- Home
- Standard 12
- Physics
Consider an initially pure $M$ gm 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.)
$\frac{{M\,{N_A}}}{T}$
$\frac{{0.693\,M\,{N_A}}}{T}$
$\frac{{0.693\,M\,{N_A}}}{{AT}}$
$\frac{{2.303\,M\,{N_A}}}{{AT}}$
Solution
(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{{M{N_A}}}{A}$
$\therefore$ ${\left| {\frac{{dN}}{{dt}}} \right|_{t = 0}} = \frac{{M{N_A}\lambda }}{A} = \frac{{0.693\,M{N_A}}}{{AT}}$
