A radioactive source (Phosphorus-32) has an initial activity 10 µCi, its half life is 14 days. What time elapses before the activity falls to 1.25 µCi?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 42 days
The formula for radioactive decay,
$A=A_1\left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}$
where,
$A_0$ = Initial Activity = $10μCi$
$A$ = Final Activity = $1.25μCi$
$T_{1/2}$ = half-life = 14 days
$⇒\frac{A}{A_0}\left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}$
$⇒\frac{1.25}{10}\left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}$
$⇒t=\frac{29.1116}{0.693}≃42days$