age of mineral\n", + "\n", + "##given values\n", + "t=4.5*10**9.;##half life in years\n", + "M1=238.;##atomic mass of Uranium in g\n", + "m=.093;##mass of lead in 1 g of uranium in g\n", + "NA=6.02*10**26;##avogadro's number\n", + "M2=206.;##atomic mass of lead in g\n", + "\n", + "##calculation\n", + "n=NA/M1;##no of nuclei in 1 g of uranium sample\n", + "n1=m*NA/M2;##no of nuclei in m mass of lead\n", + "c=n1/n;\n", + "k=.693/t;##decay constant\n", + "T=(1/k)*math.log(1+c);\n", + "print'%s %.3e %s'%('age of mineral in years is',T,'');" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "age of mineral in years is 6.627e+08 \n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex8-pg440" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "##Example 15.8\n", + "##calculation of age of wooden piece\n", + "\n", + "##given values\n", + "t=5730.;##half life of C14 in years\n", + "M1=50.;##mass of wooden piece in g\n", + "A1=320.;##activity of wooden piece (disintegration per minute per g)\n", + "A2=12.;##activity of living tree\n", + "\n", + "##calculation\n", + "k=.693/t;##decay constant\n", + "A=A1/M1;##activity after death\n", + "\n", + "T=(1/k)*math.log(A2/A);\n", + "print'%s %.2f %s'%('age of mineral in years is',T,'');" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "age of mineral in years is 5197.59 \n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex9-pg443" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "##Example 15.9\n", + "##calculation of energy released\n", + "\n", + "##given values\n", + "M1=10.016125;##atomic mass of Boron in amu\n", + "M2=13.007440;##atomic mass of C13 in amu\n", + "M3=4.003874;##atomic mass of Helium in amu\n", + "mp=1.008146;##mass of proton in amu\n", + "amu=931.;##amu in MeV\n", + "\n", + "##calculation\n", + "Q=(M1+M3-(M2+mp))*amu;##total binding energy in M\n", + "print'%s %.2f %s'%('Binding energy per nucleon in MeV is',Q,'');\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Binding energy per nucleon in MeV is 4.11 \n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex10-pg444" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "##Example 15.10\n", + "##calculation of crosssection\n", + "\n", + "##given values\n", + "t=.01*10**-3;##thickness in m\n", + "n=10**13.;##no of protons bombarding target per s\n", + "NA=6.02*10**26.;##avogadro's number\n", + "M=7.;##atomic mass of lithium in kg\n", + "d=500.;##density of lithium in kg/m**3\n", + "n0=10**8.;##no of neutrons produced per s\n", + "##calculation\n", + "n1=d*NA/M;##no of target nuclei per unit volume\n", + "n2=n1*t;##no of target nuclei per area\n", + "A=n0/(n*n2);\n", + "print'%s %.3e %s'%('crosssection(in m^2) for this reaction is',A,'');" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "crosssection(in m^2) for this reaction is 2.326e-29 \n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex11-pg450" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "##Example 15.11\n", + "##calculation of final energy \n", + "\n", + "##given values\n", + "B=.4;##max magnetic field in Wb/m**2\n", + "c=3*10**8.;\n", + "e=1.6*10**-19.;\n", + "d=1.52;##diametre in m\n", + "r=d/2.;\n", + "\n", + "##calculation\n", + "E=B*e*r*c;##E=pc,p=mv=Ber\n", + "print'%s %.3e %s'%('final energy of e(in J) is',E,'');\n", + "E1=(E/e)/10**6;\n", + "print'%s %.2f %s'%('final energy of e (in MeV) is',E1,'');" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "final energy of e(in J) is 1.459e-11 \n", + "final energy of e (in MeV) is 91.20 \n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex12-pg459" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "##Example 15.12\n", + "##calculation of amount of fuel\n", + "\n", + "##given values\n", + "P=100*10**6.;##power required by city\n", + "M=235.;##atomic mass of Uranium in g\n", + "e=20/100.;##conversion efficiency\n", + "NA=6.02*10**26.;##avogadros number\n", + "E=200*10**6*1.6*10**-19;##energy released per fission\n", + "t=8.64*10**4.;##day in seconds\n", + "\n", + "\n", + "##calculation\n", + "E1=P*t;##energy requirement\n", + "m=E1*M/(NA*e*E);##no of nuclei N=NA*m/M,energy released by m kg is N*E,energy requirement=e*N*E\n", + "print'%s %.2f %s'%('amount of fuel(in kg) required is',m,'');" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "amount of fuel(in kg) required is 0.53 \n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex13-pg459" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "##Example 15.13\n", + "##calculation of power output\n", + "\n", + "##given values\n", + "M=235.;##atomic mass of Uranium in kg\n", + "e=5/100.;##reactor efficiency\n", + "m=25/1000.;##amount of uranium consumed per day in kg\n", + "E=200*10**6*1.6*10**-19;##energy released per fission\n", + "t=8.64*10**4.;##day in seconds\n", + "NA=6.02*10**26.;##avogadros number\n", + "\n", + "##calculation\n", + "n=NA*m/M;##no of nuclei in 25g\n", + "E1=n*E;##energy produced by n nuclei\n", + "E2=E1*e;##energy converted to power\n", + "P=E2/t;##power output in Watt\n", + "print'%s %.2f %s'%('power output in MW is',P/10**6,'');" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "power output in MW is 1.19 \n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex14-pg460" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "##Example 15.14\n", + "##calculation of power developed\n", + "\n", + "##given values\n", + "M=235.;##atomic mass of Uranium in kg\n", + "m=20.4;##amount of uranium consumed per day in kg\n", + "E=200*10**6*1.6*10**-19;##energy released per fission\n", + "t=3600*1000.;##time of operation\n", + "NA=6.02*10**26;##avogadros number\n", + "\n", + "##calculation\n", + "n=NA*m/M;##no of nuclei in 20.4kg\n", + "E1=n*E;##energy produced by n nuclei\n", + "P=E1/t;##in Watt\n", + "print'%s %.2f %s'%('power developed in MW is',P/10**6,'');" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "power developed in MW is 464.52 \n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex15-pg464" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "##Example 15.15\n", + "##calculation of amount of dueterium consumed\n", + "\n", + "##given values\n", + "M1=2.01478;##atomic mass of Hydrogen in amu\n", + "M2=4.00388;##atomic mass of Helium in amu\n", + "amu=931.;##amu in MeV\n", + "e=30/100.;##efficiency\n", + "P=50*10**6.;##output power\n", + "NA=6.026*10**26.;##avogadro number\n", + "t=8.64*10**4.;##seconds in a day\n", + "\n", + "##calculation\n", + "Q=(2*M1-M2)*amu;##energy released in a D-D reaction in MeV\n", + "O=Q*e*10**6/2.;##actual output per dueterium atom in eV\n", + "n=P/(O*1.6*10**-19);##no of D atoms required\n", + "m=n*M1/NA;##equivalent mass of D required per s\n", + "X=m*t;\n", + "\n", + "print'%s %.2f %s'%('Deuterium requirement per day in kg is',X,'');\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Deuterium requirement per day in kg is 0.03 \n" + ] + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file -- cgit