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diff --git a/backup/Modern_Physics_version_backup/Chapter12.ipynb b/backup/Modern_Physics_version_backup/Chapter12.ipynb new file mode 100755 index 00000000..904fea62 --- /dev/null +++ b/backup/Modern_Physics_version_backup/Chapter12.ipynb @@ -0,0 +1,632 @@ +{ + "metadata": { + "name": "Chapter12", + "signature": "sha256:a11a7c1ba314b90500bccd87a2a9d038a55aefe70077541468dd4c2624276146" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 12:Nuclear Structure and Reactivity" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.1, Page 375" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#initiation of variable\n", + "Z=2;A=4;N=A-Z; # Given values\n", + "\n", + "#result\n", + "print\"The following method of representing atoms is followed throughout the chapter\\n\\t\\t x,y,z\\n where x=atomic number y=mass number z= Neutron Number S=symbol of the atom\\n\\n\"\n", + "print\"The helium can be reperesented as He-- \",Z,A,N;\n", + "\n", + "#part b\n", + "Z=50.0;N=66.0;A=Z+N; # Given values and standard formulae\n", + "print\"The Tin can be reperesented as Sn-- \",Z,A,N;\n", + "\n", + "\n", + "#part c\n", + "A=235;N=143;Z=A-N;\n", + "print\"The Uranium can be reperesented as U-- \",Z,A,N;" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The following method of representing atoms is followed throughout the chapter\n", + "\t\t x,y,z\n", + " where x=atomic number y=mass number z= Neutron Number S=symbol of the atom\n", + "\n", + "\n", + "The helium can be reperesented as He-- 2 4 2\n", + "The Tin can be reperesented as Sn-- 50.0 116.0 66.0\n", + "The Uranium can be reperesented as U-- 92 235 143\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.2, Page 377" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#initiation of variable\n", + "r0=1.2; #standard value.\n", + "A=12.0; \n", + "r= r0*A**(1.0/3);\n", + "\n", + "#result\n", + "print\"The value of mean radius for C in fm is\",round(r,3);\n", + "\n", + "#part2\n", + "A=70.0; #given value\n", + "r= r0*A**(1.0/3);\n", + "\n", + "#result\n", + "print\"The value of mean radius for C in fm is\",round(r,3);\n", + "\n", + "#part3\n", + "A=209;\n", + "r= r0*A**(1.0/3);\n", + "\n", + "#result\n", + "print\"The value of mean radius for C in fm is\",round(r,3);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of mean radius for C in fm is 2.747\n", + "The value of mean radius for C in fm is 4.946\n", + "The value of mean radius for C in fm is 7.121\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.3, Page 379" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#initiation of variable\n", + "from math import pi\n", + "m=1.67*10**-27; r0=1.2*10**-15; v=4*pi*(r0**3)/3.0 #standard values of mass radius and volume\n", + "\n", + "#calculation\n", + "p=m/v; #density \n", + "\n", + "#result\n", + "print\"Density of typical nucleus in kg/m3 is %.0e\" %p;\n", + "\n", + "#part 2\n", + "r0=0.01;v=4*pi*(r0**3)/3.0;p=2.0*10**17; #/hypothetical values\n", + "m1=p*v; \n", + "\n", + "#result\n", + "print\"The mass of the hypothetical nucleus would be in Kg %.0e\" %m1;\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Density of typical nucleus in kg/m3 is 2e+17\n", + "The mass of the hypothetical nucleus would be in Kg 8e+11\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.4, Page 380" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#initiation of variable\n", + "N=30.0;Z=26.0;A=56.0;Mn=1.008665;Mp=1.007825;m=55.934939;c2=931.5; #given values and constants for case-1\n", + "B=((N*Mn)+(Z*Mp)-(m))*c2; #binding energy(per nucleon)\n", + "\n", + "#result\n", + "print\"Binding energy per nucleon for 26,56Fe30 in MeV is\",round(B/A,3);\n", + "\n", + "#part 2\n", + "N=146.0;Z=92.0;A=238.0;Mn=1.008665;Mp=1.007825;m=238.050785;c2=931.5; #given values and constants for case-2\n", + "B=((N*Mn)+(Z*Mp)-(m))*c2; #binding energy(per nucleon)\n", + "\n", + "#result\n", + "print\"Binding energy per nucleon for 26,56Fe30 in MeV is\",round(B/A,3);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Binding nergy per nucleon for 26,56Fe30 in MeV is 8.79\n", + "Binding nergy per nucleon for 26,56Fe30 in MeV is 