Q. The half-life of $At^{215}$ is 100$\mu s.$ If a sample contains 215 mg of $At^{215}$, the activity of the sample initially is
A) $10^{2}Bq$
B) $3\ \times \ 10^{10}\ Bq$
C) $4.17\ \times \ 10^{24}\ Bq$
D) $1.16\ \times 10^{5}\ \ Bq$
Explanation
(c) $T = \ 100\ \mu s\ = \ 10^{- 4}\ s$
$\ \ \ \ \ \ \ \ \lambda = \frac{0.6931\ }{T} = \frac{0.6931\ }{10^{- 4}} = 0.6931 \times 10^{4}s^{- 1}$
Number of atoms in 215 mg
$\ \ \ \ \ \ \ \ \ \ = \frac{0.6931\ \times 10^{23}}{215} \times 215 \times 10^{- 3}$
$\therefore\ N = 6.023\ \times 10^{20}$
$Activity,\ \frac{dN}{dt}\ = \lambda N$
$\ \ \ \ = \ 0.6931\ \times \ 10^{4}\ \times \ 6.023\ \times \ 10^{20}$
$\ \ \ \ \ \ \ \ = \ 4.17\ \times \ 10^{24}Bq$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.