# 18: Radioactive decay¶

## Example number 18.1, Page number 347¶

In [1]:
#importing modules
import math
from __future__ import division

#Variable declaration
N0=1;   #assume

#Calculation
f=(N0/2)/N0;   #fraction after t1/2
f1=(N0/4)/N0;   #fraction after 2 half lives
f2=(N0/(2**5))/N0;   #fraction after 5 half lives
f3=(N0/(2**10))/N0;   #fraction after 10 half lives

#Result
print "fraction after 2 half lives is",f1
print "fraction after 5 half lives is",f2
print "fraction after 10 half lives is",f3

fraction after 2 half lives is 0.25
fraction after 5 half lives is 0.03125
fraction after 10 half lives is 0.0009765625


## Example number 18.2, Page number 348¶

In [10]:
#importing modules
import math
from __future__ import division

#Variable declaration
thalf=2.7*24*60*60;   #half life(s)
m=1*10**-6;   #mass(gm)
Na=6.02*10**23;   #avagadro number(atoms/mol)
M=198;    #molar mass(g/mol)
t=8*24*60*60;

#Calculation
lamda=0.693/thalf;    #decay constant(per sec)
N=m*Na/M;    #number of nuclei(atoms)
A0=lamda*N;    #activity(disintegrations per sec)
A=A0*math.exp(-lamda*t);    #activity for 8 days(decays per sec)

#Result
print "decay constant is",round(lamda*10**6,2),"*10**-6 per sec"
print "activity is",round(A0/10**9,2),"*10**9 disintegrations per sec"
print "activity for 8 days is",round(A/10**9,2),"*10**9 decays per sec"

decay constant is 2.97 *10**-6 per sec
activity is 9.03 *10**9 disintegrations per sec
activity for 8 days is 1.16 *10**9 decays per sec


## Example number 18.3, Page number 348¶

In [23]:
#importing modules
import math
from __future__ import division

#Variable declaration
thalf=5570*365*24*60*60;   #half life(s)
dNbydt=3.7*10**10*2*10**-3;   #number of decays per sec
m=14;
Na=6.02*10**23;   #avagadro number(atoms/mol)

#Calculation
lamda=0.693/thalf;    #decay constant(per sec)
N=dNbydt/lamda;    #number of atoms
mN=m*N/Na;   #mass of 2mCi(g)

#Result
print "mass of 2mCi is",round(mN*10**4,2),"*10**-4 g"

mass of 2mCi is 4.36 *10**-4 g


## Example number 18.5, Page number 353¶

In [12]:
#importing modules
import math
from __future__ import division

#Variable declaration
thalf=1.25*10**9;   #half life(yr)
r=10.2;   #ratio of number of atoms

#Calculation
a=1+r;
lamda=0.693/thalf;    #decay constant(per yr)
t=math.log(a)/lamda;   #time(yr)

#Result
print "the rock is",round(t/10**9,2),"*10**9 yrs old"

the rock is 4.36 *10**9 yrs old


## Example number 18.6, Page number 356¶

In [14]:
#importing modules
import math
from __future__ import division

#Variable declaration
mU=232.037131;    #atomic mass of U(u)
mHe=4.002603;   #atomic mass of He(u)
E=931.5;    #energy(MeV)
KE=5.32;   #kinetic energy of alpha particle(MeV)

#Calculation
mTh=mU-mHe-(KE/E);   #atomic mass of Th(u)

#Result
print "atomic mass of Th is",round(mTh,5),"u"

atomic mass of Th is 228.02882 u


## Example number 18.7, Page number 359¶

In [16]:
#importing modules
import math
from __future__ import division

#Variable declaration
E=931.5;    #energy(MeV)
mX=11.011433;   #mass of 11C(u)
mXdash=11.009305;  #mass of 11B(u)
me=0.511;

#Calculation
Q=(E*(mX-mXdash))-(2*me);    #Q value for decay(MeV)

#Result
print "maximum energy is",round(Q,2),"MeV.minimum energy is zero"

maximum energy is 0.96 MeV.minimum energy is zero


## Example number 18.8, Page number 359¶

In [18]:
#importing modules
import math
from __future__ import division

#Variable declaration
mK=39.963999;    #mass of K(u)
mAr=39.962384;   #mass of Ar(u)
E=931.5;    #energy(MeV)

#Calculation
Q=(mK-mAr)*E;   #kinetic energy of neutrino(MeV)

#Result
print "kinetic energy of neutrino is",round(Q,3),"MeV"

kinetic energy of neutrino is 1.504 MeV


## Example number 18.9, Page number 360¶

In [20]:
#importing modules
import math
from __future__ import division

#Variable declaration
mN=12.018613;    #mass of N(u)
mC=12;   #mass of C(u)
me=0.000549;   #mass of me(u)
E=931.5;    #energy(MeV)
Egamma=4.43;   #energy of emitted gamma ray(MeV)

#Calculation
Q=(mN-mC-(2*me))*E;   #Q value(MeV)
Emax=Q-Egamma;   #maximum kinetic energy(MeV)

#Result
print "maximum kinetic energy is",round(Emax,2),"MeV"

maximum kinetic energy is 11.89 MeV