# chapter 10:applications of radioactivity¶

## example 10.1;pg no: 133¶

In :
#cal of  value of numerical constant
#intiation of all variables
# Chapter 10
print"Example 10.1, Page:133  \n \n"
# Given:
E=4.8;# in MeV
M=128;# molecular weight of I
#Formula:
# Er=(536*E^2)/M
# Solution:
Er=(536*E**2)/M;
print"The energy of recoil of Iodine atom in eV=", Er

Example 10.1, Page:133

The energy of recoil of Iodine atom in eV= 96.48


## example 10.2;pg no: 133¶

In :
#cal of energy of recoil of Zinc atom
#intiation of all variables
# Chapter 10
print"Example 10.2, Page:133  \n \n"
# Given:
E=1.1;# in MeV
M=65;# molecular weight of zinc
#Formula:
# Er=(536*E^2)/M
# Solution:
Er=(536*E**2)/M;
print"The energy of recoil of Zinc atom in (eV)=", round(Er,2)

Example 10.2, Page:133

The energy of recoil of Zinc atom in (eV)= 9.98


## example 10.3;pg no: 133¶

In :
#cal of  solubility of BaSO4
#intiation of all variables
# Chapter 10
print"Example 10.3, Page:133  \n \n"
# Given:
M1=137.32;# moleular wt of barium
M2=32.;# molecular weight of sulphur
M3=16.;# molecular wt of oxygen
M4=233.32;# molecular wt of BaSO4
ai=40000.;# specific initial activity in counts min^-1 mg^-1
af=187./20.;# specific final activity in counts min^-1 0.1ml^-1

# Formula:
# (1)S1=(af/ai)*(10/M) in moles/lit
# (2)S2=(af/ai)*(10^4) in mg/lit
# Solution:
S1=(af/ai)*(10./M4); # solubility of BaSO4 in moles/lit
S2=(af/ai)*10**4;# solubility of BaSO4 in mg/lit
print"\n The solubility of BaSO4 in (moles/lit) is =",S1
print"\n \n The solubility of BaSO4 in (mg/lit) is =",round(S2,3)

Example 10.3, Page:133

The solubility of BaSO4 in (moles/lit) is = 1.00184296245e-05

The solubility of BaSO4 in (mg/lit) is = 2.337


## example 10.4;pg no: 133¶

In :
#cal of surface area of 1 gm of precipitate sample
#intiation of all variables
# Chapter 10
print"Example 10.4, Page:133  \n \n"
# Given:
conc=4.;# 4 mg per 1 l
a1=1600.;# labelled solution of PbSO4
a2=900.;# activity of filtrate (in solution)
M=303.2;# molecular wt of PbSO4
l=6.022*10**23;
# Solution:
y=20.*4./1000.;# 20 ml will contain y mg
z=y*a2/a1;# final amount of PbSO4 in solution
a3=a1-a2;# activity on surface
# Let the total PbSO4 on surfaceof precipitate be x
# Asumming exchange equilibrium is established we have
x=a3*z/a2;# in mg
molecules=x*10**-3*(l)/M;
# Give that surface area of 1 PbSO4 = 18.4*10^-16 cm^2
A=molecules*18.4*1.**-16;
print"The surface area of 1 gm of precipitate sample in (cm^2/gm)=",A

Example 10.4, Page:133

The surface area of 1 gm of precipitate sample in (cm^2/gm)= 1.27907915567e+18


## example 10.5;pg no: 134¶

In :
#cal of volume of blood in the patient
#intiation of all variables
# Chapter 10
print"Example 10.5, Page:134  \n \n"
# Given:
ai=14000;# counts per min per 0.1 cm^3, initial activity of blood
Si=1.4*10**5;# c min^-1 cm^-3, initial specific activity
a=250;# 250 net counts in 10 min, this implies 25 net counts in a min
# Formula: Si/Sr = V
# Solution:
V=Si/25;# total blood in the patient in cm^3
V1=V/1000;# volume in lit
print"The volume of blood in the patient in (lit)=",V1

Example 10.5, Page:134

The volume of blood in the patient in (lit)= 5.6


## example 10.6;pg no: 134¶

In :
#cal of percentage of penicillin in mixture
#intiation of all variables
# Chapter 10
print"Example 10.6, Page:134  \n \n"
# Given:
y=5;# in mg labelled sample
V=2000;# volume of mixture in ml
# Formula: x=y*(Si-Sf)/Sf
#Solution:
Si=20000/(5*5);# initial specific activity in counts min^-1 mg^-1
Sf=3000/(.6*10);# final specific activity in counts min^-1 mg^-1
x=y*(Si-Sf)/Sf;# in mg for V amount of volume
#% of penicillin in mixture
p=x*100/V;
print"The percentage of penicillin in mixture is =",p

Example 10.6, Page:134

The percentage of penicillin in mixture is = 0.15


## example 10.7;pg no: 135¶

In :
#cal of  percentage of iodine in mixture
#intiation of all variables
# Chapter 10
print"Example 10.7, Page:135  \n \n"
# Given:
y=2.;# in ml labelled sample
V=1000.;# volume of mixture in ml
BC=100./20.;# 100 counts for 20 min
# Formula: x=y*(Si-Sf)/Sf
#Solution:
Si=(2500.-BC)/(2.);# initial specific activity in counts min^-1 mg^-1
Sf=(600.-BC)/(3.);# final specific activity in counts min^-1 mg^-1
x=y*(Si-Sf)/Sf;# in mg for V amount of volume
#% of iodine in mixture
i=x*100./V;
print"The percentage of iodine in mixture is = ",round(i,3)
#NOte: Backward counts are taken to be 100 counts for 10 min in the solution given in textbook

