## Example 3.2.1, Page 124¶

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

The number of curie in 10^10 Bq : 0.27 Ci


## Example 3.2.2, Page 125¶

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

The activty of 10g of Th-232 : 4.102e+04 dps


## Example 3.3.3, Page 125¶

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

nThe mass of radioactive substance : 3.05e-08 g


## Example 3.2.4, Page 125¶

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

The activity of 1kg of U-238 : 3.34e-04 Ci


## Example 3.2.5, Page 126¶

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

The amount of radioactive substance decayed : 0.5795 mg


## Example 3.2.6, Page 127¶

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

The half life of radioactive substance  : 3.65 days


## Example 3.2.7, Page 127¶

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 )

Mass of 1-Ci sample of Ra-226   = 1.023e-03 kg and
Mass of 1-Rf sample of Ra-226  =  2.77e-08 kg


## Example 3.2.8, Page 128¶

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)


The weight of sample  =  1.979 mg
Activity   =  12.12 micro Ci


## Example 3.2.9, Page 129¶

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

The activity of K-40 : 0.287 micro Ci


## Example 3.2.10, Page 130¶

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

The Power produced by 1.667e+012 dps : 1.4 W


## Example 3.3.1, Page 131¶

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]);

Reaction-I:
Pb(82) - (None) --> Bi(83)
Bi(83) - (None) --> Tl(81)
Tl(81) - (None) --> Pb(82)

Reaction-II:
U(92) - (None) --> Th(90)
Th(90) - (None) --> Pa(91)
Pa(91) - (None) --> U(92)

Reaction-III:
Bi(83) - (None) --> Pa(84)
Pa(84) - (None) --> Pb(82)


## Example 3.3.2, Page 132¶

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)

Total mass of products  = 237.9985706 amu
Decrease in mass   =  0.0522154 amu and
Energy evolved   = 48.6 MeV


## Example 3.4.1, Page 133¶

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])



Reaction-I:
A(90,238) - (Alpha) --> B(88,234)
B(88,234) - (Beta minus) --> C(89,234)
C(89,234) - (Alpha) --> D(87,230)
D(87,230) - (Beta minus) --> E(88,230)


## Example 3.4.2, Page 133¶

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

Number of half-lives : 6.64


## Example 3.4.3, Page 133¶

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)

Alpha decay   =  6.49e-05 s^-1
Beta decay    = 1.26e-04 s^-1


## Example 3.4.4, Page 134¶

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

Half life of U(92,234): 4.5e+09 years


## Example 3.5.2, Page 135¶

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

Kinetic energy of alpha particle  5.16 MeV


## Example 3.5.3, Page 136¶

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

The barrier height faced by alpha particle : 31.2 MeV


## Example 3.5.4, Page 136¶

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

The coulomb barrier  faced by alpha particle : 3.36 MeV


## Example 3.5.5, Page 137¶

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

The K.E. of proton to penetrate the berrier of H nucleus : 1.2 MeV


## Example 3.6.1, Page 138¶

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

Mass of N-14 nucleus : 14.007518 amu


## Example 3.6.3, Page 139¶

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

Number of proton decays per year,: 33.5 decays/year


## Example 3.7.1, Page 141¶

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)

Energies of two gamma rays emitted : 1173 KeV and 1332 KeV


## Example 3.7.2, Page 141¶

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)

Energies conversion for K and L-shell electrons : 624.6 KeV and 656 KeV


## Example 3.9.1, Page 143¶

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

Age of the mineral : 6.6261e+08 years


## Example 3.9.2, Page 145¶

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

Age of the boat : 9666 years


## Example 3.9.4, Page 145¶

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)

The number of nuclei at t = 0,  = 2.52e+22 nuclei
Initial activity  = 2.31e+10 disintegrations/s and
Age  of Pu-239  = 2.86e+13 years