In [1]:

```
##Exa2.1: : Page-88 (2011)
#find Weight of one Curie of RaB and Weight of one Rutherford of RaB
T = 26.8*60; ## Half life of the substance, s
C = 3.7e+010; ## One curie, disintegration per sec
N = 6.022137e+026; ## Avogadro number, per kmol
m = 214.; ## Molecular weight of RaB, kg/kmol
R = 1e+006; ## One Rutherford, disintegration per sec.
W_C = C*T*m/(N*0.693); ## Weight of one Curie of RaB, Kg
W_R = R*T*m/(N*0.693); ## Weight of one Rutherford of RaB, Kg
print"%s %.2e %s %.2e %s "%("\nWeight of one Curie of RaB : ",W_C," Kg"and " \nWeight of one Rutherford of RaB : ",W_R," Kg");
## Result
## Weight of one Curie of RaB : 3.051e-011 Kg
## Weight of one Rutherford of RaB : 8.245e-016 Kg
```

In [2]:

```
## Exa2.2 : : Page 88 (2011)
#find The maximum activity of Na-24 and The time needed to produced 90 percent of the maximum activity
import math
T_h = 14.8; ## Half life of Na-24, hours
Q = 1e+008; ## Production rate of Na-24, per sec
L = 0.693/T_h; ## Decay constant, per sec
t = 2.; ## Time after the bombardment, hours
A = Q/3.7e+010*1000; ## The maximum activity of Na-24, mCi
T = -1*math.log(0.1)/L; ## The time needed to produced 90% of the maximum activity, hour
N = 0.9*Q*3600./L*math.e**(-L*t); ## Number of atoms of Na-24 left two hours after bombardment was stopped
print'%s %.2f %s %.2f %s %.2f %s '%("\nThe maximum activity of Na-24 = ",A," mCi"and "\nThe time needed to produced 90 percent of the maximum activity =",T," hrs" and"\nNumber of atoms of Na-24 left two hours after bombardment was stopped = ",N,"")
## Result
## The maximum activity of Na-24 = 2.7 mCi
## The time needed to produced 90 percent of the maximum activity = 49.2 hrs
## Number of atoms of Na-24 left two hours after bombardment was stopped = 6.30e+012
```

In [1]:

```
## Exa2.3: : Page 89 (2011)
#find The activity of K-40 and disintegrations
T = 1.31e+09*365*24*60*60; ## Half life of the substance,sec
N = 6.022137e+026; ## Avogadro number.
m = 0.35*0.012*10**-2; ## Mass of K-40, Kg.
A = m*N*0.693/(T*40); ## Activity of K-40, disintegrations/sec.
print'%s %.2e %s %.3f %s '%("\nThe activity of K-40 = ",A," "and "disintegrations/sec = ",A/3.7e+004," micro-curie");
## Result
## The activity of K-40 = 1.061e+004 disintegrations/sec = 0.287 micro-curie
```

In [4]:

```
## Exa2.4 : : Page 89 (2011)
#find The age of the boat
import math
T = 5568; ## Half life of the C-14,years
D = 0.693/T; ## Disintegration constant, years^-1.
N_0 = 15.6/D; ## Activity of fresh carbon, dpm .gm
N = 3.9/D; ## Activity of an ancient wooden boat,dpm.gm.
t = 1/(D)*math.log(N_0/N); ## Age of the boat, years
print'%s %.2e %s'%("\nThe age of the boat : ",t," years")
## Result
## The age of the boat : 1.114e+004 years
```

In [5]:

```
## Exa2.5 : : Page 90 (2011)
#find The activity of U-234
import math
m_0 = 3e-06;## Initial mass of the U-234, Kg
A = 6.022137e+026; ##Avagadro's number, atoms
N_0 = m_0*A/234.; ## Initial number of atoms
T = 2.50e+05; ## Half life, years
D= 0.693/T; ## Disintegration constant
t = 150000; ## Disintegration time, years
m = m_0*math.e**(-D*t); ## Mass after time t,Kg
activity = m*D/(365.*24.*60.*60.)*A/234.; ## Activity of U-234 after time t,dps
print'%s %.1f %s %.2f %s'%("\nThe activity of U-234 after ",t,""and" yrs = ",activity," disintegrations/sec");
## Result
## The activity of U-234 after 150000 yrs = 4.478e+005 disintegrations/sec
```

