$\frac{1}{16}$th of the initial amount of a certain radio-active isotope remains undecayed after two hours.The half life of the isotope would be:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 30 minutes
Using Decay formula,
$N=N_0\left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}$
N = Remaining Amount
$N_0$ = Initial Amount
$∵\frac{N}{N_0}=\frac{1}{16}$
time elapsed, t = 2 hours
$\left(\frac{1}{2}\right)^4=\left(\frac{1}{2}\right)^{\frac{2}{T_{1/2}}}$
$4=\frac{2}{T_{1/2}}⇒T_{1/2}=0.5\,hours$