Chapter9-Radiation Protection

Ex1-pg470

In [1]:
## Example 9.1
import math


## Given data
e = 1.6*10**(-19);                         ## Electronic charge in couloumb(coul)
X = 1*10**(-3)/3600.;                       ## Exposure rate in terms of R/sec
## According to the definition of Roentgen, 1 R = 2.58*10^(-7) coul/g 
R = 2.58*10**(-7);
## From standard table
## There is 0.001293 g of air per 1 cm^3 at 1 atmospheric pressure at 0 C 
density_air = 0.001293;
## Calculation
IR = (X*R*density_air)/e;
## Result
print'%s %.2f %s'%(" \n Rate of ions produced from gamma ray interaction = ",IR," ions/cm^3-sec");
 
 Rate of ions produced from gamma ray interaction =  579.16  ions/cm^3-sec

Ex2-pg475

In [2]:
## Example 9.2
import math


## According to the definition of radiation absorbed dose(rad), 1 rad/sec = 100 ergs/g-sec
## Given data
D = 5.*10**(-3)/100.;                            ## Absorbed dose in terms of rad/sec
## Expressing absorbed dose rate in SI units
## 1 Gray(Gy) = 100 rad 
D_dot = D*3600./100.;
## Using data from Table 9.2
Q = 1.;                                        ## Quality factor for gamma rays for tissue
## Calculation
H_dot = D_dot*Q;
print'%s %.2f %s'%(' \n Absorbed dose rate in a tissue = ',D_dot*1000,' mGy/hr \n');
print'%s %.2f %s'%(' \n Dose equivalent rate in a tissue = ',H_dot*1000,' mSv/hr \n');
 
 Absorbed dose rate in a tissue =  1.80  mGy/hr 

 
 Dose equivalent rate in a tissue =  1.80  mSv/hr 

Ex3-pg495

In [3]:
## Example 9.3
import math


## Given data
H = 25;                                       ## Equivalent dose in rem
age = 30;                                     ## Age of worker in years
exp_age = 77;                                 ## Average age upto which a person lives in years
## Using data from Table 9.6
## Bone cancer
rc_bone = 0.2;                                ## Risk coefficient per 10^6 rem/year
lp_bone = 10;                                 ## Latent period in years
## Probability of dying from bone cancer 
p_bone=(H*rc_bone*(exp_age-(lp_bone+age)))/10**6;

## Breast cancer
rc_breast = 1.5;                               ## Risk coefficient per 10^6 rem/year
lp_breast = 15.;                                ## Latent period in years
## Probability of dying from breast cancer 
p_breast = (H*rc_breast*(exp_age-(lp_breast+age)))/10**6;

## Leukemia
rc_leukemia = 1.;                              ## Risk coefficient per 10^6 rem/year
lp_leukemia = 2.;                              ## Latent period in years
## Probability of dying from leukemia 
p_leukemia = (H*rc_leukemia*(exp_age-(lp_leukemia+age)))/10**6;

## Lung cancer
rc_lung = 1.;                                  ## Risk coefficient per 10^6 rem/year
lp_lung = 15.;                                 ## Latent period in years
## Probability of dying from lung cancer
p_lung = (H*rc_lung*(exp_age-(lp_lung+age)))/10**6;

## Pancreatic cancer
rc_pancreas = 0.2;                             ## Risk coefficient per 10^6 rem/year
lp_pancreas = 15.;                              ## Latent period in years
## Probability of dying from Pancreatic cancer
p_pancreas = (H*rc_pancreas*(exp_age-(lp_pancreas+age)))/10**6;

## Stomach cancer
rc_stomach = 0.6;                              ## Risk coefficient per 10^6 rem/year
lp_stomach = 15.;                               ## Latent period in years
## Probability of dying from stomach cancer
p_stomach = (H*rc_stomach*(exp_age-(lp_stomach+age)))/10**6;

## Rest of alimentary cancer
rc_ali = 0.2;                                  ## Risk coefficient per 10^6 rem/year
lp_ali = 15.;                                   ## Latent period in years
## Probability of dying from rest of alimentary cancer
p_ali = (H*rc_ali*(exp_age-(lp_ali+age)))/10**6;

## Thyroid cancer
rc_thy = 0.43;                                  ## Risk coefficient per 10^6 rem/year
lp_thy = 10.;                                    ## Latent period in years
## Probability of dying from thyroid cancer
p_thy = (H*rc_thy*(exp_age-(lp_thy+age)))/10**6;

