# chapter 11:elements of radiation chemistry¶

## example 11.1;pg no: 144¶

In :
#cal of molecule,mass,linear absorption coefficient
#intiation of all variables
# Chapter 11
print"Example 11.1, Page:144  \n \n"
#Given:
density1=1.000;# for water in CGS units
density2=0.789;# for ethanol in CGS units
density3=0.793;# for methanol in CGS units
ue=0.211;# electron absorption coefficent in barn per electron
# 1 b=10^(-24) cm^2
#solution: (a) Water
z1=10;# no. of electrons
a1=18;# atomic mass of water
uw=ue*z1;#molecule absorption coefficient
umw=(uw*l*10**-24)/a1;#mass absorption coefficient
ulw=umw*density1;# linear absorption coefficient
print"\n The molecule absorption coefficient of water in b/molecule is = ",round(uw,3)
print"\n The mass absorption coefficient of water in cm^2/g is = ",round(umw,3)
print"\n The linear absorption coefficient of water in cm^-1 is = ",round(ulw,3)
#solution: (b) ethanol
z2=26;# no. of electrons
a2=46;# atomic mass of water
ueth=ue*z2;#molecule absorption coefficient
umeth=(ueth*l*10**-24)/a2;#mass absorption coefficient
uleth=umeth*density2;# linear absorption coefficient
print"\n \n The molecule absorption coefficient of ethanol in b/molecule is =",round(ueth,3)
print"\n The mass absorption coefficient of ethanol in cm^2/g is = ",round(umeth,3)
print"\n The linear absorption coefficient of ethanol in cm^-1 is = ",round(uleth,3)

#solution: (c) methanol
z3=18;# no. of electrons
a3=32;# atomic mass of water
umet=ue*z3;#molecule absorption coefficient
ummet=(umet*l*10**-24)/a3;#mass absorption coefficient
ulmet=ummet*density3;# linear absorption coefficient
print"\n \n The molecule absorption coefficient of methanol in b/molecule is =",round(umet,3)
print"\n The mass absorption coefficient of methanol in cm^2/g is =",round(ummet,3)
print"\n The linear absorption coefficient of methanol in cm^-1 is = ",round(ulmet,3)

Example 11.1, Page:144

The molecule absorption coefficient of water in b/molecule is =  2.11

The mass absorption coefficient of water in cm^2/g is =  0.071

The linear absorption coefficient of water in cm^-1 is =  0.071

The molecule absorption coefficient of ethanol in b/molecule is = 5.486

The mass absorption coefficient of ethanol in cm^2/g is =  0.072

The linear absorption coefficient of ethanol in cm^-1 is =  0.057

The molecule absorption coefficient of methanol in b/molecule is = 3.798

The mass absorption coefficient of methanol in cm^2/g is = 0.071

The linear absorption coefficient of methanol in cm^-1 is =  0.057


## example 11.2;pg no: 145¶

In :
#cal of  molecule,mass absorption coefficient
#intiation of all variables
# Chapter 11
print"Example 11.2, Page:145  \n \n"
#Given:
# for benzene in CGS units
density1=.879;
lac1=.06014; # linear absorption coefficient
# for cyclohexane in CGS units
density2=0.779;
lac2=.05656; # linear absorption coefficient
ue=0.211;# electron absorption coefficent in barn per electron
# 1 b=10^(-24) cm^2
#solution: (a)Benzene
a1=78;# atomic mass of benzene
mac1=lac1/density1; #mass absorption coefficient
mb=(mac1*a1)/(l*10**-24); #molecule absorption coefficient of benzene
print"\n The molecule absorption coefficient of benzene in b/molecule is =",round(mb,3)
print"\n The mass absorption coefficient of benzene in cm^2/g is = ",mac1
#solution: (b)cyclohexane
a2=84;# atomic mass of cyclohexane
mac2=lac2/density2; #mass absorption coefficient
mc=(mac2*a2)/(l*10**-24); #molecule absorption coefficient of cyclohexane
print"\n \n The molecule absorption coefficient of cyclohexane in b/molecule is = ",round(mc,3)
print"\n The mass absorption coefficient of cyclohexane in cm^2/g is = ",round(mac2,3)