In [6]:

```
## Exa2.6 : : Page 90 (2011)
#find The number of alpha decays in Th-232
import math
A = 6.022137e+023; ##Avagadro's number, atoms
N_0 = A/232.; ## Initial number of atoms
t = 3.150e+07; ## Decay time, sec
D = 1.58e-018; ## Disintegration constant,sec^-1
N = D*t*N_0; ## Number of alpha decays in Th-232
print'%s %.2e %s'%("\nThe number of alpha decays in Th-232 = ", N,"");
## Result
## The number of alpha decays in Th-232 = 1.29e+011
```

In [7]:

```
## Exa2.7 : : Page 90 (2011)
#find The maximum possible age of the earth crust
import math
T_238 = 4.5e+09;## Half life of U-238, years
T_235 = 7.13e+08; ## Half life of U-238, years
lambda_238 = 0.693/T_238; ## Disintegration constant of U-238, years^-1
lambda_235 = 0.693/T_235; ## Disintegration constant of U-235, years^-1
N = 137.8; ## Abundances of U-238/U-235
t = math.log(N)/(lambda_235 - lambda_238);## Age of the earth's crust, years
print'%s %.2e %s'%("\nThe maximum possible age of the earth crust = ",t," years");
## Result
## The maximum possible age of the earth crust = 6.022e+009 years
```

In [8]:

```
## Exa2.8 : : Page 91 (2011)
#find The number of half lives in radon-222
import math
N = 10; ## Number of atoms left undecayed in Rn-222
n = math.log(10)/math.log(2); ## Number of half lives in Ra-222
print'%s %.2f %s'%("\nThe number of half lives in radon-222 = ", n,"");
## Result
## The number of half lives in radon-222 = 3.322
```

In [9]:

```
## Exa2.9 : : Page 91 (2011)
#find The weight of Po-210 after one year and The initial activity of the material
import math
M_Po = 209.9829; ## Mass of Polonium, g
M_Pb = 205.9745; ## Mass of lead, g
A = 6.22137e+023; ## Avogadro's number
M_He = 4.0026; ## Mass of alpha particle, g
C = 3e+08; ## Velocity of light, m/s
T = 138*24*3600; ## Half life, sec
P = 250; ## Power produced, joule/sec
Q = (M_Po-M_Pb-M_He)*931.25; ## disintegration energy, MeV
D = 0.693/T; ## Disintegration constant, per year
N = P/(D*Q*1.60218e-013); ## Number of atoms, atom
N_0 = N*math.e**(1.833); ## Number of atoms present initially, atom
W = N_0/A*210; ## Weight of Po-210 after one year, g
A_0 = N_0*D/(3.7e+010); ## Initial activity, curie
print'%s %.2f %s %.2e %s'%("\nThe weight of Po-210 after one year = ",W," g"and" \nThe initial activity of the material = ",A_0," curies");
## Result
## The weight of Po-210 after one year = 10.49 g
## The initial activity of the material = 4.88e+004 curies
```

In [10]:

```
## Exa2.10 : : Page 91 (2011)
#find The decay constant for total emission and The decay constant for beta_decay lambda_b and The decay constant for beta_decay lambda_a
import math
lambda_t = 0.693/(60.5*60);## Total decay constant, per sec
lambda_a = 0.34*lambda_t;## Decay constant for alpha_decay, per sec
lambda_b = 0.66*lambda_t;## Decay constant for beta_decay, per sec
print'%s %.2e %s'%("\nThe decay constant for total emission = ",lambda_t," /sec");
print'%s %.2e %s'%("\nThe decay constant for beta_decay lambda_b = ",lambda_b," /sec");
print'%s %.2e %s'%("\nThe decay constant for alpha_decay lambda_a = ",lambda_a," /sec");
## Result
## The decay constant for total emission = 1.91e-004 /sec
## The decay constant for beta_decay lambda_b = 1.26e-004 /sec
## The decay constant for alpha_decay lambda_a = 6.49e-005 /sec
```

