In [1]:

```
#Variable declaration
Bq = 1./3.7e+010; # Number of curie in one Bq, Ci
#Calculations
N = 10**10*Bq; # The number of curie in 10^10 Bq, Ci
#Result
print "The number of curie in 10^10 Bq : %4.2f Ci"%N
```

In [2]:

```
#Variable declaration
lambda_232 = 1.58e-018; # Decay constant, s^-1
N = 2.596e+022; # Number of atoms in 10g Th-232
#Calculations
A = N*lambda_232; # The activity of 10g of Th-232, dps
#Result
print "The activty of 10g of Th-232 : %5.3e dps"%A
```

In [3]:

```
#Variable declaration
A = 3.7e+010; # Activity of 1Ci sample, dps
t = 1608; # Half life of radioactive substance, s
#Calculations
N = 6.023e+023/214; # Number of atoms in 1g of substance having atomic mass 214
lamda = 0.6931/t; # Decay constant, s^-1
m = A/(lamda*N); # The mass of radoiactive substance, g
#Result
print "nThe mass of radioactive substance : %4.2e g"%m
```

In [6]:

```
#Variable declaration
t = 1.419e+017; # Half life of U-238, s
N = 6.023e+023/238; # Number of atoms in 1g of U-238
#Calculations
lamda = 0.6931/t; # Decay constant, s^-1
A = (lamda*N)*1000/(3.7e+010); # The activity of 1kg of U-238, Ci
#Result
print "The activity of 1kg of U-238 : %4.2e Ci"%A
```

In [22]:

```
import math
#Variable declaration
t_h = 1600; # Half life of radioactive material, years
t = 2000; # Totaltime, years
#Calculations
lamda = 0.6931/t_h; # Decay constant, years^-1
m0 = 1; # The mass of radioactive substance at t0, mg
m = m0* math.exp(-(lamda*t)); # Ratio of total number of atoms and number of atoms disintegrat, mg
A = 1-m; # The amount of radioactive substance decayed, mg
#Result
print "The amount of radioactive substance decayed : %6.4f mg"%A
```

In [7]:

```
import math
#Variable declaration
t = 10; # Total period of radioactive material, days
#Calculations
lamda = math.log(6.6667)/10; #Decay constant, day^-1
t_h = 0.6931/(lamda); # Half life of radioactive substance, days
#Result
print "The half life of radioactive substance : %4.2f days"%t_h
```

In [8]:

```
#Variable declaration
t_h = 1620*31536000; # Half life of Ra-226, S
D = 0.6931/t_h; # Decay constant, S^-1
A_Ci = 3.7e+010; # Activity, Ci
#Calculations
N_Ci = A_Ci/D; # Number of atoms decayed
m = 0.226; # Mass of 6.023e+023 atoms, kg
M_Ci = m*N_Ci/6.023e+023; # Mass of 1-Ci sample of Ra-226, kg
A_rf = 10**6; # Activity, Rf
N_rf = A_rf/D; # Number of atoms decayed
M_rf = m*N_rf/6.023e+023; # Mass of 1-Rf sample of Ra-226, kg
#Result
print "Mass of 1-Ci sample of Ra-226 = %5.3e kg and \n Mass of 1-Rf sample of Ra-226 = %4.2e kg "%(M_Ci, M_rf )
```

In [11]:

```
import math
#Variable declaration
N_o = 7.721e+018; # Number of atoms in 3 mg of U-234
t_h = 2.5e+05; # Half life of U-234, years
T = 150000; # Total time, years
#Calculations
lamda = 0.6931/t_h; # Decay constant, year^-1
N = N_o*(math.exp(-lamda*T)); # Number of atoms left after T years
m = 234000; # Mass of 6.023e+023 atoms of U-234, mg
M = m*N/(6.023e+023); # Weight of sample left after t years,
L = 8.8e-014; # Given decay constant, S^-1
A = N*L*10**6/(3.7e+010); # Activity, micro Ci
#Result
print "The weight of sample = %5.3f mg \n Activity = %5.2f micro Ci "%(M, A)
```