Example 11.2, Page:145

The molecule absorption coefficient of benzene in b/molecule is = 8.86

The mass absorption coefficient of benzene in cm^2/g is =  0.0684186575654

The molecule absorption coefficient of cyclohexane in b/molecule is =  10.126

The mass absorption coefficient of cyclohexane in cm^2/g is =  0.073


## example 11.3;pg no: 145¶

In :
#cal of mass absorption coefficient
#intiation of all variables
# Chapter 11
print"Example 11.3, Page:145  \n \n"
#Given:
#Atomic absorption coefficients for diferent atoms in b/atom
O=1.69;
Na=2.32;
P=3.17;
Ca=4.22;
I=12.03
# 1 b=10^(-24) cm^2
#solution:
#mass absorption coefficient for the given atoms in cm^2/g
uO=(O*10**-24*6.022*10**23)/16;
uNa=(Na*10**-24*6.022*10**23)/23;
uP=(P*10*8-24*6.022*10**23)/31;
uCa=(Ca*10**-24*6.022*10**23)/40;
uI=(I*10**-24*6.022*10**23)/127;
# The mass absorption coefficient for the given substance is the sum of the mass absorption coefficients of the atoms present, each atom being weighted in proportion to its mass in the molecule.

#(a) NaI(A1=150)
u1=(uNa*23+uI*127)/150;
#(b) NaIO3 (A2=198)
u2=(uNa*23+uI*127+uO*48)/198;
#(c) Ca(PO3)2 (A3=198)
u3=(uCa*40+uP*62+uO*96)/198;
#(d) Ca3(PO4)2 (A4=310)
u4=(uCa*120+uP*62+uO*128)/310;#cal of  molecule,mass absorption coefficient
#intiation of all variables
# Chapter 11
print"Example 11.2, Page:145  \n \n"
#Given:
# for benzene in CGS units
density1=.879;
lac1=.06014; # linear absorption coefficient
# for cyclohexane in CGS units
density2=0.779;
lac2=.05656; # linear absorption coefficient
ue=0.211;# electron absorption coefficent in barn per electron
# 1 b=10^(-24) cm^2
#solution: (a)Benzene
a1=78;# atomic mass of benzene
mac1=lac1/density1; #mass absorption coefficient
mb=(mac1*a1)/(l*10**-24); #molecule absorption coefficient of benzene
print"\n The molecule absorption coefficient of benzene in b/molecule is =",round(mb,3)
print"\n The mass absorption coefficient of benzene in cm^2/g is = ",mac1
#solution: (b)cyclohexane
a2=84;# atomic mass of cyclohexane
mac2=lac2/density2; #mass absorption coefficient
mc=(mac2*a2)/(l*10**-24); #molecule absorption coefficient of cyclohexane
print"\n \n The molecule absorption coefficient of cyclohexane in b/molecule is = ",round(mc,3)
print"\n The mass absorption coefficient of cyclohexane in cm^2/g is = ",round(mac2,3)
print"The mass absorption coefficient of NaI in cm^2/g is =",u1
print"\n \n The mass absorption coefficient of NaIO3 in cm^2/g is = ",u2
print"\n \n The mass absorption coefficient of Ca(PO3)2 in cm^2/g is = ",u3
print"\n \n The mass absorption coefficient of Ca3(PO4)2 in cm^2/g is =",u4

Example 11.3, Page:145

Example 11.2, Page:145

The molecule absorption coefficient of benzene in b/molecule is = 8.86

The mass absorption coefficient of benzene in cm^2/g is =  0.0684186575654

The molecule absorption coefficient of cyclohexane in b/molecule is =  10.126

The mass absorption coefficient of cyclohexane in cm^2/g is =  0.073
The mass absorption coefficient of NaI in cm^2/g is = 0.0576104666667

