In [1]:

```
#cal of value of avagadro constant
#intiation of all variables
# Chapter 4
print"Example 4.1, Page:55 \n \n"
#Given:
t1=1600;# in year
a=11.6*10**17;# atoms
# Solution:
k=0.693/t1;# year^-1
L=(a*226)/k;# atomic mass of Radon is 226
print"The value of avagadro constant in 10^23 atoms per mole=",round(L/10**23,2)
```

In [2]:

```
#cal of specific activity
#intiation of all variables
# Chapter 4
print"Example 4.2, Page:55 \n \n"
# Given:
t1=1.3*10**9;# in years
w=0.0119;# wt %
# Solution:
N=(w*6.022*10**23)/(40*100);
k=(0.693*60)/(t1*3.16*10**7);
sa=N*k;# specific activity
print"The specific activity in dis min^-1 g^-1=",round(sa)
```

In [3]:

```
#cal of no. of atoms,mass of Na,P,Ra
#intiation of all variables
# Chapter 4
print"Example 4.3, Page:55 \n \n"
# Given:
L=6.022*10**23;
# Solution:
# 1 mCi= 3.7*10^7 dis/s
k1=0.693/(15*3600);
N1=3.7*10**7/k1;
m1=(24*N1*10**10)/L;
print" The no. of atoms of Na(24) are =",round(N1,2)
print" The mass of Na(24) is * 10^-10 g",round(m1,2)
k2=0.693/(14.3*24*3600);
N2=3.7*10**7/k2;
m2=(32*N2*10**9)/L;
print"The no. of atoms of P(32) are =",round(N2,2)
print"The mass of P(32) is * 10^-9 g",round(m2,2)
k3=0.693/(1600*3.16*10**7);
N3=3.7*10**7/k3;
m3=(226*N3*10**3)/L;
print"The no. of atoms of Ra(226) are =",round(N3,2)
print"The mass of Ra(226) is * 10^-3 g",round(m3,2)
```

In [4]:

```
#cal of activity
#intiation of all variables
# Chapter 4
print"Example 4.4, Page:56 \n \n"
# Given:
t1=12.3;# in yrs
L=6.022*10**23;
# Solution:
k=.693/(t1*3.16*10**7);# in s^-1
A=(2*L)/(2.24*10**4);# no. of atoms
a1=A*k;# dis per s
a=a1/(3.7*10**10);# activity in Ci/cm^3
print"The activity in Ci/cm^3 =",round(a,2)
```

In [5]:

```
#cal of ratio of C14/C12 in atmosphere
#intiation of all variables
# Chapter 4
print"Example 4.5, Page:56 \n \n"
# Given:
t=5736;# in years
Nk=16.1;# dis/min
L=6.022*10**23;
# Solution:
k=(0.693*60)/(t*3.16*10**7);
N1=Nk/k;# atoms per g for C14
N2=L/12;#
r=(N1*10**12)/N2;# ratio of C14/C12 in atmosphere
print"The ratio of C14/C12 in atmosphere in 10^-12 is =",round(r,2)
```

In [6]:

```
#cal of Number of alpha decays and number of beta decays
#intiation of all variables
# Chapter 4
print"Example 4.7, Page:57 \n \n"
# Given:
dA = 206-238;
dA_Beta=0;
dA_Alpha = -4;
dZ_Alpha = -2;
dZ_Beta = 1;
nBeta=0; #random initialisation
dZ = 82 -92;
# Solution:
nAlpha = (dA- (dA_Beta* nBeta))/dA_Alpha;
nBeta = (dZ- (dZ_Alpha * nAlpha))/dZ_Beta;
print"Number of alpha decays = and number of beta decays = ",nAlpha,nBeta
```

In [7]:

```
#cal of energy
#intiation of all variables
# Chapter 4
print"Example 4.8, Page:57 \n \n"
# Given:
E1=0.059;
E2=2.5;
E3=1.33;
Ei=0;
Ef=0;
# Solution:
# delta E for 1,2 & 3 photon
dE1=E1-Ei;
dE2=E2-E3;
dE3=E3-Ef;
# delta I for 1,2 & 3 photon
dI1=2-5;
dI2=4-2;
dI3=2-0;
# EL/ML for 1,2 & 3 photon
ELML1=3+1+1;
ELML2=2+1+1;
ELML3=2+1+1;
print"\n For first photon, dE1= MeV, dI1=, since EL/ML1= & (L+PI+PF) is odd, M3",dE1,dI1,ELML1
print"\n For second photon, dE2= MeV, dI2=,since EL/ML2= & (L+PI+PF) is even, E2",dE2,dI2,ELML2
print"\n For third photon, dE3= MeV, d3I=, since EL/ML3= & (L+PI+PF) is even, E2",dE3,dI3,ELML3
```

