• Tags: successive-decay, equilibrium, chain
• Difficulty: Advanced

When nucleus A decays to B, which decays to C: $dN_B$/dt = $lambda_A$$N_A$ - $lambda_B$$N_B$. In secular equilibrium (t_$\frac{1}{2}$ of A >> t_$\frac{1}{2}$ of B), the activity of B equals the activity of A: $lambda_A$$N_A$ = $lambda_B$$N_B$. This means $N_A$/$N_B$ = $\frac{lambda_B}{lambda_A}$ = t_\frac{1/2,A}{t_}(1/2,B). In a long decay chain (like U-238 series), all intermediate products reach secular equilibrium, and all activities are equal. The total activity of the chain equals n times the activity of the parent (where n is the number of radioactive members). For the common JEE problem of two successive decays: if A decays to B with half-life $t_1$, and B decays to C (stable) with half-life $t_2$, the number of B nuclei initially increases, reaches a maximum when $lambda_A$$N_A$ = $lambda_B$$N_B$, then decreases. If both decay constants are given, solve the differential equation or use the Bateman equations.