7.57\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.5, Page 382" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#initiation of variable\n", + "from math import exp\n", + "t12=2.7*24*3600; #converting days into seconds\n", + "w=0.693/t12; #lambda\n", + "\n", + "#result\n", + "print\"The decay constant in sec is %.2e\" %w; \n", + "\n", + "#partb\n", + "print\"The decay constant is equal to probability of decay in one second hence %.2e\" %w;\n", + "\n", + "#partc\n", + "m=10**-6;Na=6.023*10**23; M=198.0; #given values and constants\n", + "N=m*Na/M; #number of atoms in the sample \n", + "Ao=w*N; #activity\n", + "\n", + "#result\n", + "print\"The activity was found out to be in Ci is %.2e\" %Ao,\"=0.244Ci\"; \n", + "\n", + "#partd\n", + "t=7*24*3600.0; #given time\n", + "A=Ao*exp(-w*t); #activity\n", + "\n", + "#result\n", + "print\"The activity after one week was found out to be in decays/sec %.1e\" %A;" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The decay constant in sec is 2.97e-06\n", + "The decay constant is equal to probability of decay in one second hence 2.97e-06\n", + "The activity was found out to be in Ci is 9.04e+09 =0.244Ci\n", + "The activity after one week was found out to be in decays/sec 1.5e+09\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.6, Page 384" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#initiation of variable\n", + "t1=4.55*10**9;t2=7.04*10**8; #given values of time at 2 different instants\n", + "\n", + "#calculation\n", + "age=t1/t2;\n", + "r=2**age;\n", + "\n", + "#result\n", + "print \"The original rock hence contained\",round(r,3),\"Na atoms of 235U where Na is the Avagadro''s Number=6.023*10^23\";" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The original rock hence contained 88.222 Na atoms of 235U where Na is the Avagadro''s Number=6.023*10^23\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.7, Page 385" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#initiation of variable\n", + "m236Ra=226.025403;\n", + "m222Rn=222.017571;\n", + "m4He=4.002603;c2=931.5; #mass of various elements and c2=c^2\n", + "\n", + "#calculation\n", + "Q=(m236Ra-m222Rn-m4He)*c2;#Q of the reaction\n", + "A=226.0 \n", + "K=((A-4)/A)*Q; #kinetic energy\n", + "\n", + "#result\n", + "print\"The kinetic energy of the alpha particle in Mev is\",round(K,3);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The kinetic energy of the alpha particle in Mev is 4.785\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.8, Page 387" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#initiation of variable\n", + "m226Ra=226.025403; #mass of various elements\n", + "m212Pb=211.991871;\n", + "m14c=14.003242;\n", + "c2=931.5; #value of c^2\n", + "\n", + "#calculation\n", + "Q=(m226Ra-m212Pb-m14c)*c2; #Q of the reaction\n", + "\n", + "#result\n", + "print\"The value of Q for 14c emission in MeV is\",round(Q,3);\n", + "print\"The probability of 14c emission is 10^-9 times that of an alpha particle since the energy barrier for 14c emission is nearly 3 times higher and thicker.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of Q for 14c emission in MeV is 28.215\n", + "The probability of 14c emission is 10^-9 times that of an alpha particle since the energy barrier for 14c emission is nearly 3 times higher and thicker.\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.9, Page 389" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#initiation of variable\n", + "m23Ne=22.994465; #mass of various elements\n", + "m23Na=22.989768;\n", + "c2=931.5; #value of c^2\n", + "\n", + "#calculation\n", + "Q=(m23Ne-m23Na)*c2; #Q of the reaction\n", + "\n", + "#result\n", + "print \"Hence the maximum kinetic energy of the emitted electrons in MeV is\",round(Q,3);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Hence the maximum kinetic energy of the emitted electrons in MeV is 4.375\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.10, Page 390" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#initiation of variable\n", + "m40K=39.963999; #mass of various particles\n", + "m40Ca=39.962591;\n", + "c2=931.5; #value of c^2 in MeV\n", + "\n", + "#calculation\n", + "Qb1=(m40K-m40Ca)*c2; #Q value of the