In [8]:

```
#cal of energy
#intiation of all variables
# Chapter 4
print"Example 4.9, Page:58 \n \n"
# Given:
E=2.5; # in MeV
# Solution:
# 1 Mev/atom=96.32GJ/mole
E1=E*96.32# GJ/mole
E2=0.1*E1;# for 0.1 mole
print"The energy that would be released for 0.1 mole of Co will be =(GJ)",E2
```

In [9]:

```
#cal of total energy dissipated
#intiation of all variables
# Chapter 4
print"Example 4.10, Page: \n \n"
# Given
E=2.5;# in MeV
# Solution:
k=0.693/(5.27*3.16*10**7);# decay constant
A=k*0.1*6.022*10**23;# atoms/s
A1=3.6*10**3*A;# atoms /hr
E1=A1*E*1.6*10**-13*10**-3;#Energy in KJ/hr
print"The total energy dissipate per hour is = (KJ)",E1
```

In [11]:

```
#cal of net mass loss
#intiation of all variables
# Chapter 4
print"Example 4.11, Page:59 \n \n"
# Given:
Ma=4.;#mass of alpha particle
Mr=228.;# mass of Th
Ea=4.; #in MeV
# Solution:
Er=(Ma/Mr)*Ea;# energy of recoil
Et=Ea+Er;# total energy of transition
dM=Et/931.;# net mass loss in u
print"The net mass loss is = u",round(dM,4)
```

In [12]:

```
#cal of value of avagadro constant
#intiation of all variables
# Chapter 4
print"Example 4.12, Page:59 \n \n"
# Given:
Ma=4.;#mass of alpha particle
Mr1=222.;# mass of
Mr2=208.;
Ea1=4.863;
Ea2=6.082;
# Solution:
Er1=(Ma/Mr1)*Ea1;
Et1=Ea1+Er1;
print"For Ra emitting alpha"
print"\tEnergy of recoil is (MeV)",round(Er1,4)
print"\tTotal transition energy is (MeV)",round(Et1,4)
Er2=(Ma/Mr2)*Ea2;
Et2=Ea2+Er2;
print"For Bi emitting alpha"
print"\tEnergy of recoil is (MeV)",round(Er2,4)
print"\tTotal transition energy is (MeV)",round(Et2,4)
```

In [13]:

```
#cal of Kinetic Energy and velocity
#intiation of all variables
# Chapter 4
print"Example 4.13, Page:60 \n \n"
# Given:
import math
dm=0.006332;# in u
ma=4.;
mCm=244.;
# Solution:
E=dm*931.;# in MeV
KE=E*(ma/mCm); # in MeV
v=math.sqrt((2.*KE*1.6*10.**-13.)/(240.*1.6605*10.**-27.));
print"The Kinetic Energy and velocity are MeV and 10^5 m/s respectively",round(KE,4),round(v/10**5,2)
```

In [14]:

```
#cal of rate of energy emission and Penfold empirical equation
#intiation of all variables
# Chapter 4
print"Example 4.14, Page:60 \n \n"
import math
# Given:
E0=1.7;# in MeV
# Solution:
# For E0<2.5 MeV; using Katz and Penfold empirical equation we have
R1=412*((E0)**(1.265-0.0954*math.log(E0)));# mg/cm^2
# Using feather's relation we have
R2=530*E0-106;# mg/cm^2
print"The range in Al for beta radiation is mg/cm^2 using Katz and Penfold empirical equation and mg/cm^2 using feathers relation.",round(R1,1),round(R2)
```

In [15]:

```
#cal of emissions
#intiation of all variables
# Chapter 4
print"Example 4.15, Page:61 \n \n"
# Solution:
L1=(5.5-3.5);# Case 1
L2=2-0;# Case 2
L3=1.5-.5;# Case 3
ELML1=1+0+2;
ELML2=1+1+2;
ELML3=0+1+1;
print"The order for Case 1 is and emission is type M2,Case 2 is and emission is type E2,Case 3 is and emission is type E1,Case 4 is not possible.",L1,ELML1,L2,ELML2,L3,ELML3
```

