At some instant, a radioactive sample S1 having an activity 5 μCi has twice the number of nuclei as another sample S2 which has an activity of 10 μCi. The half lives of S1 and S2 are?

For sample 1 we can write, N1 = N01e⁻λ₁t
N2 = N02e⁻λ₂t
Given that at a particular time t, N1 = 2N2
Also given that
For sample 1, A1 = -dN1/dt = N01λ1e⁻λ₁t = 5μCi
A2 = -dN2/dt = N02λ2e⁻λ₂t = 10μCi
Solving above equations we get,
λ1/λ2 = 1/4
half life1/half life2 = (ln2/λ1) × (λ2/ln2) = 4
TS1 = 4TS2
⇒ TS1 = 20 years and TS2 = 5 years is the possible answer.