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

When nucleus A decays to B, which decays to C: dN_B/dt = lambda_AN_A - lambda_BN_B. In secular equilibrium (t_(1/2) of A >> t_(1/2) of B), the activity of B equals the activity of A: lambda_AN_A = lambda_BN_B. This means N_A/N_B = lambda_B/lambda_A = t_(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_AN_A = lambda_BN_B, then decreases. If both decay constants are given, solve the differential equation or use the Bateman equations.