Let N be the number of nuclei at time t, then net rate of increase of nuclei at instant t is,z

  $\frac{dN}{dt}=\alpha -\lambda N$    (where α = rate of production of nuclei)

∴                                    ${\int }_0^N\frac{dN}{\alpha -\lambda N}={\int }_0^{t}dt$

∴                                    $N=\frac{\alpha }{\lambda }\left(1-e^{-\lambda t}\right)$                                                   ...(i)

Rate of decay at this instant      R = λN = α ( 1 – e^{– λt})

Hence, rate of release of energy at this time = R (energy released in each decay)

                                             $=\alpha \left(1-e^{-\lambda t}\right)\left(200\right)\text{MeV}/\text{s}$

Substituting the values, we have

                    $\text{rate of release of energy}=1000\left(1-e^{-\frac{0\text{.}693}{1620}\times 3240}\right)\left(200\right)$

= 1.5 × 10⁵ MeV/s