## All other type of cancer
rc_other = 1.;                                   ## Risk coefficient per 10^6 rem/year
lp_other = 15.;                                  ## Latent period in years
## Probability of dying from all other type of cancer
p_other = (H*rc_other*(exp_age-(lp_other+age)))/10**6;

## Calculation
p = p_bone+p_breast+p_leukemia+p_lung+p_pancreas+p_stomach+p_ali+p_thy+p_other;
## Result
print'%s %.2f %s'%('\n Probability that the worker will die from cancer = ',p,' \n');

## The value obtained is different from the value given in the textbook. This is because of approximation of individual probabilities in the textbook.
 Probability that the worker will die from cancer =  0.01  

Ex4-pg496

In [4]:
## Example 9.4
import math

H = 1.;                              ## Equivalent dos in rem
n = 10**6;                           ## Population
## Given data

## Using the data of number of expected deaths of leukemia per 10^6 people from Table 9.9
## In utero age group
frac_utero = 0.011;                 ## Fraction of population
riskyr_utero = 10.;                  ## Risk years
riskcoef_utero = 15.;                ## Risk coefficient
## Number of deaths in utero is given by
deaths_utero = frac_utero*riskyr_utero*riskcoef_utero;

## In 0-0.99 age group
frac_0_099 = 0.014;                 ## Fraction of population
riskyr_0_099 = 25.;                  ## Risk years
riskcoef_0_099 = 2.;                 ## Risk coefficient
## Number of deaths in 0-0.99 age group is given by
deaths_0_099 = frac_0_099*riskyr_0_099*riskcoef_0_099;

## In 1-10 age group
frac_1_10 = 0.146;                  ## Fraction of population
riskyr_1_10 = 25.;                   ## Risk years
riskcoef_1_10 = 2.;                  ## Risk coefficient
## Number of deaths in 1-10 age group is given by
deaths_1_10=frac_1_10*riskyr_1_10*riskcoef_1_10;

## In 11-20 age group
frac_11_20 = 0.196;                 ## Fraction of population
riskyr_11_20 = 25.;                  ## Risk years
riskcoef_11_20 = 1.;                 ## Risk coefficient
## Number of deaths in 11-20 age group is given by
deaths_11_20=frac_11_20*riskyr_11_20*riskcoef_11_20;

## In 21-30 age group
frac_21_30 = 0.164;                 ## Fraction of population
riskyr_21_30 = 25.;                  ## Risk years
riskcoef_21_30 = 1.;                 ## Risk coefficient
## Number of deaths in 21-30 age group is given by
deaths_21_30=frac_21_30*riskyr_21_30*riskcoef_21_30;

## In 31-40 age group
frac_31_40 = 0.118;                 ## Fraction of population
riskyr_31_40 = 25.;                  ## Risk years
riskcoef_31_40 = 1.;                 ## Risk coefficient
## Number of deaths in 31-40 age group is given by
deaths_31_40=frac_31_40*riskyr_31_40*riskcoef_31_40;

## In 41-50 age group
frac_41_50 = 0.109;                 ## Fraction of population
riskyr_41_50 = 25.;                  ## Risk years
riskcoef_41_50 = 1.;                 ## Risk coefficient
## Number of deaths in 41-50 age group is given by
deaths_41_50 = frac_41_50*riskyr_41_50*riskcoef_41_50;

## In 51-60 age group
frac_51_60 = 0.104;                 ## Fraction of population
riskyr_51_60 = 22.5;                ## Risk years
riskcoef_51_60 = 1.;                 ## Risk coefficient
## Number of deaths in 51-50 age group is given by
deaths_51_60 = frac_51_60*riskyr_51_60*riskcoef_51_60;

## In 61-70 age group
frac_61_70 = 0.08;
riskyr_61_70 = 15.1;
riskcoef_61_70 = 1.;                 ## Risk coefficient
## Number of deaths in 61-70 age group is given by
deaths_61_70=frac_61_70*riskyr_61_70*riskcoef_61_70;

## In 71-80 age group
frac_71_80 = 0.044;                 ## Fraction of population
riskyr_71_80 = 9.1;                 ## Risk years
riskcoef_71_80 = 1.;                 ## Risk coefficient
## Number of deaths in 71-80 age group is given by
deaths_71_80 = frac_71_80*riskyr_71_80*riskcoef_71_80;

