Chapter 12:Nuclear Transformations

Example no:12.2,Page no:425

In [20]:
#Variable declaration
Thalf= 3.82  #half-life in days, d
Lambda= 0.693/Thalf  #decay constant
p= 0.6  # 60.0 percent of sample

#Calculation
import numpy
import math
No= numpy.poly([1])  #Number of undecayed nuclei, at time t=0
N= (1.0-p)*No  #Number of undecayed nuclei, at time t
t= (1.0/Lambda)*(np.log((N/No)))  #decay time in days, d
t= t*(-1.0)  

#Result
print"The decay time for Radon is: ",round(t[0],2),"d"

The decay time for Radon is:  5.05 d

Example no:12.3,Page no:427

In [2]:
#Variable declaration 
Thalf= 3.82  #half-life in days, d
Lambda= 0.693/(Thalf*86400.0)  #decay constant, s**(-1)
Wradon= 1.00  #weight of sample, mg
MRadon= 222.0  #atomic mass of sample, u

#Calculation
N= Wradon*(10**(-6))/(MRadon*(1.66*(10**(-27))))  #number of atoms
R= Lambda*N  #decays/sec
R_tbq=round(R/10**12,2)       #in TBq
R_Ci=R_tbq*27.15              #Ci  Calories


#Result
print"The activity of the sample is: %.2g"%R,"decays/sec=",R_tbq,"TBq=",round(R_Ci),"Ci"
 

The activity of the sample is: 5.7e+12 decays/sec= 5.7 TBq= 155.0 Ci

Example no:12.4,Page no:427

In [13]:
#Variable declaration
Ro= 155  #initial activity, Ci
Lambda= 2.11*(10**(-6))  #decay constant, s**(-1)
t= 7  #days

#Calculation
import math
t= t*86400  #converting to s
R= Ro*((math.exp(-(Lambda*t))))  #final activity, Ci

#Result
print"The activity after one week is: ",round(R),"Ci"

The activity after one week is:  43.0 Ci

Example no:12.5,Page no:428

In [12]:
#Variable declaration 
R= 13.0  #present activity,
Ro= 16.0  #activity of live wood
Thalf= 5760.0  #half life of radiocarbon, y

#Calculation
Lambda= 0.693/(Thalf)  #decay constant, y**(-1)
t= (1.0/Lambda)*(math.log(Ro/R))  #age of sample, y

#Result
print"The age of the wooden sample is:  %.2g"%t,"y"

The age of the wooden sample is:  1.7e+03 y

Example no:12.6,Page no:432

In [5]:
#Variable declaration 
Thalf1= 2.5*(10**5)  #half-life of U-234, y
AtomicRatio= 1.8*(10**4)  #atomic ratio of u-238 and U-234 in the sample

#Calculation
Thalf2= AtomicRatio*Thalf1  #using Eqn12.9

#Result
print"The half-life of Uranium-238  is:  %.2g"%Thalf2,"years" 

The half-life of Uranium-238  is:  4.5e+09 years

Example no:12.7,Page no:433

In [6]:
#Variable declaration 
Npolonium= 84  #atomic number of polonium
Nalpha= 2  #atomic number of alpha particle
Z= Npolonium-Nalpha  #atomic number of daughter nuclide
Mpolonium= 209.9829  #mass number of polonium, u
Malpha= 4.0026  #mass number of alpha particle, u

#Calculation
A= Mpolonium-Malpha  #mass number of daughter nuclide
# The daughter nuclide has atomic number:    
#   82.  

