Q38 / 54 · JEE Advanced 2026 (Paper 2) · Physics
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:
- Correct answerA$\dfrac{\alpha E_0}{m s}(1 - e^{-\lambda t})$
- B$\dfrac{\alpha E_0}{m s}(e^{\lambda t} - 1)$
- C$\dfrac{\lambda E_0}{m s}(1 - e^{-\lambda t})$
- D$\dfrac{E_0}{m s}(\alpha - \lambda e^{-\lambda t})$
Solution
Let $N$ be the number of active nuclei at time $t$. The rate of change of the number of nuclei is given by
$\dfrac{dN}{dt} = \alpha - \lambda N$
Integrating with the initial condition $N = 0$ at $t = 0$:
$\int_{0}^{N} \dfrac{dN}{\alpha - \lambda N} = \int_{0}^{t} dt$
$-\dfrac{1}{\lambda} \ln\left(\dfrac{\alpha - \lambda N}{\alpha}\right) = t$
$N = \dfrac{\alpha}{\lambda} (1 - e^{-\lambda t})$
The rate of decay of the nuclide at time $t$ is
$A = \lambda N = \alpha (1 - e^{-\lambda t})$
Since each decay produces energy $E_0$, the rate of energy production is
$\dfrac{dE}{dt} = A E_0 = \alpha E_0 (1 - e^{-\lambda t})$
This energy is used to heat the liquid. The rate of heat absorption is
$\dfrac{dQ}{dt} = m s \dfrac{dT}{dt}$
Equating the rate of energy production to the rate of heat absorption:
$m s \dfrac{dT}{dt} = \alpha E_0 (1 - e^{-\lambda t})$
$\dfrac{dT}{dt} = \dfrac{\alpha E_0}{m s} (1 - e^{-\lambda t})$
Answer: $\dfrac{\alpha E_0}{m s}(1 - e^{-\lambda t})$
$\dfrac{dN}{dt} = \alpha - \lambda N$
Integrating with the initial condition $N = 0$ at $t = 0$:
$\int_{0}^{N} \dfrac{dN}{\alpha - \lambda N} = \int_{0}^{t} dt$
$-\dfrac{1}{\lambda} \ln\left(\dfrac{\alpha - \lambda N}{\alpha}\right) = t$
$N = \dfrac{\alpha}{\lambda} (1 - e^{-\lambda t})$
The rate of decay of the nuclide at time $t$ is
$A = \lambda N = \alpha (1 - e^{-\lambda t})$
Since each decay produces energy $E_0$, the rate of energy production is
$\dfrac{dE}{dt} = A E_0 = \alpha E_0 (1 - e^{-\lambda t})$
This energy is used to heat the liquid. The rate of heat absorption is
$\dfrac{dQ}{dt} = m s \dfrac{dT}{dt}$
Equating the rate of energy production to the rate of heat absorption:
$m s \dfrac{dT}{dt} = \alpha E_0 (1 - e^{-\lambda t})$
$\dfrac{dT}{dt} = \dfrac{\alpha E_0}{m s} (1 - e^{-\lambda t})$
Answer: $\dfrac{\alpha E_0}{m s}(1 - e^{-\lambda t})$
Hence, the correct answer is option: 'A'.