Question

Energy levels $A,B$ and $C$ of a certain atom, correspond to increasing value of energy, i.e, $E_A<E_B<E_C$. If ${\lambda }_1,{\lambda }_2\text{and}{\lambda }_3$ are the wavelengths of radiations corresponding to transitions $C\to B$, $B\to A$ and $C\to A$ respectively, which one of the following relations is correct?

• A. ${\lambda }_3={\lambda }_1+{\lambda }_2$
• B. ${\lambda }_3=\frac{{\lambda }_1{\lambda }_2}{{\lambda }_1+{\lambda }_2}$
• C. ${\lambda }_1+{\lambda }_2+{\lambda }_3=0$
• D. ${\lambda }_3^2={\lambda }_1^2+{\lambda }_2^2$

Answer 1

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

For transition of an electron,
$\frac{1}{\lambda }=R(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

According to the given information,

$\frac{1}{{\lambda }_1}=R(\frac{1}{n_C^2}-\frac{1}{n_B^2})---\left(1\right)$

$\frac{1}{{\lambda }_2}=R(\frac{1}{n_B^2}-\frac{1}{n_A^2})---\left(2\right)$

$\frac{1}{{\lambda }_3}=R(\frac{1}{n_C^2}-\frac{1}{n_A^2})---\left(3\right)$

Adding $\left(1\right)$ and $\left(2\right)$, we get,
$\frac{1}{{\lambda }_1}+\frac{1}{{\lambda }_2}=R(\frac{1}{n_C^2}-\frac{1}{n_B^2})+R(\frac{1}{n_B^2}-\frac{1}{n_A^2})$
$=R(\frac{1}{n_C^2}-\frac{1}{n_A^2})$

$\therefore \frac{1}{{\lambda }_1}+\frac{1}{{\lambda }_2}=\frac{1}{{\lambda }_3}$

$\Rightarrow {\lambda }_3=\frac{{\lambda }_1{\lambda }_2}{{\lambda }_1+{\lambda }_2}$

Hence, $\left(B\right)$ is the correct answer.