Nuclear Engineering

Theory of Successive Transformation and Radioactive Equilibrium

Any radioactive atom decays to its daughter atom following its characteristics decay constant. The daughter atom, if radioactive, decays to third generation and in this manner a chain of decay propagates, which eventually stops by the formation of a stable nucleus. Let us consider a radioactive nuclide, symbolized by a subscript p (Parent nucleus) to decay into another radioactive nuclide (subscript d: daughter nucleus); which again decays to a stable end product (subscript 3). Let λ_p, λ_d be the decay constants of nuclides p and d respectively, and N_p, N_d and N₃ be the number of atoms of the three kinds at any instant. The second nuclide would be formed at the rate of λ_pN_pby the decay of the parent atom, and would decay at the rate of λ_dN_d. In the same manner, the atoms of the end product will appear at the rate λ_dN_d by the decay of nuclide, but being stable will not decay. Therefore,¹



Solving Eq. (m4.13), we have

where, is the number of atoms of nuclide p at t = 0. From (m4.14) using (m4.16), we have


Multiplying both sides by and then integrating, we get

where, A is the constant of integration. Now, at t =0,N_d = N⁰_d = a constant. So, from Eqn. (m4.17),


From above equations,

where, B is the integration constant. Now at t =0,N₃ = N⁰₃, so that we have from Eqn. 6.17,

In practical cases, as only the first nuclide is present at t = 0, we can put N⁰_d = N⁰₃ and from Eqns. 6.16 to 6.18, we get


The decay of the parent nuclide and the growth and decay of the daughter nuclide are shown in Fig. m4.2.

FIGURE m4.2 Decay of the parent nuclide and growth and decay of the first daughter nuclide with time
Example m4.2: Show that if N⁰_p be the number of parent atoms initially and γ_p and γ_dthe disintegration constants of parent and daughter respectively, the time at which the number of daughter atoms is maximum is given by

Solution:
From the relation m4.22, we get, , where, = No. of daughter atoms at any time t. For to be maximum, the condition is = 0
⇒ = 0. Denoting t as tm corresponding to maxima, we can write



Radioactive Equilibrium
When we consider a radioactive series, it is important to understand the equilibrium conditions of all the nuclei in the respective chain. Applying the mathematical condition of equilibrium, i.e., time derivative of any quantity is equal to zero, we can write from

Where 1, 2, —- n and represents parent, daughter and so on upto n^th nucleus. It is clear from Eqn. (m4.25) that the first equation is not valid as it implies λ₁ to be zero. Therefore, the mathematical condition of equilibrium cannot be applied here rigorously. However, some conditions can be set in to attain a state very close to equilibrium. Depending on the conditions, two equilibrium states can be defined:

Secular Equilibrium (λ₁<< other λ’s)
In this case the first equation of 6.22 can be considered to certain approximation and hence we can write,

or
As in this case λ₁ 0 and also λ₁<<λ₂, 0. Therefore, from Eqn. m4.2

After a long time , and from Eqn. m4.26, it can be written as

This is known as secular equilibrium.

Transient Equilibrium (λ₁<λ₂ but (T_1/2)₁ is not very long)

In this case λ₁ ^~ 0 does not hold good. When t becomes large, , Therefore from Eqn. 6.19

This indicates that daughter decays with the same half life of the parent. This is basically the condition of transient equilibrium. From equations m4.1 and m4.28, we can write


Example m4.3: Calculate the amount of Ra-226 in secular equilibrium with 5 kg of pure U-238, given the half life of ²²⁶Ra and ²³⁸U as 1620 years and 4.5 x 10⁹ years respectively.
Solution:
Let x gm be the required amount of ²²⁶Ra in secular equilibrium with 5 kg of pure ²³⁸U. Now,
5 kg of pure U-238 = atoms of U
x gm of Ra-226 = atoms of Ra
From the condition of secular equilibrium

x = 1.7 mg

Nuclear Engineering

Nuclear Engineering

Curriculum

Theory of Successive Transformation and Radioactive Equilibrium

Modal title