Q. The half-life of a radioactive substance is 3.6 days. How much of $20\ mg$ of this radioactive substance will remain after 36 days
A) $0.0019\ mg$
B) $1.019\ mg$
C) $1.109\ mg$
D) $0.019\ mg$
Explanation
(d) $M = M_{0}\left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}} = 20 \times \left( \frac{1}{2} \right)^{\frac{36}{3.6}} = 20 \times \left( \frac{1}{2} \right)^{10} = 0.019mg$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.