Define "third-life" in a similar way to "half-life," and determine the "third- life" for a nuclide that has a half-life of 31.4 years.

Half-life is defined as the time taken for a substance to reduce to half its initial amount. Similarly, we can define "third-life" as the time taken for a substance to reduce to one-third of its initial amount.

Let's consider a substance that has a half-life of T years. After this time, the remaining amount of the substance will be half of the initial amount. To obtain the third-life, we need to find the time it takes for the substance to reduce to one third of its initial amount. To do this, let's look at the general equation that governs this kind of decay: \( N_t = N_0 \times \frac{1}{2}^{\frac{t}{T_{1/2}}} \) Where: - \(N_t\) is the remaining amount of the substance after t years. - \(N_0\) is the initial amount of the substance. - \(T_{1/2}\) is the half-life of the substance. To determine the third-life (\(T_{1/3}\)), we need to modify the equation so that it represents the remaining amount of the substance equal to one-third of the initial amount. Therefore, the equation becomes: \( N_t = N_0 \times \frac{1}{3}^{\frac{t}{T_{1/3}}} \) Now, we have to find a relationship between \(T_{1/2}\) and \(T_{1/3}\) to solve the problem.

First, let's find the ratio of the initial amount (\(N_0\)) to the remaining amount (\(N_t\)) at the half-life: \( \frac{N_0}{N_t} = \frac{1}{2} \) Similarly, let's find the ratio of the initial amount (\(N_0\)) to the remaining amount (\(N_t\)) at the third-life: \( \frac{N_0}{N_t} = \frac{1}{3} \) Now let's equate both expressions in terms of the half-life and third-life: \( \frac{1}{2}^{\frac{t}{T_{1/2}}} = \frac{1}{3}^{\frac{t}{T_{1/3}}} \) To achieve a relationship between \(T_{1/2}\) and \(T_{1/3}\), let's take the natural logarithm of both sides: \( \frac{t}{T_{1/2}} \times \ln{\frac{1}{2}} = \frac{t}{T_{1/3}} \times \ln{\frac{1}{3}} \) Divide both sides by t: \( \frac{\ln{\frac{1}{2}}}{T_{1/2}} = \frac{\ln{\frac{1}{3}}}{T_{1/3}} \) Now, we can obtain the relationship: \( T_{1/3} = \frac{\ln{\frac{1}{3}}}{\ln{\frac{1}{2}}} \times T_{1/2} \)

Using the given half-life of the nuclide, T = 31.4 years, let's find the third-life (T_{1/3}): \( T_{1/3} = \frac{\ln{\frac{1}{3}}}{\ln{\frac{1}{2}}} \times 31.4 \approx 20.0 \) years So, the third-life for the nuclide with a half-life of 31.4 years is approximately 20.0 years.