The mass absorption coefficient of NaIO3 in cm^2/g is =  0.0590642626263

The mass absorption coefficient of Ca(PO3)2 in cm^2/g is =  -1.45987878788e+23

The mass absorption coefficient of Ca3(PO4)2 in cm^2/g is = -9.32438709677e+22


## example 11.4;pg no: 146¶

In :
#cal of molecule,mass absorption coefficient
#intiation of all variables
# Chapter 11
print"Example 11.4, Page:146  \n \n"
#Given:
density1=3.123;# for KI in CGS units
density2=3.168;# for KIO3 in CGS units
ue=0.211;# electron absorption coefficent in barn per electron
# 1 b=10^(-24) cm^2
#Atomic absorption coefficients for diferent atoms in b/atom
K=ue*19;
I=ue*53;
O=ue*8;
#mass absorption coefficient for the given atoms in cm^2/g
uK=(6.022*10**23*K*10**-24)/39;
uI=(6.022*10**23*I*10**-24)/127;
uO=(6.022*10**23*O*10**-24)/16;
#solution: (a)KI
uKI=K+I;#molecular absorption coefficient
umKI=(39*uK+127*uI)/166;#mass absorption coefficient
ulKI=umKI*density1;# linear absorption coefficient
print"\n The molecular absorption coefficient of KI in b/molecule is = ",round(uKI,3)
print"\n The mass absorption coefficient of KI in cm^2/g is = ",round(umKI,3)
print"\n The linear absorption coefficient of KI in cm^-1 is = ",round(ulKI,3)
#solution: (b)KIO3
uKIO4=(K+I+O*4);#molecule absorption coefficient
umKIO4=(39*uK+127*uI+64*uO)/230;#mass absorption coefficient
ulKIO4=umKIO4*density2;# linear absorption coefficient
print"The molecular absorption coefficient of KIO4 in b/molecule is = ",round(uKIO4,3)
print"The mass absorption coefficient of KIO4 in cm^2/g is = ",round(umKIO4,3)
print"The linear absorption coefficient of KIO4 in cm^-1 is = ",round(ulKIO4,3)

Example 11.4, Page:146

The molecular absorption coefficient of KI in b/molecule is =  15.192

The mass absorption coefficient of KI in cm^2/g is =  0.055

The linear absorption coefficient of KI in cm^-1 is =  0.172
The molecular absorption coefficient of KIO4 in b/molecule is =  21.944
The mass absorption coefficient of KIO4 in cm^2/g is =  0.057
The linear absorption coefficient of KIO4 in cm^-1 is =  0.182


## example 11.5;pg no: 147¶

In :
#cal of thickness of copper nedded
#intiation of all variables
# Chapter 11
print"Example 11.5, Page:147  \n \n"
#Given:
import math
i1=4000;# initial intensity of radiaton
i2=2000;# final intensity of radiation
density1=8.96;# density of copper
ue=0.211;# electron absorption coefficent in barn per electron
# 1 b=10^(-24) cm^2
#solution:
uCu=ue*29;#atomic absorbtion coefficient in b/atom
umCu=(6.022*10**23*uCu*10**-24)/63; # mass absorbtion coefficient in cm^2/g
ulCu=umCu*density1;# linear absorption coefficient in cm^-1
# we know that, i2=i1*exp(ulCu*x)
x=math.log(i1/i2)/(ulCu);# thickness of the copper plate
print"\n The thickness of copper nedded to reduce the intensity of the radiation in cm is =",round(x,3)

Example 11.5, Page:147

The thickness of copper nedded to reduce the intensity of the radiation in cm is = 1.323


## example 11.6;pg no: 147¶

In :
#cal of biological efective dose for gamma particles
#intiation of all variables
# Chapter 11
print"Example 11.6, Page:147  \n \n"
#Given:
# Relative biological effectiveness(RBE)
a7=10;# for alpha partical
tn=2.5;# for thermal neutrons
g=1;# for gamma radiation
rd=0.6;#radiation dose in gray
#Formulas
#1.The Rontgen equivalent mammal (REM)=RBE*rads
#2.The sievert is the SI unit for REM= RBE*grays
#3. 1 gray(Gy)=100 rads
#Solution:
# part (a) alpha particles
a1=a7*rd# biological efective dose in Sv
a2=a1*100;# biological efective dose in rem
print"\n The biological efective dose for alpha particles in Sv is = ",a1
print"\n The biological efective dose for alpha particles in rem is = ",a2
# part (b) thermal neutrons
tn1=tn*rd# biological efective dose in Sv
tn2=tn1*100;# biological efective dose in rem
print"\n \n The biological efective dose for thermal neutrons in Sv is = ",tn1
print"\n The biological efective dose for thermal neutrons in rem is = ",tn2
# part (c) gamma particles
g1=a6*rd# biological efective dose in Sv
g2=g1*100;# biological efective dose in rem
print"\n \n The biological efective dose for gamma particles in Sv is = ",g1
print"\n The biological efective dose for gamma particles in rem is = ",g2

