Question

1. the half life of a radioactive isotope is 30 seconds. in what period of time would the activity of the sample be reduced to 1/16 of the original activity?

Explanation:

Step1: Recall the formula for radioactive decay

The formula for the amount of a radioactive substance remaining after \(n\) half - lives is \(N = N_0\times(\frac{1}{2})^n\), where \(N\) is the final amount, \(N_0\) is the initial amount. We want to find \(n\) when \(N=\frac{1}{16}N_0\). So, \(\frac{1}{16}N_0=N_0\times(\frac{1}{2})^n\). Dividing both sides by \(N_0\) (since \(N_0
eq0\)), we get \(\frac{1}{16}=(\frac{1}{2})^n\). Since \(\frac{1}{16}=\frac{1}{2^4}=(\frac{1}{2})^4\), then \(n = 4\).

Step2: Calculate the time

Given the half - life \(T = 30\) seconds. The total time \(t=n\times T\). Substituting \(n = 4\) and \(T = 30\) seconds, we have \(t=4\times30\) seconds.

Answer:

\(120\) seconds