Q. The radioactivity of a certain radioactive element drops to $1/64$ of its initial value in $30\ seconds$. Its half life is
A) $2\ seconds$
B) $4\ seconds$
C) $5\ seconds$
D) $6\ seconds$
Explanation
(c) $N = N_{0}\left( \frac{1}{2} \right)^{t/T} \Rightarrow \frac{N_{0}}{64} = N_{0}\left( \frac{1}{2} \right)^{30/T} \Rightarrow T = \frac{30}{6} = 5s$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.