The decay constant of a radio isotope is $\lambda$. If $A_1$ and $\mathrm{A}_2$ are its activities at times…

The decay constant of a radio isotope is $\lambda$. If $A_1$ and $\mathrm{A}_2$ are its activities at times $\mathrm{t}_1$ and $\mathrm{t}_2$ respectively, the number of nuclei which have decayed during the time $\left(\mathrm{t}_1-\mathrm{t}_2\right)$
  1. $\mathrm{A}_1 \mathrm{t}_1-\mathrm{A}_2 \mathrm{t}_2$
  2. $A_1-A_2$
  3. $\frac{\left(A_1-A_2\right)}{\lambda}$
  4. $\lambda\left(A_1-A_2\right)$

Solution

$\begin{aligned} \mathrm{A}_1 & =\lambda \mathrm{N}_1 \\ \mathrm{~A}_2 & =\lambda \mathrm{N}_2 \\ \mathrm{~N}_1-\mathrm{N}_2 & =\left[\frac{\mathrm{A}_1-\mathrm{A}_2}{\lambda}\right] \end{aligned}$

Asked in: NEET 2010 (Mains)

Practice more Structure of Atoms and Nuclei questions on Aicharya