Q. What is the disintegration constant of radon if the number of its atoms diminishes by 18%in 24 h?
A) $2.1\ \times \ 10^{- 3}\ s^{- 1}$
B) $2.1\ \times \ 10^{- 4}\ s^{- 1}$
C) $2.1\ \times 10^{- 5}\ s^{- 1}$
D) $2.1\ \times 10^{- 6}\ s^{- 1}$
Explanation
(d) Undisintegrated part
$\ \ \ \ \ \ \ \ \ \frac{N}{N_{0}} = (100 - 18)\% = 82\%$
Using relation $N = N_{0}(e^{- \lambda t})$
$\ \ \ \ \ \ \ \frac{82}{100} = \ e^{- (24\ \times 60\ \times \ 60\lambda)}$
$\therefore 24\ \times \ 60\ \times \ 60\ \times \ \lambda = \log\left( \frac{100}{82} \right)$
or$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lambda = 2.1\ \times \ 10^{- 6}\ s^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.