What will be the half life a radioactive isotope whose $(\frac{1}{16})th$ of initial amount remains unchanged after 2 hours?
Answer & explanation
Correct answer: option 3
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
$\frac{1}{16}=\left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}$
$⇒\frac{t}{T_{1/2}}=4$
$T_{1/2}=\frac{1}{2}hr=30min$