{
"cells": [
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"source": [
"# Chapter 27:The Atomic Nucleus"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex27.1:pg-1242"
]
},
{
"cell_type": "code",
"execution_count": 1,
"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": [
" #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": [
" #Example 27_2\n",
"import math \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": [
" #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": [
" #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": [
" #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": 8,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The number of dis integrations per sec=\n",
"-36195210827.7\n"
]
}
],
"source": [
" #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": 10,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The fraction of uranium remains undecayed today is= 0.54\n"
]
}
],
"source": [
" #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": 11,
"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": [
" #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": 12,
"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": [
" #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": 14,
"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": [
" #Example 27_10\n",
"import math \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": 15,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The activity of sr=\n",
"5.25130010147e+12\n",
"Bq\n"
]
}
],
"source": [
" #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": 16,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The age of the axe handle is t= 27959.0 years\n"
]
}
],
"source": [
" #Example 27_12\n",
"import math \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": 17,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The energy released in the reaction E= 163.0 MeV\n"
]
}
],
"source": [
" #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"
]
}
],
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