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Given:
Half-life, $T_{1/2} = 12.5\text{ years}$
Total time elapsed, $t = 25\text{ years}$

Let the initial number of atoms be $N_0$, and the remaining number of atoms after time $t$ be $N$.

The number of half-lives elapsed is:
$n = \frac{t}{T_{1/2}} = \frac{25}{12.5} = 2$

The remaining number of atoms is given by:
$N = N_0 \left(\frac{1}{2}\right)^n$
$N = N_0 \left(\frac{1}{2}\right)^2 = \frac{N_0}{4} = 0.25 N_0$

Therefore, one-fourth (or $25\%$) of the original atoms will remain.