Question

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:

• A. $\frac{\alpha E_0}{ms}(1 - e^{-\lambda t})$
• B. $\frac{\alpha E_0}{ms}(e^{\lambda t} - 1)$
• C. $\frac{\lambda E_0}{ms}(1 - e^{-\lambda t})$
• D. $\frac{E_0}{ms}(\alpha - \lambda e^{-\lambda t})$

Hint:

In radioactive growth problems, the number of nuclei eventually reaches a steady state $N_{steady} = \alpha/\lambda$. At $t \to \infty$, the rate of temperature rise becomes constant: $\frac{\alpha E_0}{ms}$.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The nuclide is being produced at a constant rate $\alpha$ and simultaneously decays. The energy released from decays heats a liquid. We need to find the power (rate of energy release) and use it to find the rate of change of temperature ($dT/dt$).

Step 2: Key Formula or Approach:

• Rate of change of nuclei: $\frac{dN}{dt} = \text{Production rate} - \text{Decay rate} = \alpha - \lambda N$

• Activity (Decay rate): $R = \lambda N$

• Power released: $P = R \times E_0$

• Heating formula: $P = ms\frac{dT}{dt}$

Step 3: Detailed Explanation:

• Set up and solve the differential equation for the number of nuclei $N(t)$: \[ \frac{dN}{dt} = \alpha - \lambda N \implies \int_0^N \frac{dN}{\alpha - \lambda N} = \int_0^t dt \] \[ -\frac{1}{\lambda} \ln(\alpha - \lambda N) \Big|_0^N = t \implies N(t) = \frac{\alpha}{\lambda}(1 - e^{-\lambda t}) \]
• Calculate the activity $R$ (number of decays per second): \[ R = \lambda N = \alpha(1 - e^{-\lambda t}) \]
• The total energy released per second (Power $P$) is: \[ P = R \cdot E_0 = \alpha E_0 (1 - e^{-\lambda t}) \]
• Relate power to the rate of increase in temperature: \[ P = ms \frac{dT}{dt} \implies \frac{dT}{dt} = \frac{\alpha E_0}{ms}(1 - e^{-\lambda t}) \]

Step 4: Final Answer:
The rate of increase in temperature is $\frac{\alpha E_0}{ms}(1 - e^{-\lambda t})$.