In a state of radioactive steady-state equilibrium between a parent nuclide A and its daughter nucli — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

At radioactive equilibrium, the decay rates are equal: $N_A \lambda_A = N_B \lambda_B$. Since $\lambda = 0.693 / T_{1/2}$, this becomes $N_A / T_{1/2(A)} = N_B / T_{1/2(B)}$, rearranging directly to $N_A / N_B = T_{1/2(A)} / T_{1/2(B)}$.