Consider an initially pure M gm sample of_ A{ | vedclass

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(c) $N = {N_0}{e^{ - \lambda t}}$ $ \Rightarrow \left| {\frac{{dN}}{{dt}}} \right| = {N_0}\lambda {e^{ - \lambda t}}$
Initially at $t = 0$, ${\left|

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$\frac{{0.693\,M\,{N_A}}}{{AT}}$

Vedclass · Answer

(c) $N = {N_0}{e^{ - \lambda t}}$ $ \Rightarrow \left| {\frac{{dN}}{{dt}}} ight| = {N_0}\lambda {e^{ - \lambda t}}$
Initially at $t = 0$, ${\left| {\frac{{dN}}{{dt}}} ight|_{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}$
$	herefore$ ${\left| {\frac{{dN}}{{dt}}} ight|_{t = 0}} = \frac{{M{N_A}\lambda }}{A} = \frac{{0.693\,M{N_A}}}{{AT}}$

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.)

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