Step-by-step Solution

1. Relating Power to Radioactive Decay

   The power generated by the nuclear plant is proportional to the rate of radioactive decay (activity) of the fuel. The activity itself is proportional to the number of radioactive nuclei present at any given time. Therefore, the power output of the plant decreases over time as the fuel decays.

   Let $P_0$ be the initial electrical power available from the plant at time $t=0$. The power available at any time $t$, denoted as $P(t)$, follows the law of radioactive decay:

   $P(t) = P_0 \left(\frac{1}{2}\right)^{t/T}$

   where $T$ is the half-life of the radioactive material.

2. Determining the Village's Power Requirement

   The problem states that at the beginning ($t=0$), the total power requirement of the village, let's call it $P_{village}$, is 12.5% of the electrical power available from the plant at that time ($P_0$).

   $P_{village} = 12.5\% \text{ of } P_0$

   Converting the percentage to a fraction:

   $12.5\% = \frac{12.5}{100} = \frac{1}{8}$

   So, the power requirement of the village is constant:

   $P_{village} = \frac{P_0}{8}$

3. Finding the Maximum Operational Time

   The plant can meet the village's power needs as long as the power it generates, $P(t)$, is greater than or equal to the village's requirement, $P_{village}$.

   $P(t) \ge P_{village}$

   The maximum time the plant can operate, let's call it $t_{max}$, is the point where the power generated by the plant drops to a level that is just equal to the village's requirement.

   $P(t_{max}) = P_{village}$

   Substituting the expressions for $P(t)$ and $P_{village}$:

   $P_0 \left(\frac{1}{2}\right)^{t_{max}/T} = \frac{P_0}{8}$

4. Solving for the Time

   We can cancel $P_0$ from both sides of the equation (assuming $P_0 \neq 0$):

   $\left(\frac{1}{2}\right)^{t_{max}/T} = \frac{1}{8}$

   We know that $8 = 2^3$, so $\frac{1}{8} = \frac{1}{2^3} = \left(\frac{1}{2}\right)^3$.

   Substituting this back into the equation:

   $\left(\frac{1}{2}\right)^{t_{max}/T} = \left(\frac{1}{2}\right)^3$

   By comparing the exponents on both sides, we get:

   $\frac{t_{max}}{T} = 3$

   $t_{max} = 3T$

5. Determining the value of n

   The problem states that the maximum period is $nT$ years. We found that the maximum period is $t_{max} = 3T$.

   Comparing the two expressions:

   $nT = 3T$

   Therefore, the value of $n$ is 3.