In [11]:

```
## Exa2.13 : : Page 93 (2011)
#find The half life of Pu-239
import math
M_A = 4.; ## Mass of alpha particle, amu
M_U = 235.; ##Mass of U-235, amu
M_P = 239.; ## Mass of P-239, amu
Amount = 120.1; ## quantity of P-239, g
E_A = 5.144; ## Energy of emitting alpha particles, Mev
E_R = (2.*M_A)/(2.*M_U)*E_A; ## The recoil energy of U-235, Mev
E = E_R + E_A; ## The energy released per disintegration, Mev
P = 0.231; ## Evaporation rate, watt
D = P/(E*1.60218e-013); ## Disintegration rate, per sec
A = 6.022137e+023; ## Avagadro's number, atoms
N = Amount/M_P*A; ## Number of nuclei in 120.1g of P-239
T = 0.693/(D*3.15e+07)*N; ## Half life of Pu_239, years
print'%s %.2e %s'%("\nThe half life of Pu-239 = ",T," years");
## Result
## The half life of Pu-239 = 2.42e+004 years
```

In [12]:

```
## Exa2.14 : : Page 93 (2011)
#find The disintegration rate of Au-199
import math
T_h_1 = 2.7*24*3600; ## Half life of Au-198, sec
T_h_2 = 3.15*24*3600; ## Half life of Au-199, sec
S_1 = 99e-028; ## Crossection for first reaction, Sq.m
S_2 = 2.6e-024; ## Crossection for second reaction, Sq.m
I = 1e+018; ## Intensity of radiation, per Sq.m per sec
L_1 = I*S_1; ## Decay constant of Au-197, per sec
L_2 = 0.693/T_h_1+I*S_2; ## Decay constant of Au-198, per sec
L_3 = 0.693/T_h_2; ## Decay constant of Au-199, per sec
N_0 = 6.022137e+023; ## Avogadro number
N_1 = N_0/197.; ## Initial number of atoms of Au-197
t = 30.*3600.; ## Given time, sec
p = (math.exp(-L_1*t))/((L_2-L_1)*(L_3-L_1));
q = (math.exp(-L_2*t))/((L_1-L_2)*(L_3-L_2));
r = (math.exp(-L_3*t))/((L_1-L_3)*(L_2-L_3));
N3 = N_1*L_1*L_2*(p+q+r);
N_199 = N3;
L = L_3*N_199; ## Disintegration rate of Au-199, per sec
print'%s %.2e %s'%("\nThe disintegration rate of Au-199 = ", L,"");
## Result
## The disintegration rate of Au-199 = 1.9e+012 (Wrong answer in the textbook)
```

In [13]:

```
## Exa2.15 : : Page 94 (2011)
#find The maximum activity of Na-24 and The activity after a continuous bombardment
import math
Y = 110e-03;## Yield of Na-24, mCi/hr
T = 14.8;## Half life of Na-24, hours
t = 8;## Time after which activity to be compute, hours
D = 0.693/T;## Disintegration constant, hours^-1
A = 1.44*Y*T;## Maximum activity of Na-24, Ci
A_C = A*(1-math.e**(-D*t));## Activity after a continuous bombardment, Ci
Activity = A_C*(math.e**(-D*t));## Activity after 8hours, Ci
print'%s %.2f %s %.2f %s %.2f %s '%("\nThe maximum activity of Na-24 = ",A," Ci"and "\nThe activity after a continuous bombardment = ",A_C," Ci"and "\nThe activity after 8hours = ",Activity," Ci")
## Result
## The maximum activity of Na-24 = 2.344 Ci
## The activity after a continuous bombardment = 0.7324 Ci
## The activity after 8hours = 0.50360 Ci
```

