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?

1.25 μCi

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

$A(t)=10μCi\left(\frac{1}{2}\right)^{\frac{42}{14}}$

$=1.25 μCi$