## Age greater than 80
frac_80 = 0.02;                     ## Fraction of population
riskyr_80 = 4.5;                    ## Risk years
riskcoef_80 = 1.;                    ## Risk coefficient
## Number of deaths with age greater than 80 years is given by
deaths_80=frac_80*riskyr_80*riskcoef_80;

## Calculation
total_deaths = deaths_utero+deaths_0_099+deaths_1_10+deaths_11_20+deaths_21_30+deaths_31_40+deaths_41_50+deaths_51_60+deaths_61_70+deaths_71_80+deaths_80;
## Result
print'%s %.2f %s'%(" \n Number of cases or deaths expected from leukemia = ",total_deaths," \n");
 
 Number of cases or deaths expected from leukemia =  28.36  

Ex5-pg498

In [5]:
## Example 9.5
import math


## Given data
H_year = 5.;                                   ## Equiavelnt dose per year in rem
start_age = 18.;                               ## Initial age of the worker in years
ret_age = 68.;                                 ## Retirement age of the worker in years
## Using data from Table 9.6 with respect to Bone cancer
latent_period = 10.;                           ## Latent period in years
plateau_period = 30.;                          ## Plateau period in years
rc_bone = 0.2;                                ## Risk coefficient per 10^6 rem/year

n = ret_age-(start_age+latent_period);        ## Number of years of accumulated dose
H = n*H_year;                                 ## Total equivalent dose in rem
## Calculation
p_bone = (H*rc_bone*plateau_period)/10**6;
## Result
print'%s %.2e %s'%(" \n The probability of dying from bone cancer = ",p_bone," \n");
 
 The probability of dying from bone cancer =  1.20e-03  

Ex6-pg513

In [6]:
## Example 9.6
import math

## Given data
E =2. ;                                        ## Energy of gamma radiation in MeV
X_dot = 1.;                                    ## Exposure rate in mR/hour
## Using the data from Table II.5
## Let mu_a/rho of air at 2 Mev be denoted as mu_rho
mu_rho = 0.0238;                              ## Ratio of absorption coefficient to sensity of air in cm^2/g
## Calculation
I = X_dot/(E*mu_rho*0.0659);
print'%s %.2f %s'%(" \n The beam intensity of gamma radiation required = ",math.ceil(I)," gamma rays/cm^2-sec \n");
 
 The beam intensity of gamma radiation required =  319.00  gamma rays/cm^2-sec 

Ex7-pg515

In [7]:
## Example 9.7
import math

## Given data
phi = 2.4*10**5;                               ## Flux in x-rays/cm^2-sec
## From Figure 9.9
## To receive an exposure rate of 1 mR/hr at 50 keV, the flux is 8*10^3 x-rays/cm^2-sec
phi_eq = 8*10**3;                              ## Equivalent flux in x-rays/cm^2-sec
X_dot_eq = 1.;                                 ## Equivalent Exposure rate in mR/hr
X_dot = (phi*X_dot_eq)/phi_eq;                ## Exposure rate of the operator in mR/hr
## From Figure 9.10 at 50 kV energy, the energy dependent function is
f_bone = 3.3;
f_muscle = 0.93;
f_fat = 0.9;
## Using data from Table 9.2
Q = 1.;                                        ## Quality factor for x-rays
## Calculation
D_dot_bone = X_dot*f_bone*Q;                  ## Dose equivalent rate in bone
D_dot_muscle = X_dot*f_muscle*Q;              ## Dose equivalent rate in muscle
D_dot_fat = X_dot*f_fat*Q;                    ## Dose equivalent rate in fat
## Result
print'%s %.2f %s'%(" \n Dose equivalent rate in bone = ",math.ceil(D_dot_bone)," mrem/hour \n");
print'%s %.2f %s'%(" \n Dose equivalent rate in muscle = ",math.ceil(D_dot_muscle)," mrem/hour \n");
print'%s %.2f %s'%(" \n Dose equivalent rate in fat = ",math.ceil(D_dot_fat)," mrem/hour \n");
 
