Q. In a radioactive substance at $t = 0$, the number of atoms is $8 \times 10^{4}$. Its half life period is $3\ years$. The number of atoms $1 \times 10^{4}$ will remain after interval
A) $9\ years$
B) $8\ years$
C) $6\ years$
D) $24\ years$
Explanation
(a) By formula $N = N_{0}\left( \frac{1}{2} \right)^{t/T}$ or $10^{4} = 8 \times 10^{4}\left( \frac{1}{2} \right)^{t/3}$
Or $\left( \frac{1}{8} \right) = \left( \frac{1}{2} \right)^{t/3}$ or $\left( \frac{1}{2} \right)^{3} = \left( \frac{1}{2} \right)^{t/3} \Rightarrow 3 = \frac{t}{3}$
Hence $t = 9\ years$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.