Question

the half - life of a radioactive element, such as carbon - 14, is the time it takes for half of the atoms of the element to decay. how many half - life periods have passed if a fossil contains $\frac{1}{16}$ of the original amount of carbon - 14? 3 4 5 6

Explanation:

Step1: Set up the formula

The formula for the amount of a radioactive substance remaining after \(n\) half - life periods is \(A = A_0\times(\frac{1}{2})^n\), where \(A\) is the final amount, \(A_0\) is the initial amount, and \(n\) is the number of half - life periods. We know that \(A=\frac{1}{32}A_0\).

Step2: Substitute into the formula

Substitute \(A=\frac{1}{32}A_0\) into \(A = A_0\times(\frac{1}{2})^n\). We get \(\frac{1}{32}A_0=A_0\times(\frac{1}{2})^n\). Since \(A_0
eq0\) (there was an initial amount of carbon - 14), we can divide both sides of the equation by \(A_0\). So, \(\frac{1}{32}=(\frac{1}{2})^n\).

Step3: Rewrite \(\frac{1}{32}\) as a power of \(\frac{1}{2}\)

We know that \(32 = 2^5\), so \(\frac{1}{32}=\frac{1}{2^5}=(\frac{1}{2})^5\).

Step4: Solve for \(n\)

Since \(\frac{1}{32}=(\frac{1}{2})^n\) and \(\frac{1}{32}=(\frac{1}{2})^5\), then \(n = 5\).

Answer:

5