 Dose equivalent rate in bone =  99.00  mrem/hour 

 
 Dose equivalent rate in muscle =  28.00  mrem/hour 

 
 Dose equivalent rate in fat =  27.00  mrem/hour 

Ex8-pg518

In [8]:
## Example 9.8
import math
#calculate the

## Given data
phi_n = 20.;                               ## Given neutron flux in neutrons/cm^2-sec
## From Figure 9.12
## To receive an dose equivalent rate of 1 mrem/hr, the fast neutron flux is 7 neutrons/cm^2-sec
phi_n_eq = 7.;
D_dot_eq = 1.; 
D_dot_n = (phi_n*D_dot_eq)/phi_n_eq;      ## Dose rate due to fast neutron flux in mrem/hr
phi_th = 300.;                             ## Given thermal flux in neutrons/cm^2-sec
## From Figure 9.12
## To receive an dose equivalent rate of 1 mrem/hr, the thermal flux is 260 neutrons/cm^2-sec
phi_th_eq = 260.;
D_dot_th = (phi_th*D_dot_eq)/phi_th_eq;  ## Dose rate due to thermal neutron flux in mrem/hr
D_dot = D_dot_n+D_dot_th;               ## Total dose rate in mrem/hr
print("\n The permitted weekly dose is 100 mrem \n");
D_dot_perm = 100.;
## Calculation
t = D_dot_perm/D_dot;
print'%s %.2f %s'%(" \n The time of exposure upto a safe level = ",t," hour \n");
## The answer given in the textbook is wrong. This is because of wrong computation of total dose rate
 The permitted weekly dose is 100 mrem 

 
 The time of exposure upto a safe level =  24.93  hour 

Ex9-pg520

In [9]:
## Example 9.9
import math


## Given data
fluence = 10**8;                               ## Given fluence neutrons/cm^2
## From Figure 9.12
## To receive an dose equivalent rate of 1 mrem/hr, the fast neutron flux is 7 neutrons/cm^2-sec
phi_eq = 7.;                                  ## Equivalent flux in neutrons/cm^2-sec
D_eq = 1.;                                    ## Equivalent dose rate in mrem/hr
## 1 hour = 3600 seconds
fluence_eq = phi_eq*3600.;                    ## Equivalent fluence in neutrons/cm^2
## Calculation 
D = (fluence*D_eq)/fluence_eq;
## Result
print'%s %.2f %s'%(" \n Dose received due to exposure of accelerator source = ",D," mrem \n");
## The answer given in textbook is approximated to a nearest value.
 
 Dose received due to exposure of accelerator source =  3968.25  mrem 

Ex10-pg525

In [10]:
## Example 9.10
import math


## Given data
M = 20.;                               ## Mass of organ in grams

## a)
## Using the data from Table 9.15
T_12 = 8.04;                          ## Radiological half life of Iodine-131 in days
T_12_b = 138.;                         ## Biological half life of Iodine-131 in days
lambd = 0.693/T_12;                  ## Radiological decay constant of Iodine-131 in days^-1
lambda_b = 0.693/T_12_b;              ## Biological decay constant of Iodine-131 in days^-1
lambda_e = lambd+lambda_b;           ## Equivalent decay constant in days^-1
## Using the data from Table 9.15
zeta = 0.23;                          ## Effective energy equivalent in MeV
q = 0.23;                             ## The fraction of Iodine-131 that goes by inhalation
## Calculation
DCF = (51.1*zeta*q)/(M*lambda_e);
## Result
print'%s %.2f %s'%(" \n The dose commitment factor by inhalation = ",DCF," rem/ucurie \n");

## b) 
breathing_rate = 2.32*10**(-4);        ## Normal breathing rate in m^3/sec
time = 2*3600.;                        ## Time of radiation exposure in seconds
I_conc = 2*10**(-9);                   ## Iodine-131 concentration
C0 = breathing_rate*time*I_conc;      ## Total intake of Iodine-131 by inhalation 
## Calculation
H = C0*(DCF*10**6);                    ## Using DCF in micro-curie
## Result
print'%s %.2f %s %.2f %s '%("  \n The dose commitment to thyroid = ",H,"" and " rem = ",H*1000," mrem \n");
 
