Chemistry - Atomic Structure Question with Solution | TestHub

Explanation Question: Question is asking about difference in wavelength of first and last line of Lyman series for ${\mathrm{H}}^{+}$.

Concept: This question is based on atomic line spectrum.

Solution: Wave length $=\frac{1}{\text{ Wave number }}$

Wave number for a line in spectrum

$\bar{\nu }=R{z}^2\left[\frac{1}{{\mathrm{n}}_1^2}-\frac{1}{{\mathrm{n}}_2^2}\right]$

$\mathrm{Z}=2$ Lyman series ${\mathrm{n}}_1=1$

First line ${\mathrm{n}}_2=2$, last line ${\mathrm{n}}_2=\infty$

${\bar{\nu }}_{{\mathrm{L}}_{\mathrm{I}}}={\mathrm{R}}_{{\mathrm{H}}^2}{2}^2\left[\frac{1}{1^2}-\frac{1}{2^2}\right]=4{\mathrm{R}}_{\mathrm{H}}\times \frac{3}{4}\Rightarrow {\lambda }_{{\mathrm{L}}_{\mathrm{I}}}=\frac{1}{3{\mathrm{R}}_{\mathrm{H}}}$ ${\bar{\nu }}_{{\mathrm{L}}_{\text{Last }}}={\mathrm{R}}_{\mathrm{H}}{2}^2\left[\frac{1}{1^2}-\frac{1}{{\infty }^2}\right]=4{\mathrm{R}}_{\mathrm{H}}\Rightarrow {\lambda }_{{\mathrm{L}}_{\text{Last }}}=\frac{1}{4{\mathrm{R}}_{\mathrm{H}}}$

So,

$${\lambda }_{L_I}-{\lambda }_{L_{Last}}=\frac{1}{3R_H}-\frac{1}{4R_H}$$

$$=\frac{1}{12R_H}$$

Final Answer: option (A)