A nuclear reactor starts producing a radioactive nuclide X from t = 0, at a constant rate of alpha p

Q: Why is the rate of decay $\lambda N$ and not just $\lambda$?

The decay constant $\lambda$ represents the probability of decay per unit time for a *single* nucleus. If you have $N$ nuclei, the total number decaying per second is $\lambda \times N$.

Q: Why did we integrate from $0$ to $N$ and $0$ to $t$?

The problem states that the reactor *starts* producing the nuclide at $t=0$. This implies that initially, at $t=0$, there are no nuclei of $X$ present, so $N(0) = 0$.

Q: Why is the power $P = \lambda N E_0$ and not $\alpha E_0$?

Energy is released only when a nucleus *decays*, not when it is produced. The rate of decay is $\lambda N$, and each decay gives $E_0$ energy. The production rate $\alpha$ only tells us how fast $X$ is being created.

Q: What happens to the temperature rate after a very long time?

As $t \to \infty$, the term $e^{-\lambda t}$ approaches $0$. The number of nuclei reaches a steady state (equilibrium) where production equals decay ($\alpha = \lambda N$). The rate of temperature increase becomes constant at $\frac{\alpha E_0}{ms}$.

Q: Could we have solved this by finding the total energy first and then differentiating?

Yes! You could find the total number of decayed nuclei by integrating $\lambda N dt$, multiplying by $E_0$ to get total heat $Q$, and then using $Q = ms \Delta T$. Differentiating this with respect to time gives the exact same result, but equating power directly is much faster.

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:

Options:

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

Sigma Percentile · Accepted Answer

The problem presents a fascinating intersection of modern physics and thermodynamics. We have a nuclear reactor acting as a source of a radioactive nuclide, which in turn acts as a heat source for a liquid. Our ultimate goal is to find the rate at which the liquid's temperature increases.

### Analyzing the Setup

Imagine you are standing in front of a nuclear reactor. It is continuously churning out a radioactive nuclide, let's call it $X$, at a constant rate of $\alpha$ nuclei per second. But $X$ is unstable. As soon as it is created, it starts decaying into some other element $Y$ with a decay constant $\lambda$.

Every single time a nucleus of $X$ decays, it releases a tiny burst of energy, $E_0$. This energy doesn't just vanish; it is entirely absorbed by a liquid of mass $m$ and specific heat capacity $s$. Because there is no heat loss to the surroundings, all this nuclear energy goes directly into raising the temperature of the liquid.

### The Master Equation for Radioactive Growth

To find out how much energy is being released, we first need to know exactly how many nuclei of $X$ are present at any given time $t$. Let this number be $N$.

The population of $X$ is governed by a tug-of-war between its production and its decay. The net rate of change of the number of nuclei is the production rate minus the decay rate. Mathematically, we can write this as:

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

This is a standard linear first-order differential equation. We can solve it by separating the variables:

$$ \int_0^N \frac{dN}{\alpha - \lambda N} = \int_0^t dt $$

Integrating both sides, we get:

$$ -\frac{1}{\lambda} \ln\left(\frac{\alpha - \lambda N}{\alpha}ight) = t $$

Rearranging this to solve for $N$, we find the population of $X$ as a function of time:

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

### Power: The Bridge Between Nuclear and Thermal Physics

Now that we know $N(t)$, we can determine the rate at which energy is being produced. This rate of energy production is simply the power, $P$.

Energy is released only when a nucleus decays. The number of decays happening per second is given by the activity, which is $\lambda N$. Since each decay releases an energy $E_0$, the total power generated is:

$$ P = (	ext{Rate of decay}) 	imes (	ext{Energy per decay}) $$

$$ P = (\lambda N) 	imes E_0 $$

Substitute our expression for $N(t)$ into this power equation:

$$ P = \lambda \left[ \frac{\alpha}{\lambda}(1 - e^{-\lambda t}) ight] E_0 $$

Notice how beautifully the $\lambda$ terms cancel out! We are left with:

$$ P = \alpha E_0 (1 - e^{-\lambda t}) $$

This equation tells us that the power output starts at zero (when $t=0$) and gradually increases, eventually plateauing at a maximum value of $\alpha E_0$ as $t 	o \infty$.

### Final Calculation: Heating the Liquid

We have the power, and now we need to connect it to the liquid's temperature. According to the principles of calorimetry, the rate at which heat is absorbed by a substance is related to its rate of temperature change by the equation:

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

Here, $\frac{dT}{dt}$ is exactly what we are looking for—the rate of increase in the temperature of the liquid.

Equating our two expressions for power, we get:

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

Finally, isolating $\frac{dT}{dt}$, we arrive at our answer:

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

This elegant result perfectly captures the physics of the situation. The temperature rises at a rate that mirrors the buildup of the radioactive nuclide, eventually reaching a steady state where the liquid heats up at a constant maximum rate.

Select Answer:

Visualized Solution

\text{System Overview}

\text{Rate of Change of } N

\text{Solving for } N(t)

\text{Rate of Energy Production (Power)}

\text{Substituting } N(t) \text{ in Power}

\text{Calorimetry Equation}

\text{Rate of Temperature Increase}

\text{What if...}

The Sigma Insight: Radioactivity

Analyzing the Setup

The Master Equation for Radioactive Growth

Power

Final Calculation

Similar Questions

If $f$ denotes the ratio of the number of nuclei decayed ( $N_{d}$ ) to the number of nuclei at $t = 0$ ( $N_{0}$ ), then for a collection of radioactive nuclei, the rate of change of $f$ with respect to time is given as [ $λ$ is the radioactive decay constant]

Two radioactive substances $A$ and $B$ have decay constants $5 λ$ and $λ$ , respectively. At $t = 0$ , a sample has the same number of the two nuclei. The time taken for the ratio of the number of nuclei to become $(\frac{1}{e})^{2}$ will be

A radioactive nucleus $A$ with a half-life $T$ , decays into a nucleus $B$ . At $t = 0$ , there is no nucleus $B$ . After sometime $t$ , the ratio of the number of $B$ to that of $A$ is $0.3$ . Then, $t$ is given by

The decay constant of a radioactive sample is $λ$ . The half-life and mean-life of the sample are respectively given by

Two radioactive materials $A$ and $B$ have decay constants $10 λ$ and $λ$ , respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of $A$ to that of $B$ will be $1/ e$ after a time

Two radioactive materials $X_{1}$ and $X_{2}$ have decay constants $10 λ$ and $λ$ respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of $X_{1}$ to that of $X_{2}$ will be $1/ e$ after a time

A radioactive sample disintegrates via two independent decay processes having half-lives $T_{1/2}^{(1)}$ and $T_{1/2}^{(2)}$ , respectively. The effective half-life $T_{1/2}$ of the nuclei is