Indium-116 is radioactive and has a half life of 14.10 seconds. What percentage of a sample would be left after 64.8 seconds? Round your answer to 2 significant digits.
$\square \%$

Step 1 ：The decay of a radioactive substance is governed by the equation: \(N = N0 * (1/2)^{t/T}\) where: \(N\) is the final amount of the substance, \(N0\) is the initial amount of the substance, \(t\) is the time that has passed, \(T\) is the half-life of the substance.

Step 2 ：We want to find \(N/N0\), the fraction of the substance that remains after time \(t\). So we can simplify the equation to: \(N/N0 = (1/2)^{t/T}\)

Step 3 ：Substituting the given values: \(N/N0 = (1/2)^{64.8/14.10}\)

Step 4 ：Calculating this gives: \(N/N0 = 0.0061\)

Step 5 ：To convert this to a percentage, we multiply by 100: \(N/N0 = 0.0061 * 100 = 0.61%\)

Step 6 ：So, after 64.8 seconds, approximately 0.61% of the Indium-116 sample would remain. The final answer is \(\boxed{0.61}\)