Q. The half-life of radon is 3.8 days. How many radon will be left out of 1024 mg after 38 days
A) 1 mg
B) 2 mg
C) 3 mg
D) 4 mg
Explanation
(a) From Rutherford-Soddy law
$N = N_{0}\left( \frac{1}{2} \right)^{n}$
$\ \ \ \ \ \ \ \ n = \frac{38}{3.8} = 10$
The initial quantity of radon $N_{0} = 1024\ mg.$
Therefore, the mass of radon left after 10 half-lives is
$N = 1024\ \times \left( \frac{1}{2} \right)^{10} = \frac{1024}{1024}\ = \ 1mg.$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.