In [12]:

```
#Variable declaration
N = 6.324e+020; # Number of atoms in 4.2e-05 kg of K-40
#Calculations
t_h = 1.31e+09*31536000; # Half life of K-40, s
D = 0.693/t_h; # Decay constant, s^-1
A = N*D/(3.7e+010)*10**6; # Activity of K-40, microCi
#Result
print "The activity of K-40 : %5.3f micro Ci"%A
```

In [13]:

```
#Variable declaration
N = 2.87e+019; # Number of atoms in 10e-10kg of Po-210
t_h = 138*24*3600; # Half life of Po-210, s
#Calculations
D = 0.693/t_h; # Decay constant, s^-1
A = N*D; # Activity of K-40, dps
E = 5.3*1.6e-013; # Power produce by one dps, MeV
P = A*E; # Power produced by 1.667e+012 dps, W
#Result
print "The Power produced by 1.667e+012 dps : %3.1f W"%P
```

In [14]:

```
#Variable declaration
# Declare three cells (for three reactions)
R1 = [[0,0,0],[0,0,0],[0,0,0],[0,0,0]]
R2 = [[0,0,0],[0,0,0],[0,0,0],[0,0,0]]
R3 = [[0,0,0],[0,0,0],[0,0,0],[0,0,0]]
# Enter data for first cell (Reaction)
R1[0][0] = "Pb";
R1[0][1] = 82;
R1[0][2] = 211;
R1[1][0] = 'Bi';
R1[1][1] = 83;
R1[1][2] = 211;
R1[2][0] = 'Tl';
R1[2][1] = 81;
R1[2][2] = 207;
R1[3][0] = 'Pb';
R1[3][1] = 82;
R1[3][2] = 207;
# Enter data for second cell (Reaction)
R2[0][0] = "U";
R2[0][1] = 92;
R2[0][2] = 238;
R2[1][0] = 'Th';
R2[1][1] = 90;
R2[1][2] = 234;
R2[2][0] = 'Pa';
R2[2][1] = 91;
R2[2][2] = 234;
R2[3][0] = 'U';
R2[3][1] = 92;
R2[3][2] = 234;
# Enter data for third cell (Reaction)
R3[0][0] = "Bi";
R3[0][1] = 83;
R3[0][2] = 211;
R3[1][0] = 'Pa';
R3[1][1] = 84;
R3[1][2] = 211;
R3[2][0] = 'Pb';
R3[2][1] = 82;
R3[2][2] = 207;
#Calculations&Results
# Declare a function returning the type of particle emitted
def identify_particle(d_Z, d_A):
if d_Z == 2 & d_A == 4:
particle = "Alpha";
elif d_Z == -1 & d_A == 0:
particle = "Beta minus";
elif d_Z == 1 & d_A == 0:
particle = "Beta plus";
# Display emitted particles for first reaction
print "\nReaction-I:"
for i in range(0,3):
dZ = R1[i][1]-R1[i+1][1];
dA = R1[i][2]-R1[i+1][2];
p = identify_particle(dZ,dA);
print "%s(%d) - (%s) --> %s(%d)"%(R1[i][0], R1[i][1], p, R1[i+1][0], R1[i+1][1]);
# Display emitted particles for second reaction
print "\n\nReaction-II:"
for i in range(0,3):
dZ = R2[i][1]-R2[i+1][1];
dA = R2[i][2]-R2[i+1][2];
p = identify_particle(dZ,dA);
print "%s(%d) - (%s) --> %s(%d)"%(R2[i][0],R2[i][1], p, R2[i+1][0], R2[i+1][1]);
# Display emitted particles for third reaction
print "\n\nReaction-III:"
for i in range(0,2):
dZ = R3[i][1]-R3[i+1][1];
dA = R3[i][2]-R3[i+1][2];
p = identify_particle(dZ,dA);
print "%s(%d) - (%s) --> %s(%d)"%(R3[i][0], R3[i][1], p, R3[i+1][0], R3[i+1][1]);
```

In [16]:

```
#Variable declaration
M_U = 238.050786; # Atomic mass of U-238, amu
M_Pb = 205.9744550; # Atomic mass of Pb-205, amu
M_He = 4.002603; # Atomic mass of He-4, amu
M_e = 5.486e-04; # Atomic mass of electron, amu
#Calculations
M = M_Pb+(8*M_He)+(6*M_e); # Total mass of products, amu
D = M_U-M; # Decrease in mass, amu
E = D*931.47; # Energy evolved, MeV
#Results
print "Total mass of products = %1.7f amu \n Decrease in mass = %9.7f amu and \n Energy evolved = %4.1f MeV"%(M, D, E)
```

