Successive Radioactive Disintegration (Decay Chains): Setting up and solving differential equations for radioactive series decay ($A \rightarrow B \rightarrow C$) to find the time $t_{\max}$ at which the concentration of the intermediate nuclide $B$ becomes maximu

Successive Radioactive Disintegration — A → B → C

📘 Basic Concept

Consider a radioactive decay chain:

A → B → C

Here A is the parent nuclide, B is the intermediate nuclide and C is the final product. Nuclide A produces B, while B itself undergoes radioactive decay to produce C.

SUCCESSIVE RADIOACTIVE DECAY

A

B

C

→

→

Parent → Intermediate → Final Product

Important:
B is continuously produced by decay of A and simultaneously destroyed by its own decay. Therefore, the number of B nuclei first increases, reaches a maximum and then decreases.

🧮 Differential Equations

Decay of A

dN_A/dt = −λ_AN_A

Solving:

N_A(t) = N₀e^−λ_At

Production and Decay of B

B is produced due to decay of A at the rate:

Production rate = λ_AN_A

B disappears due to its own decay at the rate:

Decay rate = λ_BN_B

Therefore:

dN_B/dt = λ_AN_A − λ_BN_B

Substituting N_A:

dN_B/dt + λ_BN_B = λ_AN₀e^−λ_At

📐 Complete Solution for N_B(t)

For the initial condition:

N_B(0)=0

The solution is:

N_B(t) = [λ_AN₀/ (λ_B−λ_A)] [e^−λ_At −e^−λ_Bt]

This is the standard Bateman equation for the intermediate daughter nuclide when the initial amount of B is zero.

⭐ Derivation of t_max

At maximum concentration of B:

dN_B/dt = 0

From the differential equation:

λ_AN_A = λ_BN_B

Using the expression for N_B:

λ_Ae^−λ_At = λ_Be^−λ_Bt

Therefore:

e^{(λ_B−λ_A)t} = λ_B/λ_A

Taking natural logarithm:

(λ_B−λ_A)t = ln(λ_B/λ_A)

Hence:

t_max = ln(λ_B/λ_A) / (λ_B−λ_A)

Final Formula:

For A → B → C with initially zero B:

t_max = ln(λ_B/λ_A) / (λ_B−λ_A)

⏱️ Formula in Terms of Half-Life

Since:

λ = ln2/T_1/2

Therefore:

t_max = [T_AT_B/ (T_B−T_A)] ln(T_A/T_B)

This form is especially useful when the problem gives half-lives instead of decay constants.

⚠️ Important Special Case

If λ_A = λ_B

The normal formula contains a 0/0 form and cannot be used directly. The limiting solution must be taken.

N_B(t) = λN₀t e^−λt

For maximum:

t_max = 1/λ

Physical meaning:

B reaches maximum when its production rate from A becomes exactly equal to its own decay rate.

🎯 Fully Solved Numerical Example

Question:

A radioactive nucleus A has decay constant λ_A=0.1 min⁻¹. It decays into B, which has decay constant λ_B=0.2 min⁻¹. Initially there is no B. Find the time at which B becomes maximum.

Given:
λ_A=0.1 min⁻¹
λ_B=0.2 min⁻¹

Use:
t_max = ln(λ_B/λ_A) / (λ_B−λ_A)

Therefore:
t_max = ln(0.2/0.1)/(0.2−0.1)

= ln2/0.1

≈0.693/0.1

t_max ≈ 6.93 min

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🚀 Quick Revision

N_A=N₀e^−λ_At

dN_B/dt = λ_AN_A −λ_BN_B

N_B = λ_AN₀ / (λ_B−λ_A) [ e^−λ_At −e^−λ_Bt ]

t_max = ln(λ_B/λ_A) / (λ_B−λ_A)

Golden Rule:

At the maximum concentration of intermediate B:

Rate of production of B = Rate of decay of B

i.e.
λ_AN_A = λ_BN_B