A nuclear reactor starts producing a radioactive nuclide X from t = 0, at a constant rate of α per second. Each decay of X produces energy E₀, which is utilized to heat a liquid of mass m and specific heat s. Assuming no heat loss from the liquid and taking λ as the decay constant of X, the rate of increase in the temperature of the liquid is:

The correct answer is $\frac{\alpha E_0}{ms}(1-e^{-\lambda t})$.

Step 1: Determine the number of radioactive nuclei N at any time t.
Let $N$ be the number of active nuclei of nuclide $X$ present at time $t$.
Nuclide $X$ is produced at a constant rate of $\alpha$ per second and decays at a rate of $\lambda N$, where $\lambda$ is the decay constant.
Therefore, the rate of change of the number of nuclei is given by the differential equation:

$\frac{dN}{dt}=\alpha -\lambda N$

Rearranging and integrating from $t=0$ (where $N=0$) to time $t$:

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

$-\frac{1}{\lambda }[ln(\alpha -\lambda N){]}_0^N=t$

$ln\left(\frac{\alpha -\lambda N}{\alpha }\right)=-\lambda t$

Taking the exponential on both sides gives:

$1-\frac{\lambda N}{\alpha }=e^{-\lambda t}$

$N=\frac{\alpha }{\lambda }(1-e^{-\lambda t})$

Step 2: Calculate the rate of decay of nuclei.
The rate of decay of nuclei at any instant $t$ is given by the activity $R$:

$R=\lambda N=\lambda ·\frac{\alpha }{\lambda }(1-e^{-\lambda t})=\alpha (1-e^{-\lambda t})$

Step 3: Relate rate of energy release to rate of temperature rise.
Each decay releases energy $E_0$. Thus, the rate of heat energy released and transferred to the liquid is:

$\frac{dQ}{dt}=R·E_0=\alpha E_0(1-e^{-\lambda t})$

The heat gained by a liquid of mass $m$ and specific heat $s$ causing a temperature change $dT$ is given by $dQ=msdT$.
Therefore, the rate of energy absorption is:

$\frac{dQ}{dt}=ms\frac{dT}{dt}$

Equating the two expressions for $\frac{dQ}{dt}$:

$ms\frac{dT}{dt}=\alpha E_0(1-e^{-\lambda t})$

Solving for the rate of increase in temperature $\frac{dT}{dt}$:

$\frac{dT}{dt}=\frac{\alpha E_0}{ms}(1-e^{-\lambda t})$