What will be the half life a radioactive isotope whose $(\frac{1}{16})th$ of initial amount remains unchanged after 2 hours?

30 min

The correct answer is Option (3) → 30 min

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

where,

N → Remaining amount of substance after t

$N_0$ → initial amount

$T_{1/2}$ → half life of substance

$t$ → time elapsed

$\frac{N}{N_0}=\frac{1}{16}$ [given] for $t=2$ hours

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

$⇒\frac{t}{T_{1/2}}=4$

$T_{1/2}=\frac{1}{2}hr=30min$