Question

Atomic weight of a radioactive element is $M$ gram/mol. Radioactivity of $m$ gram of its mass is :

• A. $N_A\lambda$
• B. $\left(\frac{N_{A}m}{M}\right)\lambda$
• C. $\left(\frac{N_A}{m}\right)\lambda$
• D. $\left(\frac{N_A}{m}M\right)\lambda$

Answer 1

The correct option is B $\left(\frac{N_{A}m}{M}\right)\lambda$

$\begin{array}{c}\text{Given,}\\ \text{Atomic weight of element = Mgram / mol}\\ \text{So, no of mole =}\frac{M}{M_{atomic}}\end{array}$

$\begin{array}{c}\text{Now no of atoms}=\text{mole}\times \text{Avagardo No.}\\ N=\frac{m}{M}\times N_a\end{array}$

$\begin{array}{c}\text{We knew,}\\ \begin{array}{cc}\text{Radioactivity}& =\lambda N\\ R& =\frac{\lambda m}{M}Na\\ N& =\text{decay constant}\\ N& =\text{no of atom}\\ R& =\left(\frac{N_{a}m}{M}\right)\lambda \end{array}\end{array}$