After 4 hours, $\frac{1}{16}$ of the initial amount of a certain radioactive isotype remains undecayed. What is the half life of the isotope?

60 min

The correct answer is Option (4) → 60 min

Radioactive Decay,

$N(t)=N_0\left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}$

where,

$N(t)$ → Amount remaining after time t

$N_0$ → Initial amount

$t$ → time elapsed = 4 hour

$T_{1/2}$ → half life

$\frac{N(t)}{N_0}=\frac{1}{16}$

$\frac{1}{16}=\left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}⇒\left(\frac{1}{2}\right)^4=\left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}$

$⇒\frac{t}{T_{1/2}}=4⇒T_{1/2}=60\,min=1\,hour$