A certain radioactive isotope is a by-product of some nuclear reactors. Due to an explosion, a nuclear reactor experiences a massive leak of this radioactive isotope. Fortunately, the isotope has a very short half-life of 13 days. Estimate the percentage of the original amount of the isotope released by the explosion that remains 12 days after the explosion.

After 12 days, about $\square \%$ of the isotope remains.
(Do not round until the final answer. Then round to the nearest whole number as needed.)

Step 1 ：The formula for exponential decay is given by \(y = y_0 * e^{-kt}\), where \(y_0\) is the initial amount, \(k\) is the decay constant, and \(t\) is the time. Given that the half-life is 13 days, we can use the formula for the decay constant in terms of the half-life: \(k = \ln(2) / T_{1/2}\), where \(T_{1/2}\) is the half-life.

Step 2 ：Given that \(T_{1/2} = 13\), we can calculate \(k = 0.05331549160877513\).

Step 3 ：Substituting these values into the decay formula, we get the formula relating \(y\) to \(t\) as \(y = y_0 * e^{-0.05331549160877513t}\).

Step 4 ：To find out how much of the substance will be present in 12 days, we substitute \(t = 12\) into the formula.

Step 5 ：Given that \(y_0 = 1\) (since we're looking for a percentage), \(T_{1/2} = 13\), \(k = 0.05331549160877513\), we can calculate \(y_{12} = 0.5272924243957591\).

Step 6 ：The percentage of the isotope that remains after 12 days is approximately \(\boxed{53 \%}\).