Example 11.6, Page:147

The biological efective dose for alpha particles in Sv is =  6.0

The biological efective dose for alpha particles in rem is =  600.0

The biological efective dose for thermal neutrons in Sv is =  1.5

The biological efective dose for thermal neutrons in rem is =  150.0

The biological efective dose for gamma particles in Sv is =  0.6

The biological efective dose for gamma particles in rem is =  60.0


## example 11.7;pg no: 148¶

In :
#cal of thickness of lead nedded
#intiation of all variables
# Chapter 11
print"Example 11.7, Page:148  \n \n"
#Given:
import math
density1=11.35;# density of copper
ue=0.211;# electron absorption coefficent in barn per electron
# 1 b=10^(-24) cm^2
#solution:
uPb=ue*82;#atomic absorbtion coefficient in b/atom
umPb=(6.022*10**23*uPb*10**-24)/207.2; # mass absorbtion coefficient in cm^2/g
ulPb=umPb*density1;# linear absorption coefficient in cm^-1
# we know that, i2=i1*exp(ulCu*x)
# Case (i) from 0.1 Gy/min to 3.1 mGy/h
i1=6;# in Gy/h
i2=3.1*10**-3;#in Gy/h
x=math.log(i1/i2)/(ulPb);# thickness of the lead plate
print"\n The thickness of lead nedded to reduce the intensity of the radiation in cm is =",round(x,2)
# Case (ii) from 100 Gy/min to 0.1 mGy/h
j1=6000;# in Gy/h
j2=0.1*10**-3;# in Gy/h
y=math.log(j1/j2)/(ulPb);# thickness of the lead plate
print"\n \n The thickness of lead nedded to reduce the intensity of the radiation in cm is =",round(y,2)
# Case (iii) half thickness
z=(0.693)/ulPb;# thickness of the lead plate
print"\n \n The thickness of lead nedded to reduce the intensity of the radiation in cm is =",round(z,2)

Example 11.7, Page:148

The thickness of lead nedded to reduce the intensity of the radiation in cm is = 13.26

The thickness of lead nedded to reduce the intensity of the radiation in cm is = 31.38

The thickness of lead nedded to reduce the intensity of the radiation in cm is = 1.21


## example 11.8;pg no: 148¶

In :
#cal of  Z/A value for Carbontetrachloride,acetic acid,Cyclohexane
#intiation of all variables
# Chapter 11
print"Example 11.8, Page:148  \n \n"
# To find:
#(a)Z/A value for Carbontetrachloride
#(b)Z/A value for acetic acid
#(c)Z/A value for Cyclohexane
#Formula:
#(Z/A) for  compound = (summation of Z of all the atoms)/ molecular weight of compound
#solution:
# Part (a) CCl4
z1=(6.+4.*17.)/154.;# Z/A value for Carbontetrachloride
z2=(2.*6.+4.*1.+2.*8.)/60.;# Z/A value for acetic acid
z3=(6.*6.+12.*1.)/84.;# Z/A value for Cyclohexane
print"\n The Z/A value for Carbontetrachloride is = ",round(z1,3)
print"\n \n The Z/A value for acetic acid is = ",round(z2,3)
print"\n \n The Z/A value for Cyclohexane is = ",round(z3,3)

Example 11.8, Page:148

The Z/A value for Carbontetrachloride is =  0.481

The Z/A value for acetic acid is =  0.533

The Z/A value for Cyclohexane is =  0.571


## example 11.9;pg no: 149¶

In :
#cal of dose absorbed by sodium periodate
#intiation of all variables
# Chapter 11
print"Example 11.9, Page:149  \n \n"
#Given:
dose1=2.15# in Gy/min
# Z values
Na=11.;
I=53.;
O=8.;
# A values
mNa=23.;
mI=127.;
mO=16.;
z2=0.553;#Z/A for fricke solution
# Solution:
z1=(11.+53.+8.*4.)/(127.+23.+16.*4.);# Z/A value for sodium periodate(Z/A NaIO4)
# Formula : D(NaIo4)*(Z/A Fricke)=D(Fricke)*(Z/A NaIO4)
dose2=(dose1*z1)/z2;# in Gy/min
# for 3 hours
Dose=dose2*180.;# in Gy
print"The dose absorbed by sodium periodate in 3 h is = (Gy)",round(Dose,2)
# Note: There is correction in the value of (Z/A NaIO4) calculated in the book. The actual value comes out to be 0.44859

