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?
Answer & explanation
Correct answer: option 4
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
$\frac{N(t)}{N_0}=\frac{1}{16}$
$\frac{1}{16}=\left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}⇒\left(\frac{1}{2}\right)^4=\left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}$
$⇒\frac{t}{T_{1/2}}=4⇒T_{1/2}=60\,min=1\,hour$