Question

A radionuclide with half-life $1620 \, s$ is produced in a reactor at a constant rate of $1000$ nuclei per second. During each decay energy, $200 \, MeV$ is released. If the production of radionuclides started at $t \, = \, 0$ , then the rate of release of energy at $t \, = \, 3240 \, s$ is

• A. $1.5\times 10^{5}MeVs^{- 1}$
• B. $1.5\times 10^{2}MeVs^{- 1}$
• C. $2.5\times 10^{2}MeVs^{- 1}$
• D. $3.5\times 10^{5}MeVs^{- 1}$

Answer 1

Answer: (A)

Let N be the number of nuclei at time t, then net rate of increase of nuclei at instant t is,z
$\frac{ d N ⁡}{ d ⁡ t} = \alpha - \lambda N ⁡$ (where α = rate of production of nuclei)
$ \begin{array}{l} \int_{0}^{N} \frac{d N}{\alpha-\lambda N}=\int_{0}^{t} d t \\ N=\frac{\alpha}{\lambda}\left(1-e^{-\lambda t}\right) \end{array} $
Rate of decay at this instant $R =\lambda N =\alpha\left(1- e ^{-\lambda t}\right)$
Hence, rate of release of energy at this time $= R$ (energy released in each decay)
$ =\alpha\left(1-e^{-\lambda t}\right)(200) MeV / s $
Substituting the values, we have $ \text { rate of release of energy }=1000\left(1-e^{-\frac{0.693}{1620} \times 3240}\right)(200) $
$ =1.5 \times 10^{5} MeV / s $