Plot of number (N) of decayed atoms versus activity (A) of a radioactive substance is

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13.Nuclei

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The plot of the number $(N)$ of decayed atoms versus activity $(A)$ of a radioactive substance is

A

B

C

D

Solution

(d) $N = {N_0}{e^{{ – ^{\lambda t}}}}$

and $A = {A_0}{e^{ – \lambda t}}$$ = \lambda {N_0}{e^{ – \lambda t}}$

$\therefore$ $N_{decayed} = N_0 -N = N_0 -N_0{e^{ – \lambda t}}$

==> $N_{decayed} = {N_0} – \frac{A}{\lambda }$

This is equation of straight line with negative slope.

Standard 12

Physics

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