Question

The ratio of minimum wavelengths of Lyman and Balmer series will be

• A. $1.25$
• B. $5$
• C. $4$
• D. $0.25$

Answer 1

The correct option is D $0.25$

Using Rydberg’s formula
$\frac{1}{\lambda }=R[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$

Minimum wavelength is formed when the electron falls from infinity, so,

$\frac{1}{{\lambda }_{Lyman}}=R[\frac{1}{n_1^2}-\frac{1}{{\infty }^2}]=\frac{R}{n_1^2}$

$\therefore \frac{1}{{\lambda }_{Lyman}}=\frac{R}{1^2}=R$

Similarly, $\frac{1}{{\lambda }_{Balmer}}=\frac{R}{2^2}=\frac{R}{4}$

$\therefore \frac{{\lambda }_{lyman}}{{\lambda }_{balmer}}=\frac{1}{4}=0.25$

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


$\begin{array}{c}\text{Why this question?}\\ \text{It is important to know Lyman, Balmer series for NEET, especially the ones involving calculation of lambda using Rydberg's formula.}\end{array}$