Discussion
A nuclear reactor starts producing a radioactive nuclide \(X\) from \(t = 0\), at a constant rate of \(\alpha\) per second. Each decay of \(X\) produces energy \(E_{0}\), which is utilized to heat a liquid of mass \(m\) and specific heat \(s\). Assuming no heat loss from the liquid and taking \(\lambda\) as the decay constant of \(X\), the rate of increase in the temperature of the liquid is:
Answer: A
The rate is \(\frac{\alpha E_{0}}{m\,s}(1 - e^{-\lambda t})\).
The number \(N\) of nuclei of \(X\) grows by production and falls by decay:
\[ \begin{aligned} \frac{dN}{dt} &= \alpha - \lambda N,\quad N(0) = 0 \\ N &= \frac{\alpha}{\lambda}\left(1 - e^{-\lambda t}\right) \end{aligned} \]
The activity is \(\lambda N = \alpha(1 - e^{-\lambda t})\) decays per second, each releasing \(E_{0}\). With no heat loss this power goes into the liquid, \(P = ms\,\frac{dT}{dt}\), so \(\frac{dT}{dt} = \frac{\alpha E_{0}}{ms}(1 - e^{-\lambda t})\).
The physics checks the shape: at \(t = 0\) there are no nuclei, so the heating rate starts at zero; later the activity saturates at the production rate \(\alpha\). The \((e^{\lambda t} - 1)\) form grows without limit, which would need activity to exceed production. The \(\lambda E_{0}\) form saturates at the wrong value, since activity levels off at \(\alpha\), not \(\lambda\). The \((\alpha - \lambda e^{-\lambda t})\) form is not zero at \(t = 0\).
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