In [14]:

```
## Exa2.16 : : Page 94 (2011)
#find The energy of beta rays emitted per gram of tissue
import math
A_0 = 3.7e+07; ## Initial activity, disintegrations per sec
T = 12.6; ## Half life of I-130, hours
t = 24*3600; ## time for dose absorbed calculation,sec
E = 0.29*1.6e-06; ## Average energy of beta rays, ergs
m = 2; ## Mass of iodine thyroid tissue, gm
l = 0.693/(T*3600); ## Disintegration constant, sec^-1
N_0 = A_0/l; ## Initial number of atoms
N = N_0*(1-math.e**(-l*t)); ## Number of average atoms disintegrated
E_A = N*E; ## Energy of beta rays emitted, ergs
E_G = E_A/(2*97.00035); ## Energy of beta rays emitted per gram of tissue, REP
print'%s %.2f %s'%("\nThe energy of beta rays emitted per gram of tissue = ",E_G," REP");
## Result
## The energy of beta rays emitted per gram of tissue = 4245.0 REP
```

In [15]:

```
## Exa2.18 : : Page 95 (2011)
#find The activity of Au-198 and The maximum amount of Au-198 produced
import math
N_0 = 6.022137e+023; ## Avagadro number
d = 0.02; ## Thickness of the foil, cm
R = 19.3; ## Density of Au,g/cc
N_1 = d*R/197.*N_0; ## Initial number of Au-197 nuclei per unit area of foil,cm^-2
T_H = 2.7*24.*3600.; ## Half life of Au-198,sec
L = math.log(2)/T_H; ## Decay constant for Au-198,sec^-1
I = 10**12; ## Intensity of neutron beam,neutrons/cm^2/sec
S = 97.8e-024; ## Cross section for reaction,cm^-2
t = 5*60.; ## Reaction time,s
A = S*I*N_1*(1-math.e**(-L*t)); ## Activity of Au-198,cm^-2sec^-1
N_2 = S*I*N_1/L; ## The maximum amount of Au-198 produced,cm^-2
print'%s %.2e %s %.2e %s '%("\nThe activity of Au-198 = ",A," per Sq.cm per sec"and "\nThe maximum amount of Au-198 produced = ",N_2," per Sq.cm");
## Result
## The activity of Au-198 = 1.028e+008 per Sq.cm per sec
## The maximum amount of Au-198 produced = 3.88e+016 per Sq.cm
```

In [16]:

```
## Exa2.19 : : Page 95 (2011)
#find The power released by the gm molecule of Pu-238 and The time in which power reduces to 1/8 time of its initial value
import math
N_0 = 6.022137e+023; ## Avagadro number
T_P = 90.*365.*24.*3600.; ## Half life of Pu-238,s
L_P = 0.693/T_P ; ## Decay constant of Pu-238,s^-1
E = 5.5; ## Energy of alpha particle, MeV
P =E*L_P*N_0; ## Power released by the gm molecule of Pu-238,MeV/s
t = math.log(8)/(L_P*365.*24.*3600.); ## Time in which power reduces to 1/8 time of its initial value
print'%s %.2e %s %.1f %s '%("\nThe power released by the gm molecule of Pu-238 = ",P," MeV/s"and "\nThe time in which power reduces to 1/8 time of its initial value = ",t," yrs")
## Result
## The power released by the gm molecule of Pu-238 = 8.09e+014 MeV/s
## The time in which power reduces to 1/8 time of its initial value = 270 yrs
```

In [17]:

```
## Exa2.20 : : Page 96 (2011)
#find The required time for decay of daughter nucleus and The number of nuclei of daughter isotope
import math
N_1 = 10**20; ## Number of nuclei of parent isotopes
T_P = 10**4; ## Half life of parent nucleus,years
T_D = 20.; ## Half life of daughter nucleus,years
T = 10**4; ## Given time,years
L_P = 0.693/T_P ; ## Decay constant of parent nucleus,years^-1
L_D = 0.693/T_D ; ## Decay constant of daughter nucleus,years^-1
t_0 = math.log(0.03)/(L_P-L_D); ## Required time for decay of daughter nucleus,years
N = L_P/L_D*(math.e**(-L_P*T)-math.e**(-L_D*T))*N_1; ## Number of nuclei of daughter isotope
print'%s %.2f %s %.0e %s'%("\nThe required time for decay of daughter nucleus =",t_0," yr"and " \nThe number of nuclei of daughter isotope = ",N," ");
## Result
## The required time for decay of daughter nucleus = 101 yr
## The number of nuclei of daughter isotope = 1e+017
```