Q. $C^{14}$ has half-life 5700 year. At the end of 11400 years, the actual amount left is
A) 0.5 of original amount
B) 0.25 of original amount
C) 0.125 of original amount
D) 0.0625 of original amount
Explanation
(b) $N = \ N_{0} \times \left( \frac{1}{2} \right)^{11400/5700}$
$\ \ = \ N_{0}\left( \frac{1}{2} \right)^{2} = 0.25\ N_{0}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.