A radioactive isotope phosphorous-32 has an initial activity 10 μCi and has half life of 14 days. What is the activity of the source after 42 days?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 1.25 μCi
To calculate the activity of a radioactive isotope,
$A(t)=A_0\left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}$
$A(t)$, Current activity
$A_0$, (Initial activity) = $10μCi$
$T_{1/2}$ (Half life) = 14 days
$t$ (Elapsed time) = 42 days
$A(t)=10μCi\left(\frac{1}{2}\right)^{\frac{42}{14}}$
$=1.25 μCi$