In [ ]:

```
#cal of rate of energy emission
#intiation of all variables
# Chapter 4
print"Example 4.16, Page:61 \n \n"
# Given:
m=4*10**-3;# in gms
M=210;
E=0.34;# in MeV
# Solution:
N=(m*6.022*10**23)/M;
k=0.693/(5*24*3600);# in s^-1
A=N*k;# in dis/s
# Energy released at 0.34 MeV per dis/s will be
E1=E*A;# in MeV/s
E2=E1*1.6*10**-13;# watts
print"The rate of energy emission is W",round(E2)
```

In [16]:

```
#cal of strength in KCi
#intiation of all variables
# Chapter 4
print"Example 4.17, Page:61 \n \n"
# Given:
A=0.2506*10**15;# atoms/s re: Ex4_10
# Solution:
Strength=A/(3.7*10**10);# in kCi
S1=Strength*10**-3;# in KCi
print"The strength in KCi is ",round(S1,2)
```

In [17]:

```
#cal of half life periods
#intiation of all variables
# Chapter 4
print"Example 4.18, Page:62 \n \n"
# Given:
N1=10^24;# atoms
N2=10^16;# atoms
N3=1000;# atoms
N4=80;# atoms
# Solution:
N11=N1*0.5;# 1st half life
N12=N11/2;# 2nd half life
N13=N12/2;# 3rd half life
print"\n The 1st half life, 2nd half life, 3rd half life are respectively.",N11,N12,N13
N21=N2/2;# 1st half life
N22=N21/2;# 2nd half life
N23=N22/2;# 3rd half life
print"\n The 1st half life, 2nd half life, 3rd half life are respectively.",N21,N22,N23
N31=N3/2;# 1st half life
N32=N31/2;# 2nd half life
N33=N32/2;# 3rd half life
print"\n The 1st half life, 2nd half life, 3rd half life are respectively.",N31,N32,N33
#Radiactivity is a statistical property. Decay kinetics are reliable only when initial number is large
```

In [18]:

```
#cal of partial half life for beta and EC decay
#intiation of all variables
# Chapter 4
print"Example 4.20, Page:62 \n \n"
# Given:
t1=1.28*10**9;# in years
# Solution:
k=0.693/(1.28*10**9);
# beta deay is 88.8%
k1=0.888*k;
# EC decay is 11.2%
k2=0.112*k;
tbeta=(0.693*10**-9)/(k1);# partial half life for beta decay in Gy
tEC=(0.693*10**-9)/(k2);# partial half life for EC decay in Gy
print"The partial half life for beta decay in Gy and partial half life for EC decay in Gy.",round(tbeta,3),round(tEC,3)
```

In [19]:

```
#cal of time
#intiation of all variables
# Chapter 4
print"Example 4.21, Page:63 \n \n"
# Given:
import math
t=15.02;# in hours
# Solution:
ar=1000;# activity ratio given that 0.1% of intial activity
k=0.693/t;
t1=(math.log(ar))/k;
print"The time required will be h",round(t1)
```

In [20]:

```
#cal of time required
#intiation of all variables
# Chapter 4
print"Example 4.22, Page:63 \n \n"
# Given:
import math
t=6.01;# in hours
# Solution:
ar=100/5;# activity ratio given that 5% of intial activity
k=0.693/t;
t1=(math.log(ar))/k;
print"The time required will be h",round(t1)
```

In [22]:

```
#cal of activity for k
#intiation of all variables
# Chapter 4
print"Example 4.23, Page:64 \n \n"
# Given:
t=1.83*10**9;# in years
# Solution:
# Part (a)
k=(0.693)/(t*3.16*10**7);
k1=(0.693*10**17)/(t*3.16*10**7);# in 10^-17 s^-1
print"\n The overall decay constant will be *10^-17 s^-1",round(k1,2)
# Part (b)
a=(6.022*10**23)/40; # atoms of K(40)
A=a*k;# activity
print"\n The activity for k(40) in 10^5 beta/s",round(A/10**5,3)
# Part (c)
a1=(6.022*10**23*1.2*10**-4)/41; # atoms of K(41)
A1=a1*k;# activity
print"\n The activity for k(41) in beta/s",round(A1,2)
```

