Q. A radioactive isotope has a half-life of $T$ years. How long will it take the activity to reduce to 1% of its original value
A) $3.2T$ year
B) $4.6\ T$ year
C) $6.6\ T$ year
D) $9.2\ T$ year
Explanation
(c) $N = N_{0}\left( \frac{1}{2} \right)^{n}$
$\Rightarrow \frac{1}{100}N_{0} = N_{0}\left( \frac{1}{2} \right)^{n} \Rightarrow \frac{1}{100} = \left( \frac{1}{2} \right)^{n} \Rightarrow n = \frac{2}{\log 2}$
$\Rightarrow \frac{t}{T} = \frac{2}{\log 2} \Rightarrow t = 6.6T\ year$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.