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:

This is a growth-with-decay setup: the nuclide count changes due to constant production at rate alpha and simultaneous decay at rate lambda times N. Write that balance as a differential equation first, then convert the number of decays per second into power using E0, and finally use P = ms(dT/dt) to connect power with temperature rise.

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

Answer 2

Concept:

• The number of nuclei $N(t)$ obeys a first order linear differential equation with constant coefficients, $\frac{dN}{dt} + \lambda N = \alpha$, which can be solved directly with an integrating factor instead of separating variables.
• This equation has the same form as a capacitor charging through a resistor, so $N(t)$ rises from zero and approaches a steady value of $\alpha/\lambda$.
• Once $N(t)$ is known, the activity (decays per second) is $\lambda N(t)$, and each decay releases energy $E_0$ into the liquid, so power delivered links to temperature rise through $P = ms\frac{dT}{dt}$.

Step 1: Write the balance equation in standard linear form
Production minus decay gives $\frac{dN}{dt} = \alpha - \lambda N$, which rearranges to $\frac{dN}{dt} + \lambda N = \alpha$. This matches the standard linear form $\frac{dN}{dt} + P N = Q$ with $P = \lambda$ and $Q = \alpha$, both constants.

Step 2: Find and apply the integrating factor
The integrating factor is $IF = e^{\int \lambda \, dt} = e^{\lambda t}$. Multiplying the equation by $e^{\lambda t}$ gives $e^{\lambda t}\frac{dN}{dt} + \lambda e^{\lambda t} N = \alpha e^{\lambda t}$, and the left side is exactly $\frac{d}{dt}\left(N e^{\lambda t}\right)$.

Step 3: Integrate both sides using the initial condition
$\frac{d}{dt}\left(N e^{\lambda t}\right) = \alpha e^{\lambda t}$
Integrating from $0$ to $t$, with $N = 0$ at $t = 0$:
$N e^{\lambda t} = \frac{\alpha}{\lambda}\left(e^{\lambda t} - 1\right)$
$N(t) = \frac{\alpha}{\lambda}\left(1 - e^{-\lambda t}\right)$

Step 4: Convert nuclei count to activity, then to power
Activity (decays per second) is $R = \lambda N(t) = \alpha\left(1 - e^{-\lambda t}\right)$. Each decay releases energy $E_0$, so the power delivered to the liquid is $P = R E_0 = \alpha E_0\left(1 - e^{-\lambda t}\right)$.

Step 5: Convert power to rate of temperature rise
Using $P = ms\frac{dT}{dt}$:
$\frac{dT}{dt} = \frac{P}{ms} = \frac{\alpha E_0}{ms}\left(1 - e^{-\lambda t}\right)$

Final Answer: $\frac{\alpha E_0}{ms}\left(1 - e^{-\lambda t}\right)$