In [23]:

```
#cal of decay constant and half life
#intiation of all variables
# Chapter 4
print"Example 4.24, Page:65 \n \n"
# Given:
import math
a1=6520.;# c/min
a2=4820.;#c/min
t=2.;#min
# Solution:
k=math.log(a1/a2)/t;
t1=0.693/k;# half life
print"The decay constant is min^-1 and the half life is min",round(k,4),round(t1,4)
```

In [24]:

```
#cal of value of avagadro constant
#intiation of all variables
# Chapter 4
print"Example 4.25, Page:65 \n \n"
# Given:
import math
a=(1./32.);# activity drop of its initial value
t1=7.5;#in h case(a)
t2=64.45;# in min case(b)
# Solution:
n=math.log(a)/math.log(0.5);
t11=t1/n;# half life
t12=t2/n;# half life
print"The half life for case (a)is h and case(b) is min",t11,t12
```

In [25]:

```
#cal of proportion of U235 704 million years back
#intiation of all variables
# Chapter 4
print"Example 4.26, Page:66 \n \n"
# Given:
import math
t238=4.5*10**9;# in y
t235=7.04*10**8;# in y
a0=0.72;# atoms per cent
t=7.04*10**8;
# Solution:
k1=0.693/(t238);#decay constant for U 238
N1=(100-a0)*math.exp(k1*t);
k2=0.693/(t235);#decay constant for U 235
N2=(a0)*math.exp(k2*t);
proportion=N2/N1;
print"The proportion of U235 704 million years back in 10^-2 is " ,round(proportion*100,2)
```

In [26]:

```
#cal of The no. of atoms produced is and its mass
#intiation of all variables
# Chapter 4
print"Example 4.27, Page:67 \n \n"
# Given:
t=110;# in min
a=10;#dpmg^-1
# Solution:
k=0.693/t;
N=a/k;# atoms of F18
mass=(N*18)/6.022*10**23;
mass1=(N*18*10**20)/(6.022*10**23);# in 10^-20 grams
print"The no. of atoms produced is and its mass is *10^-20 grams",round(N/10**3,4),round(mass1,2)
```

In [27]:

```
#cal of mass in micro-grams
#intiation of all variables
# Chapter 4
print"Example 4.28, Page:67 \n \n"
# Given:
import math
t=14.3;# half life in days
# Solution:
k=0.693/(t*24*3600);
N=(3.7*10**10)/(k);# No. of atoms in 1 Ci
N1=N*(1-(math.exp(-0.693/14.3)));# atoms of S32 produced
mass=(N1*32)/(6.022*10**23);
m1=mass*10**6;# in micro grams
print"The mass in micro-grams is ",round(m1,3)
```

In [28]:

```
#cal of fraction decays and half lives
#intiation of all variables
# Chapter 4
print"Example 4.29, Page:68 \n \n"
# Given:
import math
t=3.82;# in days
# Solution:
# part(a)
days=1;
D1=(1-(math.exp(-0.693*days/t)))*100;
print"\n The fraction decayed in 1 day will be ",round(D1,2)
# part(b)
days=5;
D1=(1-(math.exp(-0.693*days/t)))*100;
print"\n The fraction decayed in 5 days will be ",round(D1,2)
# part(c)
days=10;
D1=(1-(math.exp(-0.693*days/t)))*100;
print"\n The fraction decayed in 10 days will be .",round(D1,2)
# part(d)
days=6*t;
D1=(1-(math.exp(-0.693*days/t)))*100;
print"\n The fraction decayed in 6 half lives will be .",round(D1,2)
# part(e)
n=math.log(0.001)/math.log(0.5);
print"\n Time needed for the decay of 99.9 percent is half lives i.e.days.",round(n,2),round(t*n,3)
```

In [29]:

```
#cal of decay constant and half life
#intiation of all variables
# Chapter 4
print"Example 4.30, Page:68 \n \n"
# Given:
t=2.6;# years
# Solution:
k=0.693/t;# decay constant
#part(a)
kbeta=0.89*k;
print"\n The decay constant is y^-1",round(kbeta,2)
kEC=0.11*k;
print"\n The decay constant is y^-1",round(kEC,2)
#part(b)
tbeta=0.693/kbeta;
print"\n The half life is y",round(tbeta,2)
tEC=0.693/kEC;
print"\n The half life is y",round(tEC,2)
```

