Suppose the amount of a certain radioactive substance in a sample decays from $9.60 \mathrm{mg}$ to $900 . \mu \mathrm{g}$ over a period of 4.87 minutes. Calculate the half life of the substance. Round your answer to 2 significant digits.

Step 1 ：Given that the initial amount of the substance (N0) is 9.60 mg, the final amount (N) is 900 μg (or 0.9 mg), and the time (t) is 4.87 minutes.

Step 2 ：The decay of a radioactive substance is an exponential process, and can be described by the equation: \(N = N0 \times e^{-\lambda t}\) where: N is the final amount of the substance, N0 is the initial amount of the substance, \(\lambda\) is the decay constant, and t is the time.

Step 3 ：We can rearrange this equation to solve for \(\lambda\): \(\lambda = -\ln(N/N0) / t\)

Step 4 ：Substitute the given values into the equation to find \(\lambda\): \(\lambda = -\ln(0.9/9.6) / 4.87 = 0.4860623437641924\)

Step 5 ：Once we have \(\lambda\), we can calculate the half-life (T) using the equation: \(T = \ln(2) / \lambda\)

Step 6 ：Substitute the value of \(\lambda\) into the equation to find T: \(T = \ln(2) / 0.4860623437641924 = 1.43\)

Step 7 ：Final Answer: The half-life of the substance is \(\boxed{1.43}\) minutes.