Gujarati
13.Nuclei
medium

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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