Q. Consider an initially pure $'M^{'}g$ sample of $\ ^{A}X$, an isotope that has a half life of $T$ hour. What is it’s initial decay rate $(N_{A} =$ Avogrado No.)
A) $\frac{MN_{A}}{T}$
B) $\frac{0.693MN_{A}}{T}$
C) $\frac{0.693MN_{A}}{AT}$
D) $\frac{2.303MN_{A}}{AT}$
Explanation
(c) $N = N_{0}e^{- \lambda t} \Rightarrow \left| \frac{dN}{dt} \right| = N_{0}\lambda e^{- \lambda t}$
Initially at $t = 0,\left| \frac{dN}{dt} \right|_{t = 0} = N_{0}\lambda$
Where $N_{0} =$ Initial number of undecayed atoms
$= \frac{Mass\ of\ the\ sample}{Mass\ of\ a\ single\ atom\ of\ X} = \frac{M}{A/N_{A}} = \frac{MN_{A}}{A}$
$\therefore\left| \frac{dN}{dt} \right|_{t = 0} = \frac{MN_{A}\lambda}{A} = \frac{0.693MN_{A}}{AT}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.