Question

Transition between three energy levels in a particular atom gives rise to three spectral lines of wavelengths, in increasing magnitudes as ${\lambda }_1,{\lambda }_2$ and ${\lambda }_3.$ Which one of the following equations correctly relates ${\lambda }_1,{\lambda }_2$ and ${\lambda }_3$?

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

Answer 1

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

Let the orbits of atom be $n_1$, $n_2$ & $n_3$ such that $n_1<n_2<n_3$.

$\therefore {\lambda }_3={\lambda }_{n_3\to n_2};{\lambda }_2={\lambda }_{n_2\to n_1};{\lambda }_1={\lambda }_{n_3\to n_1}$

From Rydberg's formula,

$\frac{1}{{\lambda }_3}=R{Z}^2(\frac{1}{n_2^2}-\frac{1}{n_3^1}).......\left(1\right)$

Similarly,

$\frac{1}{{\lambda }_2}=R{Z}^2(\frac{1}{n_1^2}-\frac{1}{n_2^1}).......\left(2\right)$

$\frac{1}{{\lambda }_1}=R{Z}^2(\frac{1}{n_1^2}-\frac{1}{n_3^1}).......\left(3\right)$


Adding equations $\left(1\right)\&\left(2\right)$, we get $\left(3\right)$,

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

Therefore, $\left(\text{C}\right)$ is the right option.