A radioactive sample's activity drops to $1/8$ of the initial value over $30$ years. Its half-life is:
A$30$ years, the same as the elapsed time given here
B$3.75$ years, dividing the elapsed by the ratio factor
C$10$ years, since $1/8 = (1/2)^3$ means three half-lives
D$8$ years, equal to the inverse ratio of the activity drop
Answer & Solution
Correct answer: C. $10$ years, since $1/8 = (1/2)^3$ means three half-lives
$1/8 = (1/2)^3 \Rightarrow 3$ half-lives in 30 years $\Rightarrow T_{1/2} = 10$ years.
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