Nuclei: Edge Cases and Subtle Traps (2)

Let $\lambda_A = \ln 2 / T$ and $\lambda_B = \ln 2 / (2T) = \lambda_A / 2$.

$N_B$ is maximum when $dN_B/dt = 0$, which means production rate = decay rate:

Solving the linear ODE with $N_B(0) = 0$:

Differentiate and set to zero, or use the condition above. From step 3:

Simplify:

$\lambda_A/\lambda_B = 2$. $\lambda_A - \lambda_B = \lambda_A/2 = \ln 2/(2T)$.

At $t = 2T$: $N_A = N_0 e^{-\lambda_A (2T)} = N_0 e^{-2\ln 2} = N_0 / 4$.

Final answer: $t_{max} = 2T$, $N_A(t_{max}) = N_0/4$.

Why This Works

When B is at its peak, the rate of B production from A exactly cancels B’s own decay. This is a quasi-steady-state instant, not an equilibrium — both populations keep changing.

The general formula $t_{max} = \ln(\lambda_A/\lambda_B)/(\lambda_A - \lambda_B)$ is worth memorising for JEE Advanced.

Alternative Method

Numerical: tabulate $N_A$ and $N_B$ in steps of $T/4$ and locate the peak. Slow but bulletproof.