Example 10.7, Page:135

The percentage of iodine in mixture is =  1.058


## example 10.8;pg no: 135¶

In :
#cal of percentage Cr content in the ruby
#intiation of all variables
# Chapter 10
print"Example 10.8, Page:135  \n \n"
# Given:
flux=10**12;
s=15.9*10**-24;
m1=0.5;# weight of ruby in mg
#Soluton:
a1=35000;# measured activity in c/s
a2=350000;# corrected activity in )d/s
N=a2/(flux*s*(1-0.5**(1/27.7)));
m=50*N/(6.02*10**23);
Cr=(100*m)/4.35;# total Cr in in the Ruby
crp=(Cr*100)/0.5;# % cr in the ruby
print"The percentage Cr content in the ruby is = ",round(crp,2)

Example 10.8, Page:135

The percentage Cr content in the ruby is =  0.34


## example 10.9;pg no: 136¶

In :
#cal of total mass of Ge atoms present in sample
#intiation of all variables
# Chapter 10
print"Example 10.9, Page:136  \n \n"
# Given:
flux=10**12;
s=3.28*10**-27;
hf=1;# half life in min
#Soluton:
a1=2500;# measured activity in d/s
a2=5000;# corrected activity in d/s
N=a2/(flux*s);
m=76*N/(6.02*10**23);
Cr=(100*m)/7.8;# total mass of Ge atoms (isotopic abandunce is 7.8%)
print"The total mass of Ge atoms present in sample in (mg)=",round(Cr*10**3,2)

Example 10.9, Page:136

The total mass of Ge atoms present in sample in (mg)= 2.47


## example 10.10;pg no: 136¶

In :
#cal of  activity of sample in dpm
#intiation of all variables
# Chapter 10
print"Example 10.10, Page:136  \n \n"
# Given:
import math
M=55;# wt of Mn
m=0.1;# in g
t=90;# min irradated
flux=10**6;
t1=5; #in hours
cs=13.3*10**-24# in cm^2
hl=2.58;# in hours
Na=6.022*10**23;
r=100;# in %
# Solution:
s=1-(math.exp(-.693*t/(2.58*60)));
A=(m*Na*r*flux*cs*s)/(100*M);
x=math.exp(-0.693*5/2.58);
# activity after cooling period
A1=A*x*60;# in  dpm
print"The activity of sample in dpm is = ",round(A1)

Example 10.10, Page:136

The activity of sample in dpm is =  75642.0


## example 10.11;pg no: 137¶

In :
#cal of
#intiation of all variables
# Chapter 10
print"Example 10.11, Page:137  \n \n"
# Given:
import math
t1=12.7;# in hours
a=4.5*10**-24;# in cm^2
r=69.1;#in %
cf=10;# in %
flux=10**6;
Na=6.022*10**23;
cpm=1500;# activity in counts per min
M=63;
#Solution:
dpm=cpm*100/10;
dps=dpm/60;
x=math.exp(-0.693*5/t1);
A=dps/x;
s=1-(math.exp(-.693*10/(12.7)));
w=(A*M*100)/(Na*r*a*flux*s);# gms of Cu
# given that 5g Cu in 100g alloy, for wg amount of alloy will be
Y=100*w/5;# amount of alloy ing
print"\n The weight of the sample that should be taken is =(g)",round(Y,3)

Example 10.11, Page:137

The weight of the sample that should be taken is =(g) 0.525


## example 10.12;pg no: 138¶

In :
#cal of mass of potassium bromide in the original solution
#intiation of all variables
# Chapter 10
print"Example 10.12, Page:138  \n \n"
# Given:
vi=2.5;#titrant volume
V1=10; # vol of KBr in ml
N2=0.01;#normality of AgNO3
M1=119;# mol wt of KBr
# Solution:
ai=((12500/5)-10);
af=((6000/6)-10);
# deerease in activity due to addition of titrant 2.5ml
d=ai-af;
# volume corresponding to ai for AgNO3
V2=ai*vi/d;
N1=(N2*V2)/V1;# Normality of KBr solution
m=N1*M1/100;# mass of KBr in 10 ml solution
print"\n The mass of potassium bromide in the original solution in (g)=",m

Example 10.12, Page:138

The mass of potassium bromide in the original solution in (g)= 0.0049385


## example 10.13;pg no: 138¶

In :
#cal of  age of the sample
#intiation of all variables
# Chapter 10
print"Example 10.13, Page:138  \n \n"
# Given:
import math
w=5.;# in g
ai=55.;# counts per 10 min
A0=15.8;# in dpm/g
# Solution:

cpm=55./10.;
dpm=cpm*100/10; # 10% efficient counting
sa=dpm/w;# in dpm/g
t=5730.*math.log(A0/sa)/(0.693); # Age determination
print"The age of the sample in (years)=",round(t,2)

Example 10.13, Page:138

The age of the sample in (years)= 2994.11