In [17]:

```
#Variable declaration
# Declare cell (for given reaction)
R1 = [[0,0,0,0],[0,0,0,0],[0,0,0,0],[0,0,0,0],[0,0,0,0]];
# Enter data for cell (Reaction-I)
R1[0][0] = "A";
R1[0][1] = 90;
R1[0][2] = 238;
R1[0][3] = "Alpha";
R1[1][0] = 'B';
R1[1][3] = "Beta minus";
R1[2][0] = 'C';
R1[2][3] = "Alpha";
R1[3][0] = 'D';
R1[3][3] = "Beta minus";
R1[4][0] = 'E';
# Declare a function returning the type of particle emitted
def daughter_nucleus(particle_emitted):
if particle_emitted == "Alpha":
Z = 2
A = 4;
elif particle_emitted == "Beta minus":
Z = -1
A = 0;
elif particle_emitted == "Beta plus":
Z = 1
A = 0;
return Z,A
# Display emitted particles for first reaction
print "\n\nReaction-I:"
for i in range(0,4):
[Z, A] = daughter_nucleus(R1[i][3]);
R1[i+1][1] = R1[i][1]-Z;
R1[i+1][2] = R1[i][2]-A;
print "%s(%d,%d) - (%s) --> %s(%d,%d)"%(R1[i][0], R1[i][1], R1[i][2], R1[i][3], R1[i+1][0], R1[i+1][1], R1[i+1][2])
```

In [18]:

```
import math
#Variable declaration
D = math.log(2); # Decay constant, s^-1
t = math.log(100); # Half life, s
#Calculations
n = t/D; # Number of half-lives
#Result
print "Number of half-lives : %4.2f "%n
```

In [20]:

```
#Variable declaration
H_t = 60.5*60; # Total half life period, s
#Calculations
T_d = 0.693/H_t; # Total decay constant, s^-1
A_d = 34./100*T_d; # Decay constant for alpha decays, s^-1
B_d = 66./100*T_d; # Decay constant for beta decay, s^-1
#Result
print "Alpha decay = %4.2e s^-1 \n Beta decay = %4.2e s^-1"%(A_d, B_d)
```

In [21]:

```
#Variable declaration
A_r = 1.8e+04; # Atomic ratio of U(92,238) and U(92,234)
T_238 = 2.5e+05; # Half life of U(92,238), years
#Calculations
T_234 = A_r*T_238; # Half life of U(92,234), years
#Result
print "Half life of U(92,234): %3.1e years"%T_234
```

In [23]:

```
#Variable declaration
M_239 = 239.052158; # Atomic mass of Pu-239, amu
M_235 = 235.043925; # Atomic mass of U-235, amu
M_4 = 4.002603; # Atomic mass of He-4, amu
#Calculations
Q = (M_239-M_235-M_4)*931.47; # Difference in masses, MeV
A = 241; # Mass number
K_alpha = Q*(A-4)/A; # Kinetic energy of alpha particle, MeV
#Result
print "Kinetic energy of alpha particle %5.2f MeV"%K_alpha
```

In [24]:

```
import math
#Variable declaration
Z = 88; # Atomic number of Ra-226 nucleus,
A = 226; # Atomic mass of Ra-226 nucleus
R_0 = 1.3e-015; # Distance of closest approach, m
E_0 = 8.854e-012; # Permittivity of free space, C^2/Nm^2
e = 1.6e-019; # Charge of an electron, C
#Calculations
B = 2/(1.6e-013)*(Z-2)*e**2/(4*math.pi*E_0*R_0*A**(1./3)); # The barrier height faced by alpha particle, MeV
#Result
print "The barrier height faced by alpha particle : %4.1f MeV"%B
```