Example 11.9, Page:149

The dose absorbed by sodium periodate in 3 h is = (Gy) 313.94


## example 11.10;pg no: 149¶

In :
#cal of  dose absorbed by sodium periodate
#intiation of all variables
# Chapter 11
print"Example 11.10, Page:149  \n \n"
#Given:
dose=4.06# in Gy/min
z=0.553;#Z/A for fricke solution
# Formula : D(NaIo4)*(Z/A Fricke)=D(Fricke)*(Z/A NaIO4)
# Solution:
# Part(a) Chloroform
z1=58./119.5;# Z/A value for Chloroform
dose1=dose*z1/z;
Dose1=dose1*360.;# for 6 hours # in Gy
print"\n The dose absorbed by sodium periodate in 6 h is = (Gy)",round(Dose1,3)
# Part(b) Bromoform
z2=112./253.;# Z/A value for Bromoform
dose2=dose*z2/z;
Dose2=dose2*360.;# for 6 hours # in Gy
print"\n \n The dose absorbed by sodium periodate in 6 h is = (Gy)",round(Dose2,3)
# Part(c) Iodoform
z3=166./394.;# Z/A value for Iodoform
dose3=dose*z3/z;
Dose3=dose3*360.;# for 6 hours # in Gy
print"\n \n The dose absorbed by sodium periodate in 6 h is = (Gy)",round(Dose3,2)

Example 11.10, Page:149

The dose absorbed by sodium periodate in 6 h is = (Gy) 1282.813

The dose absorbed by sodium periodate in 6 h is = (Gy) 1170.041

The dose absorbed by sodium periodate in 6 h is = (Gy) 1113.56


## example 11.11;pg no: 150¶

In :
#cal of fraction of the energy absorbed by ethanol
#intiation of all variables
# Chapter 11
print"Example 11.11, Page:150  \n \n"
# To find:
#Fraction of the energy absorbed by ethanol
#Formula:
#(Z/A) for  compound = (summation of Z of all the atoms)/ molecular weight of compound
z1=26./46.;# Z/A value for ethanl
z2=32./60.;# Z/A value for acetic acid
Deth=z1/(z1+z2);
# Note that the dose absorbed by each component is proportional to its Z/A in the total
print"The fraction of the energy absorbed by ethanol = ",round(Deth,4)

Example 11.11, Page:150

The fraction of the energy absorbed by ethanol =  0.5145


## example 11.12;pg no: 150¶

In :
#cal of value of G(HCL) in the radiolysis of CHCl3
#intiation of all variables
# Chapter 11
print"Example 11.12, Page:150  \n \n"
#Given:
density1=1.48;# for chloroform
density2=1.024;# for fricke solution
mole=30*10**-6;# moles of HCl
e=2174;#extension coefficient
OD=0.5633;
d=1;# cell path in cm
# G(Fe+3)=15.5 ions/100 eV
# Solution:
#HCl produced
m1=mole*l;
conc=OD*d/e;# in moles/l
conc1=conc*l;# molecules in 100 min
#This implies amount (conc1) of molecules/l in 10 min for CHCl3 solution will be
conc2=conc1/10;
# energy that would be absorbed by frickes solution to produce conc1 amount of molecules
e1=conc2*100/15.5;# eV/l
e2=e1/100;# eV per 10 ml
# we know that energy absorbed is proportional to its density
e3=e2*density1/density2;# in eV
#G(HCl)
g=m1*100/e3;
print"The value of G(HCL) in the radiolysis of CHCl3 is = ",round(g)

Example 11.12, Page:150

The value of G(HCL) in the radiolysis of CHCl3 is =  1242.0