Answer:
12.5 %.
Explanation:
Let's see the half-life formula:
[tex]N(t)=N_0\cdot(\frac{1}{2})^{t\text{/h}}.[/tex]Where N₀ is the initial amount, t is time, and h is the half-life.
We want to know what would be the percentage of a sample of iodine-131 that remains. This is the same that N(t)/N₀, so we have to replace the given data in the formula and multiply it by 100 because it is a percentage.
The half-life of iodine-131 is 8.0 days, and 3 half-lives are equal to 24 days (8.0 x 3 = 24):
[tex]\begin{gathered} \frac{N(t)}{N_0}=(\frac{1}{2})^{t\text{/h}}, \\ (\frac{1}{2})^{24\text{/8}}=(\frac{1}{2})^3=\frac{1}{8}=0.125. \\ We\text{ want the result in percentage, so multiplying by 100}\%,\text{ we obtain:} \\ 0.125\cdot100\%=12.5\%. \end{gathered}[/tex]The answer would be 12.5 %.