# and mass number:    
#    205.9803.  
 
Ealpha= 5.3  #energy of alpha particle, MeV
Q= Mpolonium*Ealpha/A  #disintegration energy, MeV
Mq= Q/931  #mass equivalent for Q, u
Mnuclide= Mpolonium-Malpha-Mq  #u

#Result
print"(a)The daughter nuclide has atomic number: ",Z
print"and mass number: ",round(A)
print"(b).The atomic mass of the daughter nuclide is: ",round(Mnuclide,4)

(a)The daughter nuclide has atomic number:  82
and mass number:  206.0
(b).The atomic mass of the daughter nuclide is:  205.9745

Example no:12.8,Page no:444

In [11]:
#Variable declaration 
CrossSection= 2*(10**4)  # capture cross section of Cadmium-113
CrossSection= CrossSection*(10**(-28))  # converting to m**2
Mcadmium= 112.0  #mean atomic mass of cadmium, u
density= 8.64*(10**3)  #density of cadmium sheet used, kg/m**3

#Calculation
#Part (a)
import math
t= 0.1  #hickness of sheet used, mm
t= t*10**(-3)  #converting to m
p= 12.0  #percent of Cadmium-113 in natural cadmium
u= 1.66*(10**(-27))  #atomic mass unit, kg
n= (p/100.0)*density/(Mcadmium*u)  #number of atoms, atoms/m**3
Fabsorbed= 1.0- (math.exp((-n)*(CrossSection)*(t)))  #absorbed fraction
#Part (b)
t2= (-(math.log(0.01)))/(n*CrossSection)  #required thickness, m
t2= t2*10**(3.0)  #converting to mm


#Result
print"(a).The fraction of incident beam absorbed is: ",round(Fabsorbed,2)
print"(b).The thickness required to absorb 99 percent of incident thermal neutrons is: ",round(t2,2),"mm"

(a).The fraction of incident beam absorbed is:  0.67
(b).The thickness required to absorb 99 percent of incident thermal neutrons is:  0.41 mm

Example no:12.9,Page no:445

In [8]:
#Variable declaration
CrossSection= 2.0*(10**4)  # capture cross section of Cadmium-113, b
CrossSection= CrossSection*(10**(-28))  # converting to m**2

#Calculation
n= (12.0/100.0)*(8.64*10**3)/(112.0*(1.66*10**(-27)))  #number of atoms, atoms/m**3
Lambda= 1.0/(n*CrossSection)  #mean free path, m
Lambda= Lambda*10**3  #converting to, mm

#Result
print"The mean free path is: ",round(Lambda,4),"mm"

The mean free path is:  0.0897 mm

Example no:12.10,Page no:446

In [9]:
#Variable declaration 
Thalf= 2.69  #half life of gold,d
Lambda= 0.693/(Thalf*86400.0)  #decay constant, s**(-1)
R= 200.0  #required activity, mCi
R= R*10**(-6)  #converting to Ci

#Calculation
dN= R/(Lambda/(3.70*10**(10)))  #atoms
Wgold= 10.0  #mass of foil, mg
u= 1.66*(10**(-27))  #atomic mass unit, kg
Mgold= 197.0  # u
n2= Wgold*10.0**(-6)/(Mgold*u)  #total no. of atoms
phi= 2.0*10**(16)  #flux, neutrons/m**2
CrossSection= 99.0*10**(-28)  #m**2
dT= dN/(phi*n2*CrossSection)   #s
dT_m=divmod(dT,60)
#Result
print"The irradiation period required is: ",round(dT,1),"s=",round(dT_m[0],1),"min",round(dT_m[1],1),"seconds(Approx)"
 

The irradiation period required is:  409.9 s= 6.0 min 49.9 seconds(Approx)

Example no:12.11,Page no:450

In [10]:
#Variable declaration 
m1= 14.00307  #u
m2= 4.00260  #u
m3= 1.00783  #u
m4= 16.99913  #u

#Calculation
k= m1+m2-m3-m4  # difference in total mass of reactants and products, u
Q= k*931.5  #energy exchanged, MeV
KEcm= -Q  #minimum KE needed in centre of mass system, MeV
KElab= KEcm*(m2+m1)/m1  #minimum KE in laboratory system

#Result
print"The minimum KE required by the alpha particle is: ",round(KElab,3),"MeV(APPROX)"

The minimum KE required by the alpha particle is:  1.545 MeV(APPROX)