Atoms and Nuclei

Concept:Radioactive decay and heat transfer: Rate of temperature rise depends on decay activity and energy per decay.Explanation:Let $N$ be the number of active nuclei at time $t$.The net rate of change is: $\frac{dN}{dt} = \alpha - \lambda N$, where $\alpha$ is constant production rate and $\lambda$ is decay constant.At $t=0$, $N=0$. Integrate: $\int_0^N \frac{dN}{\alpha - \lambda N} = \int_0^t dt$.Solving gives: $\ln\left(\frac{\alpha - \lambda N}{\alpha}\right) = -\lambda t$, so $\lambda N = \alpha (1 - e^{-\lambda t})$.Activity $A(t) = \lambda N = \alpha (1 - e^{-\lambda t})$ decays per second.Each decay releases energy $E_0$, so power to liquid: $\frac{dQ}{dt} = A(t) E_0 = \alpha E_0 (1 - e^{-\lambda t})$.For liquid mass $m$ and specific heat $s$, $\frac{dQ}{dt} = m s \frac{dT}{dt}$.Thus, $\frac{dT}{dt} = \frac{\alpha E_0}{m s} (1 - e^{-\lambda t})$.Answer:$\boxed{\frac{\alpha E_0}{ms}(1 - e^{-\lambda t})}$ (Option A).