Discussion
75% of 23490Th decays in t years. Its half-life is (in years):
Answer: B
The half-life is \(\frac{t}{2}\) years.
When 75% has decayed, 25% remains, which is \(\frac{1}{4} = \left(\frac{1}{2}\right)^{2}\). So two half-lives have passed in t years:
\[ 2T_{1/2} = t \Rightarrow T_{1/2} = \frac{t}{2} \]
The trap is to read 75% decayed as 75% remaining, or to take t itself as the half-life, which would leave 50% undecayed. \(\frac{t}{4}\) assumes four half-lives.
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