A nuclear power plant supplying electrical power to a village uses a radioactive material of half li

Q: Why does the power output decay exponentially?

The power output is directly proportional to the activity of the radioactive fuel, which follows the exponential decay law $A = A_0 e^{-\lambda t}$.

Q: What does $nT$ represent in the problem?

$nT$ represents the total time elapsed, expressed as a multiple $n$ of the half-life $T$.

Q: Why do we equate the available power to the required power?

The plant can supply the village as long as its output is $\ge$ the requirement. The maximum time is reached exactly when the output drops to match the requirement.

Q: How did we get $12.5\% = 1/8$?

$12.5\% = 12.5 / 100$. Dividing both numerator and denominator by $12.5$ gives $1/8$.

Q: What if the requirement was $10\%$?

You would set $(1/2)^n = 0.1$. Then $n = \log_2(10) \approx 3.32$ half-lives.

Question

A nuclear power plant supplying electrical power to a village uses a radioactive material of half life $T$ years as the fuel. The amount of fuel at the beginning is such that the total power requirement of the village is 12.5% of the electrical power available from the plant at that time. If the plant is able to meet the total power needs of the village for a maximum period of $nT$ years, then the value of $n$ is

Sigma Percentile · Accepted Answer

### The Setup: A Decaying Power Source

Imagine you are the chief engineer of a nuclear power plant. Your plant is the beating heart of a nearby village, providing the electrical power needed to keep the lights on and the machines running. However, there is a catch: your fuel is a radioactive material.

Unlike a steady stream of coal or a constant waterfall, radioactive fuel decays over time. This means the maximum power your plant can generate isn't constant; it drops exponentially. We are told that the half-life of this material is $T$ years. Every $T$ years, the available power is slashed exactly in half.

### The Village's Lifeline

The village doesn't care about half-lives; it just needs a steady, constant supply of power to survive. The problem states that the village's power requirement is exactly $12.5\%$ of the initial power the plant can produce.

Let's translate this into math. If the initial power is $P_0$, the required power is:
$$P_{	ext{req}} = 12.5\% 	ext{ of } P_0 = \frac{12.5}{100} P_0 = \frac{1}{8} P_0$$

This $\frac{1}{8} P_0$ is the critical threshold. As long as the plant's output stays above this line, the village thrives. The moment it drops below, the lights go out.

### The Intersection of Supply and Demand

We need to find the exact moment when the plant's dwindling supply perfectly matches the village's demand. Let's call this maximum time $t = nT$, where $n$ is the number of half-lives.

The law of radioactive decay tells us that the power available after $n$ half-lives is:
$$P(nT) = P_0 \left(\frac{1}{2}ight)^n$$

To find the breaking point, we equate the available power to the required power:
$$P_0 \left(\frac{1}{2}ight)^n = \frac{1}{8} P_0$$

### The Final Countdown

This equation is beautifully simple. The initial power $P_0$ cancels out from both sides, leaving us with a pure exponential equation:
$$\left(\frac{1}{2}ight)^n = \frac{1}{8}$$

To solve for $n$, we just need to express $\frac{1}{8}$ as a power of $\frac{1}{2}$. Since $2^3 = 8$, it follows that:
$$\left(\frac{1}{2}ight)^n = \left(\frac{1}{2}ight)^3$$

Comparing the exponents, the answer reveals itself:
**$n = 3$**

The plant can sustain the village for exactly 3 half-lives before the fuel must be replenished. It's a perfect demonstration of how exponential decay governs the lifespan of radioactive power sources!

Enter Numerical Value:

Visualized Solution

$P (t)$ vs $t$ Graph

Power Requirement

Intersection Point

Equating Powers

Simplifying the Equation

Final Answer

The Way Forward

The Sigma Insight: Radioactivity

The Setup

The Village's Lifeline

The Intersection of Supply and Demand

The Final Countdown

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