 The dose commitment factor by inhalation =  1.48  rem/ucurie 

  
 The dose commitment to thyroid =  0.00  4.95  mrem 
 

Ex11-pg529

In [11]:
## Example 9.11
import math


## Given data
V_W = 2200.;                           ## Volume of water inatke in terms of cm^3/day
## 1 litre = 1000 gram(g)
M = 43.*1000.;                          ## Mass of water present in standard man according to standards
## Using the data from Table 9.13
MPD = 0.1/7.;                          ## Maximum Permissible Dose (MPD) in rem/day
## Using the data from Table 9.15
zeta = 0.01;                          ## Effective energy equivalent in MeV
q = 1.;                                ## The fraction of Tritium that goes inside by ingestion
T_b = 11.9;                           ## Biological Half life of Tritium in years
lambda_b = 0.693/T_b;                 ## Biological decay constant of Tritium in years^-1 

## As biological and radiological half lives are less than 50 year intake period, the exponential term (exp(-lambda_e*50)) is neglected
## Maximum Permissible Concentration(MPC) for a 7 day or 168 hour week tritium dose 
MPC_w_168 = (lambda_b*M*MPD)/(51.1*V_W*zeta*q);
print'%s %.2f %s'%("\n Maximum Permissible Concentration(MPC) for a 7 day or 168 hour week tritium dose for occupational purpose = ",MPC_w_168," uCi/cm^3 \n");
## The exposure at work is doubled for 40 hour week as compared to 168 hour week 
## For 40 hour week, with work of 5 days out of 7 day week according to a study
MPC_w_40 = MPC_w_168*2.*(7/5.);
print'%s %.2f %s'%("\n Maximum Permissible Concentration(MPC) for a 40 hour week tritium dose for occupational purpose = ",MPC_w_40," uCi/cm^3 \n");

## By analyzing the data of Table 9.13
## The whole body dose of general public is one tenth of the occupational purpose.
MPC_w_168_gp = MPC_w_168*0.1;
print'%s %.2f %s'%("\n Maximum Permissible Concentration(MPC) for a 7 day or 168 hour week tritium dose for general public = ",MPC_w_168_gp," uCi/cm^3 \n");
## The answer of Maximum Permissible Concentration(MPC) for a 168 hour week tritium dose for general public is given wrong in the textbook.
 Maximum Permissible Concentration(MPC) for a 7 day or 168 hour week tritium dose for occupational purpose =  0.03  uCi/cm^3 


 Maximum Permissible Concentration(MPC) for a 40 hour week tritium dose for occupational purpose =  0.09  uCi/cm^3 


 Maximum Permissible Concentration(MPC) for a 7 day or 168 hour week tritium dose for general public =  0.00  uCi/cm^3 

Ex12-pg534

In [12]:
## Example 9.12
import math


## Given data
no_home = 10**6;                               ## Number of houses
no_resident = 4.;                              ## Number of residents in a home
total_time = 50.;                              ## Number of years the analysis is carried out
radon_concn_old = 1.;                          ## Radon concentration in older uninsulated homes in terms of pCi/l
radon_concn_new = 6.;                          ## Radon concentration in modern insulated homes in terms of pCi/l
time = 3500.;                                  ## Time spent in home by a person per year in hours
eq_concn = 0.5;                               ## Equilibrium concentration of 50% 
## 1 year = 24*365 hours
X_increased = eq_concn*(radon_concn_new-radon_concn_old)*(time/(24.*365.));   ## The increased exposure of radon per person

## Using the data of Radon risk assessment of United States of America
## There are nearly 100 cases of cancer per 10^6 persons at 1 pCi-year dose.
## Calculation
no_cancer = (no_home*no_resident)*total_time*(100./10**6)*X_increased;
## Result
print'%s %.2f %s'%("\n Number of additional cases of cancer from insulation of home = ",no_cancer," \n");
## There is a slight deviation in the value given in the textbook. This is because of approximation of the number of additional cases of cancer in the textbook.
 Number of additional cases of cancer from insulation of home =  19977.17  

Ex13-pg535

In [13]:
## Example 9.13
import math


## Given data
H_ext = 3.;                                    ## External dose in rem
H_wbL = 5.;                                    ## Annual whole body dose limit in rem
## Using the data from Table 9.17
## Annual Limit Intake (ALI) for inhalation of Iodine-131 is 54uCurie (Ci)
ALI = 54.;
## Calculation
I = ALI*(1.-(H_ext/H_wbL));
## Result
print'%s %.2f %s'%("\n Amount of Iodine-131 intake within safety limits = ",math.ceil(I)," uCi \n");
 Amount of Iodine-131 intake within safety limits =  22.00  uCi