Chapter 27:The Atomic Nucleus

Ex27.1:pg-1242

In [2]:
  import math  # Example 27_1
 
  
#What fraction of atomic mass of Uranium is due to its electrons
n=92        #Units in constant
mass=0.000549       #Units in u
tmass=235       #units in u
per=(n*mass)/tmass         #Units in fractions
print "The fraction of atomic mass of Uranium is due to its electrons is=",round(per,6)
The fraction of atomic mass of Uranium is due to its electrons is= 0.000215

Ex27.2:pg-1243

In [3]:
  import math  # Example 27_2
 
  
#To find the density of gold nucleus
r=6.97*10**-15             #Units in meters
a=197       #Units in u
v=(4/3.0)*math.pi*r**3       #Units in meter**3
m1=1.66*10**-27         #Units in Kg/u
mass=a*m1          #Units in Kg
p=mass/v      #Units in Kg/meter**3
print "The density of gold nucleus is p="
print p,"Kg/meter**3"
The density of gold nucleus is p=
2.30561808801e+17 Kg/meter**3

Ex27.3:pg-1243

In [4]:
  import math  # Example 27_3
 
  
#To calculate the energy required to change the mass of a system
c=3*10**8          #units in meters/sec
m=1.66*10**-27      #Units in g
e=m*c**2          #Units in J
e=e/(1.6*10**-19)*10**-6        #Units in MeV
print "The energy required to change the mass of a system is=",round(e,1)," MeV"
#In text book answer is printed wrong as e=931.5Mev the correct answer is933.7 MeV
The energy required to change the mass of a system is= 933.7  MeV

Ex27.4:pg-1243

In [5]:
  import math  # Example 27_4
 
  
#To compute the binding energy of deuterium
m1=2.014102          #Units in u
m2=0.000549      #Units in u
total=m1-m2        #Unts in u
m3=1.007276      #Units in u
m4=1.008665          #Units in u
suma=m3+m4          #Units in u
massdefect=suma-total      #units in u
e1=931.5    #Units in MeV
m5=1             #Units iin eV
e=massdefect*e1/m5         #Units in MeV
print "The binding energy of deuterium is E=",round(e,2)," MeV "
The binding energy of deuterium is E= 2.22  MeV 

Ex27.5:pg-1245

In [6]:
  import math  # Example 27_5
 
  
#To find how much of the orignal I will still present
d1=20.0       #Units in mg
d2=d1/2          #Units in mg
d3=d2/2          #Units in mg
d4=d3/2          #Units in mg
d5=d4/2          #Units in mg
d6=d5/2         #Units in mg
d7=d6/2             #Units in mg
print "After 48 days only ",round(d7,3)," mg will remain"
After 48 days only  0.313  mg will remain

Ex27.6:pg-1246

In [7]:
  import math  # Example 27_6
 
  
#To find how many radium atoms in the vial undergo decay
t1=5.1*10**10     #Units in sec
lamda=0.693/t1       #Units in sec**-1
n1=6.02*10**26          #Units in atoms/Kmol
n2=226   #Units in Kg/Kmol
m1=0.001        #Units in Kg
N=n1*m1/n2           #Units in number of atoms
deltat=1         #Units in sec
deltan=-lamda*N*deltat         #Units in Number
print "The number of dis integrations per sec="
print deltan
The number of dis integrations per sec=
-36195210827.7

Ex27.7:pg-1247

In [8]:
  import math  # Example 27_7
 
  
#To find what fraction of uranium remains undecayed today
t1=4.5*10**9     #Units in Years
lamda=0.693/t1       #Units in years**-1
t=4*10.0**9       #Units in Years
n_no=math.e**(-lamda*t)           #Units in Fractions
print "The fraction of uranium remains undecayed today is=",round(n_no,2)
The fraction of uranium remains undecayed today is= 0.54

Ex27.8:pg-1248

In [9]:
  import math  # Example 27_8
 
  
#To calculate the decay constant and half life of substance
n_no=0.9       #Units in constant
t=12     #Units in h
lamda=math.log(1/n_no)/t           #Units in h**-1
t1=round(0.693/lamda)           #Units in h
print "The decay constant is lamda=",round(lamda,7)," h**-1\n The Half life is t0.5=",round(t1)," h"
The decay constant is lamda= 0.00878  h**-1
 The Half life is t0.5= 79.0  h

Ex27.9:pg-1249

In [10]:
  import math  # Example 27_9
 
  
#To fnd the approximate energy of the emitted alpha particle
m1=222.01753         #Units in u
m3=4.00263           #Units in u
m2=218.00893          #Units in u
massloss=m1-(m2+m3)         #Units in u
e1=931.5           #Units in MeV
e=e1/massloss*10**-5       #Units in MeV
print "The approximate energy of the emitted alpha particle is E=",round(e,2)," MeV"
  #In textbook answer s printed wrong as E=5.56eV the correct answer is E=1.56 eV
The approximate energy of the emitted alpha particle is E= 1.56  MeV

Ex27.10:pg-1250

In [11]:
  import math  # Example 27_10
  
#To find the fraction of original amount still existence in earth
t1=1.41*10**10          #Units in Years
lamda=0.693/t1         #Units in year**-1
t=5*10**9        #Units in years
n_no=math.e**-(lamda*t)         #Units in constant
n_no=n_no*100      #Units in percentage
print "The percentage of original amount still remainng is N/No=",round(n_no,3)," Percent"
The percentage of original amount still remainng is N/No= 78.212  Percent

Ex27.11:pg-1251

In [12]:
  import math  # Example 27_11
 
  
#To find the activity of sr
t1=28      #units in Years
t1=t1*86400*365      #Units in sec
acti=6.022*10**26        #Units of Bq
m1=90       #Units in Kg
m2=0.001      #Units in Kg
N=(m2/m1)*acti      #Units in constant
activity=0.693*N/t1      #Units in Bq
print "The activity of sr="
print activity
print "Bq"
The activity of sr=
5.25130010147e+12
Bq

Ex27.12:pg-1252

In [13]:
  import math  # Example 27_12
 
  
#To estimate the age of the axe handle
n_no=0.034
t1=5730    #Units in Years
t=-(math.log(n_no)*t1)/0.693   #Units in Years
print "The age of the axe handle is t=",round(t)," years"
  #In textbook answer is printed wrong as t=28000 years correct answer is t=27958 years 
The age of the axe handle is t= 27959.0  years

Ex27.13:pg-1253

In [14]:
  import math  # Example 27_13
  
  
#To find the energy released in the reaction
m1=141.91635      #Units in u
m2=89.91972      #Units in u
m3=4.03466       #Units in u
n2=36           #Units in Constant
n1=56           #Units in Constant
n4=92         #units in constant
m5=236.04564            #Units in u
loss=m5-(m1+m2+m3)+n4-(n1+n2)                  #Units in u
e1=931.5          #units n MeV
energy=round(e1*loss)          #units in MeV
print "The energy released in the reaction E=",round(energy)," MeV"
The energy released in the reaction E= 163.0  MeV