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+{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 27:The Atomic Nucleus"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Ex27.1:pg-1242"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 2,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The fraction of atomic mass of Uranium is due to its electrons is= 0.000215\n"
+ ]
+ }
+ ],
+ "source": [
+ " import math # Example 27_1\n",
+ " \n",
+ " \n",
+ "#What fraction of atomic mass of Uranium is due to its electrons\n",
+ "n=92 #Units in constant\n",
+ "mass=0.000549 #Units in u\n",
+ "tmass=235 #units in u\n",
+ "per=(n*mass)/tmass #Units in fractions\n",
+ "print \"The fraction of atomic mass of Uranium is due to its electrons is=\",round(per,6)\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Ex27.2:pg-1243"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 3,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The density of gold nucleus is p=\n",
+ "2.30561808801e+17 Kg/meter**3\n"
+ ]
+ }
+ ],
+ "source": [
+ " import math # Example 27_2\n",
+ " \n",
+ " \n",
+ "#To find the density of gold nucleus\n",
+ "r=6.97*10**-15 #Units in meters\n",
+ "a=197 #Units in u\n",
+ "v=(4/3.0)*math.pi*r**3 #Units in meter**3\n",
+ "m1=1.66*10**-27 #Units in Kg/u\n",
+ "mass=a*m1 #Units in Kg\n",
+ "p=mass/v #Units in Kg/meter**3\n",
+ "print \"The density of gold nucleus is p=\"\n",
+ "print p,\"Kg/meter**3\"\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Ex27.3:pg-1243"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The energy required to change the mass of a system is= 933.7 MeV\n"
+ ]
+ }
+ ],
+ "source": [
+ " import math # Example 27_3\n",
+ " \n",
+ " \n",
+ "#To calculate the energy required to change the mass of a system\n",
+ "c=3*10**8 #units in meters/sec\n",
+ "m=1.66*10**-27 #Units in g\n",
+ "e=m*c**2 #Units in J\n",
+ "e=e/(1.6*10**-19)*10**-6 #Units in MeV\n",
+ "print \"The energy required to change the mass of a system is=\",round(e,1),\" MeV\"\n",
+ "#In text book answer is printed wrong as e=931.5Mev the correct answer is933.7 MeV\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Ex27.4:pg-1243"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 5,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The binding energy of deuterium is E= 2.22 MeV \n"
+ ]
+ }
+ ],
+ "source": [
+ " import math # Example 27_4\n",
+ " \n",
+ " \n",
+ "#To compute the binding energy of deuterium\n",
+ "m1=2.014102 #Units in u\n",
+ "m2=0.000549 #Units in u\n",
+ "total=m1-m2 #Unts in u\n",
+ "m3=1.007276 #Units in u\n",
+ "m4=1.008665 #Units in u\n",
+ "suma=m3+m4 #Units in u\n",
+ "massdefect=suma-total #units in u\n",
+ "e1=931.5 #Units in MeV\n",
+ "m5=1 #Units iin eV\n",
+ "e=massdefect*e1/m5 #Units in MeV\n",
+ "print \"The binding energy of deuterium is E=\",round(e,2),\" MeV \"\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Ex27.5:pg-1245"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 6,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "After 48 days only 0.313 mg will remain\n"
+ ]
+ }
+ ],
+ "source": [
+ " import math # Example 27_5\n",
+ " \n",
+ " \n",
+ "#To find how much of the orignal I will still present\n",
+ "d1=20.0 #Units in mg\n",
+ "d2=d1/2 #Units in mg\n",
+ "d3=d2/2 #Units in mg\n",
+ "d4=d3/2 #Units in mg\n",
+ "d5=d4/2 #Units in mg\n",
+ "d6=d5/2 #Units in mg\n",
+ "d7=d6/2 #Units in mg\n",
+ "print \"After 48 days only \",round(d7,3),\" mg will remain\"\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Ex27.6:pg-1246"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 7,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The number of dis integrations per sec=\n",
+ "-36195210827.7\n"
+ ]
+ }
+ ],
+ "source": [
+ " import math # Example 27_6\n",
+ " \n",
+ " \n",
+ "#To find how many radium atoms in the vial undergo decay\n",
+ "t1=5.1*10**10 #Units in sec\n",
+ "lamda=0.693/t1 #Units in sec**-1\n",
+ "n1=6.02*10**26 #Units in atoms/Kmol\n",
+ "n2=226 #Units in Kg/Kmol\n",
+ "m1=0.001 #Units in Kg\n",
+ "N=n1*m1/n2 #Units in number of atoms\n",
+ "deltat=1 #Units in sec\n",
+ "deltan=-lamda*N*deltat #Units in Number\n",
+ "print \"The number of dis integrations per sec=\"\n",
+ "print deltan\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Ex27.7:pg-1247"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 8,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The fraction of uranium remains undecayed today is= 0.54\n"
+ ]
+ }
+ ],
+ "source": [
+ " import math # Example 27_7\n",
+ " \n",
+ " \n",
+ "#To find what fraction of uranium remains undecayed today\n",
