Question

how many half-lives does it take for a radioactive substance to decay until only one quarter (1/4) of the original substance remains?

a. 1

b. 2

c. 3

d. 4

Explanation:

Step1: Recall half - life concept

In each half - life, the amount of radioactive substance remaining is half of the previous amount. Let the initial amount be \(N_0\) and the amount after \(n\) half - lives be \(N\). The formula for radioactive decay is \(N = N_0\times(\frac{1}{2})^n\).
We want to find \(n\) when \(N=\frac{1}{4}N_0\). So we set up the equation \(\frac{1}{4}N_0=N_0\times(\frac{1}{2})^n\). Since \(N_0
eq0\) (because there is an initial amount of the substance), we can divide both sides by \(N_0\) and get \(\frac{1}{4}=(\frac{1}{2})^n\).

Step2: Solve for \(n\)

We know that \(\frac{1}{4}=\frac{1}{2}\times\frac{1}{2}=(\frac{1}{2})^2\). So, if \((\frac{1}{2})^n = (\frac{1}{2})^2\), then by the property of exponential functions with the same base (\(y = a^x,a>0,a
eq1\), if \(a^m=a^n\) then \(m = n\)), we can conclude that \(n = 2\).

Answer:

B. 2