Question

The ratio of minimum wavelength of Lyman and Balmer series will be:

• A. 1.25
• B. 0.25
• C. 5
• D. 10

Answer 1

The correct option is B 0.25

The expression for wavelength is written as $\frac{1}{\lambda }=R{Z}^2[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$
The last line is a line of shortest wavelength or highest energy. When $n_2=\infty$, we get the last line wavelength.
$\frac{1}{\lambda }=R{Z}^2\left[\frac{1}{n_1^2}\right]$
${\lambda }_{Serieslimit}=\frac{n_1^2}{R{Z}^2}$
For Lyman series, $n_1=1$ and for Balmer series $n_1=2$.
Therefore, $\frac{{\lambda }_{Lyman}}{{\lambda }_{Balmer}}=\frac{{n_1^2}_{Lyman}}{{n_1^2}_{Balmer}}=\frac{1^2}{2^2}=\frac{1}{4}=0.25$