In [25]:

```
import math
#Variable declaration
Z_1 = 2; #Atomic number of He-4,
Z_2 = 7; # Atomic number of N-14,
A_1 = 4; # Atomis mass of He-4 nucleus
A_2 = 14; # Atomic mass of N-14 nucleus
R_0 = 1.5e-015; # Distance of closest approach, m
E_0 = 8.854e-012; # Permittivity of free space, C**2/Nm**2
e = 1.6e-019; # Charge of an electron, C
#Calculations
B = Z_1/(1.6e-013)*Z_2*e**2/(4*math.pi*E_0*R_0*(A_1**(1./3)+A_2**(1./3))); # The coulomb barrier faced by alpha particle, MeV
#Result
print "The coulomb barrier faced by alpha particle : %4.2f MeV"%B
```

In [26]:

```
#Variable declaration
R_0 = 1.2; # Distance of closest approach, m
#Calculations
E_b = 197/(R_0*137); # The K.E. of proton to penetrate the berrier of H nucleus, Mev
#Result
print "The K.E. of proton to penetrate the berrier of H nucleus : %3.1f MeV"%E_b
```

In [27]:

```
#Variable declaration
M_C = 14.007685; # Mass of C-14 nucleus, amu
E_e = 0.156/931.47; # Kinetic energy of emitted electron, amu
#Calculations
M_N = M_C-E_e; # Mass of N-14 nucleus, amu
#Result
print "Mass of N-14 nucleus : %9.6f amu"%M_N
```

In [28]:

```
#Variable declaration
N_p = 6.70e+033;# Number of protons
T_p = 10**32; # Mean life of proton, years
#Calculations
D_p = N_p/T_p*0.5; # Number of proton decays per year, decays/year
#Result
print "Number of proton decays per year,: %4.1f decays/year"%D_p
```

In [29]:

```
#Variable declaration
E_2 = 2505; # Second excited state of Ni-60, KeV
E_1 = 1332; # First excited state of Ni-60, KeV
E_0 = 0; # Ground state of Ni-60 , KeV
#Calculations
E_G_2 = E_2-E_1; # Energy of gamma rays emitted when transition from 2 to 1, KeV
E_G_1 = E_1-E_0; # Energy of gamma rays emitted when transition from 1 to 0, KeV
#Result
print "Energies of two gamma rays emitted : %d KeV and %d KeV"%(E_G_2, E_G_1)
```

In [30]:

```
#Variable declaration
E = 662; # Energy available with the nucleus, KeV
I_b_K = 37.4; # Binding energy for K-shell, KeV
I_b_L = 6.0; # Binding energy for L-shell, KeV
#Calculations
E_c_K = E-I_b_K; # Energy conversion for K-shell, KeV
E_c_L = E-I_b_L; # Energy conversion for L-shell, KeV
#Results
print "Energies conversion for K and L-shell electrons : %5.1f KeV and %d KeV"%(E_c_K, E_c_L)
```

In [31]:

```
import math
#Variable declaration
t_h = 4.5e+09; # Half life of mineral, years
#Calculations
D_c = 0.6931/t_h; # Decay constant of minerals, years^-1
N_1 = 6.023e+023/238; # Number of nuclei in 1g of Uranium
N = 6.023e+023*0.093/206; # Number of nuclei in 0.093g of lead
t = math.log(1+N/N_1)/D_c; # Age of the mineral, years
#Result
print "Age of the mineral : %6.4e years "%t
```

In [33]:

```
import math
#Variable declaration
t_h = 5760; # Half life of boat, years
D_c = 0.6931/t_h; # Decay constant of boat, years^-1
N_1 = 16.; # Number of atoms decay per min. per gram initially
N = 5; # Number of atoms decay per min per gram presently
#Calculations
t = math.log(N_1/N)*1/D_c; # Age of the boat, years
#Result
print "Age of the boat : %d years "%t
```

In [34]:

```
import math
#Variable declaration
t_h = 24000*365*24*3600; # Half life of Pu-239, s
#Calculations
D_c = 0.6931/t_h; # Decay constant of Pu-239, s^-1
N = 6.023e+023*10/239; # Number of nuclei at t = 0, nuclei
A_0 = D_c*N; # Initial activity, disintegrations/sec
A = 0.1; # Activity after time t, disintegrations/sec
t = math.log(A_0/A)*1/D_c; # Age of the Pu-239, years
#Results
print "The number of nuclei at t = 0, = %4.2e nuclei \nInitial activity = %4.2e disintegrations/s and \nAge of Pu-239 = %4.2e years "%(N, A_0, t)
```