Question

Ratio of the longest wavelength corresponding to Lyman and Balmer series, in the hydrogen spectrum, is-

• A. $\frac{9}{31}$
• B. $\frac{5}{27}$
• C. $\frac{3}{23}$
• D. $\frac{7}{29}$

Answer 1

The correct option is B $\frac{5}{27}$

As we know, $\frac{1}{\lambda }=R(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

For longest wavelength the energy difference should be minimum, and it becomes minimum when electron jumps to its immediate higher orbital.

$\frac{1}{{\lambda }_{\text{Lyman}}}=R(\frac{1}{1^2}-\frac{1}{2^2})=\frac{3R}{4}\dots \left(1\right)$

$\frac{1}{{\lambda }_{\text{Balmer}}}=R(\frac{1}{2^2}-\frac{1}{3^2})=\frac{5R}{36}\dots \left(2\right)$

From (1) and (2) we get,

${\left(\frac{{\lambda }_{\text{Lyman}}}{{\lambda }_{\text{Balmer}}}\right)}_{\text{max}}=\frac{\frac{5R}{36}}{\frac{3R}{4}}=\frac{5}{27}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.