Question

manganese - 52 has a half - life of 6 days. how many days would a scientist have to wait for the radioactivity to be 12.5% of the starting amount?
6
12
18
24

Explanation:

Step1: Understand half - life concept

The amount of a radioactive substance after $n$ half - lives is given by $A = A_0\times(\frac{1}{2})^n$, where $A_0$ is the initial amount and $A$ is the final amount. We want $A = 0.125A_0$ (since $12.5\%=\frac{12.5}{100}=0.125$).

Step2: Solve for number of half - lives $n$

Set $0.125A_0=A_0\times(\frac{1}{2})^n$. Divide both sides by $A_0$ (since $A_0
eq0$), we get $0.125 = (\frac{1}{2})^n$. Since $0.125=\frac{1}{8}=(\frac{1}{2})^3$, then $n = 3$.

Step3: Calculate total time

Given that one half - life is 6 days, and $n = 3$ half - lives. The total time $t$ is $t=n\times$ (half - life). So $t=3\times6 = 18$ days.

Answer:

18