Answer :

We start with the formula for radioactive decay:

$$
A = A_0 \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}
$$

where
- $A_0$ is the initial amount,
- $A$ is the amount remaining after time $t$, and
- $T_{1/2}$ is the half-life.

Given that a sample decays to $6.25$ g in $10.8$ days and assuming an initial amount of $100$ g, we substitute into the formula:

$$
6.25 = 100 \left(\frac{1}{2}\right)^{\frac{10.8}{T_{1/2}}}.
$$

Divide both sides by $100$:

$$
0.0625 = \left(\frac{1}{2}\right)^{\frac{10.8}{T_{1/2}}}.
$$

Notice that $0.0625$ can be written as a power of $\frac{1}{2}$:

$$
0.0625 = \frac{1}{16} = \left(\frac{1}{2}\right)^4.
$$

Thus, we equate the exponents:

$$
\frac{10.8}{T_{1/2}} = 4.
$$

Now solve for $T_{1/2}$ by multiplying both sides by $T_{1/2}$ and then dividing by $4$:

$$
T_{1/2} = \frac{10.8}{4}.
$$

This gives:

$$
T_{1/2} = 2.7 \text{ days}.
$$

So, the half-life of gold-198 is $\boxed{2.7 \text{ days}}$.

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Rewritten by : Barada