A radioactive sample decays $$\frac{7}{8}$$ times its original quantity in 15 minutes. The half-life of the sample is

We are told that a radioactive sample decays $$\frac{7}{8}$$ of its original quantity in 15 minutes. This means the remaining quantity is $$1 - \frac{7}{8} = \frac{1}{8}$$ of the original.

Using the radioactive decay formula, $$\frac{N}{N_0} = \left(\frac{1}{2}\right)^{t/t_{1/2}}$$, where $$t_{1/2}$$ is the half-life.

So $$\frac{1}{8} = \left(\frac{1}{2}\right)^{15/t_{1/2}}$$. Since $$\frac{1}{8} = \left(\frac{1}{2}\right)^3$$, we get $$\frac{15}{t_{1/2}} = 3$$.

Therefore, $$t_{1/2} = \frac{15}{3} = 5$$ minutes.

Hence, the correct answer is Option A.