reaction\n", + "\n", + "#result\n", + "print\"The Q value for -VE beta emission in Mev in\",round(Qb1,3);\n", + "\n", + "#partb\n", + "m40K=39.963999; #mass of various particles\n", + "m40Ar=39.962384;\n", + "me=0.000549;\n", + "Qb2=(m40K-m40Ar-2*me)*c2; #Q value of the reaction\n", + "\n", + "#result\n", + "print\"The Q value for +VE beta emission in Mev in\",round(Qb2,3);\n", + "\n", + "#partc\n", + "m40K=39.963999;\n", + "m40Ar=39.962384;\n", + "\n", + "#calculation\n", + "Qec=(m40K-m40Ar)*c2;\n", + "\n", + "#result\n", + "print\"The Q value for +VE beta emission in Mev in\",round(Qec,3);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The Q value for -VE beta emission in Mev in 1.312\n", + "The Q value for +VE beta emission in Mev in 0.482\n", + "The Q value for +VE beta emission in Mev in 1.504\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.11, Page 392" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#initiation of variable\n", + "Mg=12.000000; #mass of the carbon atom in amu\n", + "c2=931.5; \n", + "Eg=4.43; #given energy of gamma ray \n", + "Mex=Mg+(Eg/c2); #mass in excited state\n", + "Me=0.000549; #mass of an electron\n", + "\n", + "#calculation\n", + "Q=(12.018613-Mex-2*Me)*c2; #Q of the particle\n", + "\n", + "#result\n", + "print\"The maximum value of kinetic energy is in MeV\",round(Q,3);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The maximum value of kinetic energy is in MeV 11.885\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.12, Page 393" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#initiation of variable\n", + "m238U=238.050786; #mass of various quantities\n", + "m206Pb=205.974455;\n", + "m4He=4.002603;\n", + "c2=931.5; #constants\n", + "Na=6.023*10**23; #avagadro's number\n", + "\n", + "#calculation\n", + "Q=(m238U-m206Pb-8*m4He)*c2; \n", + "t12=(4.5)*10**9*(3.16*10**7); #half life years to seconds conversion\n", + "w=0.693/t12; # lambeda\n", + "NoD=(Na/238)*w; #number of decays\n", + "E=NoD*Q*(1.6*10**-19)*10**6; #rate of liberation of energy,converting MeV to eV\n", + "\n", + "#result\n", + "print\"Rate of energy liberation in W is %.0e\" %E;" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Rate of energy liberation in W is 1e-07\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.13, Page 395" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#initiation of variable\n", + "from math import log\n", + "R=0.5;t12=4.5*10**9; #value of radius and half-life \n", + "t1=(t12/0.693)*log(1+(1/R)); #age of rock 1\n", + "R=1.0;\n", + "t2=(t12/0.693)*log(1+(1/R)); #age of rock-2\n", + "R=2.0\n", + "t3=(t12/0.693)*log(1+(1/R)); #age of rock 3\n", + "\n", + "#result\n", + "print\"The ages of rock samples in years respectively are %.1e\"%t1,\" %.1e\" %t2,\" %.1e\" %t3;" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The ages of rock samples in years respectively are 7.1e+09 4.5e+09 2.6e+09\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.14, Page 397" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#initiation of variable\n", + "from math import log\n", + "P=2.0*10**14; V=2.0*10**-14; R=8.314; T=295.0;Na=6.023*10**23; #various constants and given values\n", + "\n", + "#calculation\n", + "n=P*V/(R*T); #ideal gas law\n", + "N=Na*n;f=10**-12 #avagadaro's number and fraction of carbon molecules\n", + "t12=5730*3.16*(10**7); #half life\n", + "A=(0.693/t12)*N*f; #activity\n", + "D1w=A*7*24*60*60; #decays per second\n", + "\n", + "#result\n", + "print\"the no. of decays per second is %.2e\" %A\n", + "print\"The no of decays per week is \",round(D1w);\n", + " \n", + " \n", + "#partb\n", + "c1=1420.0; #concentration at instant 1\n", + "c2=D1w; #concentration at instant 2\n", + "t12y=5730; #half life\n", + "t=t12y*log(c2/c1)/0.693; #age of the sample\n", + "\n", + "#result\n", + "print\"Age of the sample in years is\",round(t,3);\n", + "print\"the answer in the book is wrong\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the no. of decays per second is 3.76e-03\n", + "The no of decays pers week is 2274.0\n", + "Age of the sample in years is 3892.57\n" + ] + } + ], + "prompt_number": 15 + } + ], + "metadata": {} + } + ] +}
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