## example 4.1;pg no:55¶

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)

Example 4.1, Page:55

The value of avagadro constant in 10^23 atoms per mole= 6.05


## example 4.2;pg no:55¶

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)

Example 4.2, Page:55

The specific activity in dis min^-1 g^-1= 1813.0


## example 4.3;pg no:55¶

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)

Example 4.3, Page:55

The no. of atoms of Na(24) are = 2.88311688312e+12
The mass of Na(24) is * 10^-10 g 1.15
The no. of atoms of P(32) are = 6.59657142857e+13
The mass of P(32) is * 10^-9 g 3.51
The no. of atoms of Ra(226) are = 2.69945165945e+18
The mass of Ra(226) is * 10^-3 g 1.01


## example 4.4;pg no:56¶

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)

Example 4.4, Page:56

The activity in Ci/cm^3 = 2.59


## example 4.5;pg no:56¶

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)

Example 4.5, Page:56

The ratio of C14/C12 in atmosphere in 10^-12 is = 1.4


## example 4.7;pg no:57¶

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

Example 4.7, Page:57

Number of alpha decays = and number of beta decays =  8 6


## example 4.8;pg no:57¶

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

Example 4.8, Page:57

For first photon, dE1= MeV, dI1=, since EL/ML1= & (L+PI+PF) is odd, M3 0.059 -3 5

For second photon, dE2= MeV, dI2=,since  EL/ML2= & (L+PI+PF) is even, E2 1.17 2 4

For third photon, dE3= MeV, d3I=, since EL/ML3= & (L+PI+PF) is even, E2 1.33 2 4


## example 4.9;pg no:58¶

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

Example 4.9, Page:58

The energy that would be released for 0.1 mole of Co will be =(GJ) 24.08


## example 4.10;pg no:58¶

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

Example 4.10, Page:

The total energy dissipate per hour is = (KJ) 360.860029304


## example 4.11;pg no:59¶

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)

Example 4.11, Page:59

The net mass loss is = u 0.0044


## example 4.12;pg no:59¶

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)

Example 4.12, Page:59

For Ra emitting alpha
Energy of recoil is (MeV) 0.0876
Total transition energy is (MeV) 4.9506
For Bi emitting alpha
Energy of recoil is (MeV) 0.117
Total transition energy is (MeV) 6.199


## example 4.13;pg no:60¶

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)

Example 4.13, Page:60

The Kinetic Energy and velocity are  MeV and  10^5 m/s respectively 0.0966 2.79


## example 4.14;pg no:60¶

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)

Example 4.14, Page:60

The range in Al for beta radiation is  mg/cm^2 using Katz and Penfold empirical equation and  mg/cm^2 using feathers relation. 784.8 795.0


## example 4.15;pg no:60¶

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

Example 4.15, Page:61

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. 2.0 3 2 4 1.0 2


## example 4.16;pg no:61¶

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)


## example 4.17;pg no:61¶

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)

Example 4.17, Page:61

The strength in KCi is  6.77


## example4.18;pg no:62¶

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

Example 4.18, Page:62

The 1st half life, 2nd half life, 3rd half life are respectively. 9.0 4.5 2.25

The 1st half life, 2nd half life, 3rd half life are respectively. 13 6 3

The 1st half life, 2nd half life, 3rd half life are respectively. 500 250 125


## example 4.20;pg no:62¶

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)

Example 4.20, Page:62

The partial half life for beta decay in Gy and partial half life for EC decay in Gy. 1.441 11.429


## example 4.21;pg no:63¶

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)

Example 4.21, Page:63

The time required will be h 150.0


## example 4.22;pg no:63¶

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)

Example 4.22, Page:63

The time required will be h 26.0


## example 4.23;pg no:64¶

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)

Example 4.23, Page:64

The overall decay constant will be *10^-17 s^-1 1.2

The activity for k(40) in 10^5 beta/s 1.804

The activity for k(41) in beta/s 21.12


## example 4.24;pg no:65¶

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)

Example 4.24, Page:65

The decay constant is min^-1 and the half life is min 0.1511 4.5879


## example 4.25;pg no:65¶

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

Example 4.25, Page:65

The half life for case (a)is h and case(b) is min 1.5 12.89


## example 4.26;pg no:66¶

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)

Example 4.26, Page:66

The proportion of U235 704 million years back in 10^-2 is  1.3


## example 4.27;pg no:67¶

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)

Example 4.27, Page:67

The no. of atoms produced is and its mass is *10^-20 grams 1.5873 4.74


## example 4.28;pgno:67¶

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)

Example 4.28, Page:67

The mass in micro-grams is  0.166


## example 4.29;pg no:68¶

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)

Example 4.29, Page:68

The fraction decayed in 1 day will be  16.59

The fraction decayed in 5 days  will be  59.63

The fraction decayed in 10 days  will be . 83.7

The fraction decayed in 6 half lives will be . 98.44

Time needed for the decay of 99.9 percent is  half lives i.e.days. 9.97 38.069


## example 4.30;pg no:68¶

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)

Example 4.30, Page:68

The decay constant is y^-1 0.24

The decay constant is y^-1 0.03

The half life is y 2.92

The half life is  y 23.64


## example 4.31;pg no:69¶

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)

Example 4.31, Page:69

Decay by EC

The decay constant is  h^-1 0.02

The half life is h 30.48

Decay by beta+

The decay constant is h^-1 0.01

The half life is h 67.37

Decay by beta-

The decay constant is h^-1 0.02

The half life is h 32.82


## example 4.32;pg no:69¶

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

Example 4.32, Page:69

The proportion of U,Ra,Rn is 1: 152879.581152 4.27107329843e+11


## example 4.33;pg no:70¶

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)

Example 4.33, Page:70

The proportion of ay4 at the end of 4 hrs is 63.6


## example 4.34;pg no:71¶

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)

Example 4.34, Page:71

Activity due to La(140) at the end of 12 hrs will be dps 1111.0
Activity due to La(140) at the end of 24 d will be dps 518.0


## example 4.35;pg no:71¶

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)

Example 4.35, Page:71

The time when daughter activity reaches maximum is  and this is same when activities of both are equal. 4.48


## example 4.36;pg no:71¶

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)

Example 4.36, Page:71

The half life of Bi is days 3.84


## example 4.37;pg no:72¶

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

Example 4.37, Page:72

The residual activity in the sample in 10^5 dps= 93.75