Question

Three energy levels P, Q, R of a certain atom are such that $E_P<E_Q<E_R$. If ${\lambda }_1,{\lambda }_{2}and{\lambda }_3$ are the wavelengths of radiation corresponding to transition $R\to Q;Q\to PandR\to P$ respectively. The correct relationship between ${\lambda }_1,{\lambda }_{2}and{\lambda }_3$ is:

• A. ${\lambda }_1+{\lambda }_2={\lambda }_3$
• B. $\frac{1}{{\lambda }_3}=\frac{1}{{\lambda }_1}+\frac{1}{{\lambda }_2}$
• C. ${\lambda }_3=\sqrt{{\lambda }_1{\lambda }_2}$
• D. $\frac{2}{{\lambda }_3}=\frac{1}{{\lambda }_1}+\frac{1}{{\lambda }_2}$

Answer 1

The correct option is B $\frac{1}{{\lambda }_3}=\frac{1}{{\lambda }_1}+\frac{1}{{\lambda }_2}$

Energy is given by formula
$E=\frac{hc}{\lambda }$
For transition from $R\to P$
$E_R=\frac{hc}{{\lambda }_3}$
For transition from $Q\to P$
$E_Q=\frac{hc}{{\lambda }_2}$
For transition from $R\to Q$
$E_P=\frac{hc}{{\lambda }_1}$
$E_R=E_P+E_Q$
$\frac{hc}{{\lambda }_3}=\frac{hc}{{\lambda }_1}+\frac{hc}{{\lambda }_2}$
$\Rightarrow \frac{1}{{\lambda }_3}=\frac{1}{{\lambda }_1}+\frac{1}{{\lambda }_2}$