Q. Half lives of two radioactive substances $A$ and $B$ are respectively $20\ minutes$ and $40\ minutes$. Initially the sample of $A$ and $B$ have equal number of nuclei. After 80 minutes, the ratio of remaining number of $A$ and $B$ nuclei is
A) 1 : 16
B) 4 : 1
C) 1 : 4
D) 1 : 1
Explanation
(c) For 80 minutes, number of half lives of sample $A = n_{A} = \frac{80}{20} = 4$ and number of half lives of sample $B = n_{B} = \frac{80}{40} = 2$. Also by using $N = N_{0}\left( \frac{1}{2} \right)^{n}$
$\Rightarrow N \propto \frac{1}{2^{n}} \Rightarrow \frac{N_{A}}{N_{B}} = \frac{2^{n_{B}}}{2^{n_{A}}} = \frac{2^{2}}{2^{4}} = \frac{1}{4}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.