Q. If a radioactive substance reduces to $\frac{1}{16}$ of its original mass in $40\ days$, what is its half life
A) $10\ days$
B) $20\ days$
C) $40\ days$
D) None of these
Explanation
(a) $\frac{N}{N_{0}} = \left( \frac{1}{2} \right)^{n} \Rightarrow \frac{1}{16} = \left( \frac{1}{2} \right)^{4} = \left( \frac{1}{2} \right)^{n} \Rightarrow n = 4$
Also $n = \frac{t}{T_{1/2}} \Rightarrow T_{1/2} = \frac{40}{4} = 10days$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.