In [30]:

```
#cal of Decay by beta
#intiation of all variables
# Chapter 4
print"Example 4.31, Page:69 \n \n"
# Given:
t=12.8;# hours
# Solution:
k=0.693/t;# decay constant
#part(a)by EC
kEC=0.42*k;
print"Decay by EC"
print"\n\t The decay constant is h^-1",round(kEC,2)
tEC=0.693/kEC;
print"\n\t The half life is h",round(tEC,2)
#part(b)by beta+
kbeta1=0.19*k;
print"\nDecay by beta+"
print"\n\t The decay constant is h^-1",round(kbeta1,2)
tbeta1=0.693/kbeta1;
print"\n\t The half life is h",round(tbeta1,2)
#part(b)by beta+
kbeta2=0.39*k;
print"\nDecay by beta-"
print"\n\t The decay constant is h^-1",round(kbeta2,2)
tbeta2=0.693/kbeta2;
print"\n\t The half life is h",round(tbeta2,2)
```

In [31]:

```
#cal of proportion of U,Ra,Rn
#intiation of all variables
# Chapter 4
print"Example 4.32, Page:69 \n \n"
# Given:
tU=4.47*10**9;# y
tRa=1600;# y
tRn=3.82;# days
nU=1;
# Solution:
#under secular equilibrium we have
nRa=(tRa*365/tRn)*nU;
nRn=(tU*365/tRn)*nU;
print"The proportion of U,Ra,Rn is 1:",nRa,nRn
```

In [32]:

```
#cal of proportion of ay4
#intiation of all variables
# Chapter 4
print"Example 4.33, Page:70 \n \n"
# Given:
ax0 =1.; #assume
tx = 2.; #hrs
ty = 1.; #hrs
# Solution:
# general equation connecting Ax and Ay is
# Ax(n) = (ky * Ax(0) * (exp(-kx * t) - exp(-ky * t))/ (ky - kx)) + Ay(0) * exp(-ky * t)
ax0 = 1.;
ay4 = (ax0 * (0.693/1.) * ((1./4.)-(1./16.)))/((0.693/1.)-(0.693/2.)) + ax0 * (1./16.);
ax4 = (1.* ax0)/4.;
proportion = (ay4 * 100)/(ay4+ax4)
print"The proportion of ay4 at the end of 4 hrs is",round(proportion,1)
```

In [33]:

```
#cal of Activity due to La(140)
#intiation of all variables
# Chapter 4
print"Example 4.34, Page:71 \n \n"
#Given:
Ax0 = 2000.; #dps
#Solution:
#part a
ky = 0.693/10.;
kx = 0.693/288.;
# general equation connecting Ax and Ay is
Ax12 = (ky * Ax0 * (0.5**(1./24.) - 0.5**(1.2)))/ (ky - kx)
print"Activity due to La(140) at the end of 12 hrs will be dps",round(Ax12)
#part b
ky = 0.693/10.;
kx = 0.693/288.;
# general equation connecting Ax and Ay is
Ax24 = (ky * Ax0 * (0.5**(2) - 0.5**(57.6)))/ (ky - kx)
print"Activity due to La(140) at the end of 24 d will be dps",round(Ax24)
```

In [34]:

```
#cal of time
#intiation of all variables
# Chapter 4
print"Example 4.35, Page:71 \n \n"
# Given:
import math
t1=2.7;# h
t2=3.6;# h
# Solution:
k1=0.693/t1;
k2=.693/t2;
tmax=(math.log(k2/k1))/(k2-k1);
print"The time when daughter activity reaches maximum is and this is same when activities of both are equal.",round(tmax,2)
```

In [35]:

```
#cal of half life of Bi
#intiation of all variables
# Chapter 4
print"Example 4.36, Page:71 \n \n"
# Given:
tPo=138;# days
n=24.86;# days
# Solution:
kPo = 0.693/tPo;
# using simplification logx=2(x-1)/(x+1)
kBi=((2 * 2.303)-(n*kPo))/n;
tBi=0.693/kBi;
print"The half life of Bi is days",round(tBi,2)
```

In [37]:

```
#cal of residual activity in the sample
#intiation of all variables
# Chapter 4
print"Example 4.37, Page:72 \n \n"
# Given:
a=10*10**7;# rate
t=15;# h
# Soution:
A30=a*(1-(0.5)**(2));# dps
A45=A30*((0.5)**(3));# dps
print"The residual activity in the sample in 10^5 dps=",A45/10**5
```