+ "t1=4.5*10**9 #Units in Years\n",
+ "lamda=0.693/t1 #Units in years**-1\n",
+ "t=4*10.0**9 #Units in Years\n",
+ "n_no=math.e**(-lamda*t) #Units in Fractions\n",
+ "print \"The fraction of uranium remains undecayed today is=\",round(n_no,2)\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Ex27.8:pg-1248"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 9,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The decay constant is lamda= 0.00878 h**-1\n",
+ " The Half life is t0.5= 79.0 h\n"
+ ]
+ }
+ ],
+ "source": [
+ " import math # Example 27_8\n",
+ " \n",
+ " \n",
+ "#To calculate the decay constant and half life of substance\n",
+ "n_no=0.9 #Units in constant\n",
+ "t=12 #Units in h\n",
+ "lamda=math.log(1/n_no)/t #Units in h**-1\n",
+ "t1=round(0.693/lamda) #Units in h\n",
+ "print \"The decay constant is lamda=\",round(lamda,7),\" h**-1\\n The Half life is t0.5=\",round(t1),\" h\"\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Ex27.9:pg-1249"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 10,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The approximate energy of the emitted alpha particle is E= 1.56 MeV\n"
+ ]
+ }
+ ],
+ "source": [
+ " import math # Example 27_9\n",
+ " \n",
+ " \n",
+ "#To fnd the approximate energy of the emitted alpha particle\n",
+ "m1=222.01753 #Units in u\n",
+ "m3=4.00263 #Units in u\n",
+ "m2=218.00893 #Units in u\n",
+ "massloss=m1-(m2+m3) #Units in u\n",
+ "e1=931.5 #Units in MeV\n",
+ "e=e1/massloss*10**-5 #Units in MeV\n",
+ "print \"The approximate energy of the emitted alpha particle is E=\",round(e,2),\" MeV\"\n",
+ " #In textbook answer s printed wrong as E=5.56eV the correct answer is E=1.56 eV\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Ex27.10:pg-1250"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 11,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The percentage of original amount still remainng is N/No= 78.212 Percent\n"
+ ]
+ }
+ ],
+ "source": [
+ " import math # Example 27_10\n",
+ " \n",
+ "#To find the fraction of original amount still existence in earth\n",
+ "t1=1.41*10**10 #Units in Years\n",
+ "lamda=0.693/t1 #Units in year**-1\n",
+ "t=5*10**9 #Units in years\n",
+ "n_no=math.e**-(lamda*t) #Units in constant\n",
+ "n_no=n_no*100 #Units in percentage\n",
+ "print \"The percentage of original amount still remainng is N/No=\",round(n_no,3),\" Percent\"\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Ex27.11:pg-1251"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 12,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The activity of sr=\n",
+ "5.25130010147e+12\n",
+ "Bq\n"
+ ]
+ }
+ ],
+ "source": [
+ " import math # Example 27_11\n",
+ " \n",
+ " \n",
+ "#To find the activity of sr\n",
+ "t1=28 #units in Years\n",
+ "t1=t1*86400*365 #Units in sec\n",
+ "acti=6.022*10**26 #Units of Bq\n",
+ "m1=90 #Units in Kg\n",
+ "m2=0.001 #Units in Kg\n",
+ "N=(m2/m1)*acti #Units in constant\n",
+ "activity=0.693*N/t1 #Units in Bq\n",
+ "print \"The activity of sr=\"\n",
+ "print activity\n",
+ "print \"Bq\"\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Ex27.12:pg-1252"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 13,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The age of the axe handle is t= 27959.0 years\n"
+ ]
+ }
+ ],
+ "source": [
+ " import math # Example 27_12\n",
+ " \n",
+ " \n",
+ "#To estimate the age of the axe handle\n",
+ "n_no=0.034\n",
+ "t1=5730 #Units in Years\n",
+ "t=-(math.log(n_no)*t1)/0.693 #Units in Years\n",
+ "print \"The age of the axe handle is t=\",round(t),\" years\"\n",
+ " #In textbook answer is printed wrong as t=28000 years correct answer is t=27958 years \n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Ex27.13:pg-1253"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 14,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The energy released in the reaction E= 163.0 MeV\n"
+ ]
+ }
+ ],
+ "source": [
+ " import math # Example 27_13\n",
+ " \n",
+ " \n",
+ "#To find the energy released in the reaction\n",
+ "m1=141.91635 #Units in u\n",
+ "m2=89.91972 #Units in u\n",
+ "m3=4.03466 #Units in u\n",
+ "n2=36 #Units in Constant\n",
+ "n1=56 #Units in Constant\n",
+ "n4=92 #units in constant\n",
+ "m5=236.04564 #Units in u\n",
+ "loss=m5-(m1+m2+m3)+n4-(n1+n2) #Units in u\n",
+ "e1=931.5 #units n MeV\n",
+ "energy=round(e1*loss) #units in MeV\n",
+ "print \"The energy released in the reaction E=\",round(energy),\" MeV\"\n"
+ ]
+ }
+ ],
+ "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
+}
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