Question

Sometimes a radioactive nucleus decays into a nucleus which itself is radioactive. An example is :

\[\ce{^38Sulphur ->[half-life][= 2.48h] ^{38}Cl ->[half-life][= 0.62h] ^38Air (stable)}\]

Assume that we start with 1000 ³⁸S nuclei at time t = 0. The number of ³⁸Cl is of count zero at t = 0 and will again be zero at t = ∞ . At what value of t, would the number of counts be a maximum?

Answer 1

\[\ce{^38S ->[][2.48 h] ^38Cl ->[][0.62h] 38Ar}\]

At time t, Let ³⁸S have N₁(t ) active nuclei and ³⁸Cl have N₂(t) active nuclei.

`(dN_1)/(dt) = -λ_1N_1` = rate of formation of Cl³⁸.

Also `(dN_2)/(dt) = - λ_1N_2 + λ_1N_1`

But `N_1 = N_0e^(-λ_1t)`

`(dN_2)/(dt) = - λ_1 N_0e^(-λ_1t) - λ_2N_2`

Multiplying by `e^(λ_2t) dt` and rearranging

`e^(λ_2t) dN_2 + λ_2N_2e^(λ_2t) dt = λ_1N_0e^((λ_2 - λ_1)t) dt`

Integrating both sides.

`N_2e^(λ_2t) = (N_0λ_1)/(λ_2 - λ_1) e^((λ_2 - λ_1)t) + C`

Since at t = 0, N₂ = 0, C = `-(N_0λ_1)/(λ_2 - λ_1)`

∴ `N_2e^(λ_2t) = (N_0λ_1)/(λ_2 - λ_1) e^((λ_2 - λ_1)t) + C`

`N_2 = (N_0λ_1)/(λ_2 - λ_1) (e^(-λ.t) - e^(-λ_2t))`

For maximum count, `(dN_2)/(dt)` = 0

On solving, `t = (In  λ_1/λ_2)/(λ_1 - λ_2)`

= In  `(2.48/0.62)/(2.48 - 0.62)`

= `(In  4)/1.86`

= `(2.303 log 4)/1.86`

= 0.745 s.