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Consider an initially pure M gm sample of ^AX, an isotope that has a half life of T hour, what is its initial decay rate (N_A= Avagadro No.)

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$\frac{{M\,{N_A}}}{T}$

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

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

$\frac{{2.303\,M\,{N_A}}}{{AT}}$

answer is C.

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Detailed Solution

$N = {N_0}{e^{ - \lambda t}} \Rightarrow \left| {\frac{{dN}}{{dt}}} \right| = {N_0}\lambda {e^{ - \lambda t}}$

$Initially\,\,at\,\,t = 0,{\left| {\frac{{dN}}{{dt}}} \right|_{t = 0}} = {N_0}\lambda$

where N₀ = Initial number of undecayed atoms

$= \frac{{Mass\,of\,the\,sample}}{{Mass\,of\,a\,\sin gle\,atom\,of\,X}} = \frac{M}{{A/{N_A}}} = \frac{{M{N_A}}}{A}$

$\therefore {\left| {\frac{{dN}}{{dt}}} \right|_{t = 0}} = \frac{{M{N_A}\lambda }}{A} = \frac{{0.693\,M{N_A}}}{{AT}}$

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