Question

The ratio of the longest to the shortest wavelength in the Balmer series of Hydrogen spectrum is

• A. 25/9
• B. 17/6
• C. 9/5
• D. 5/4

Answer 1

The correct option is C

9/5

Longest wavelength$\frac{1}{{\lambda }_{\alpha }}=k(\frac{1}{2^2}-\frac{1}{\alpha })=R.\frac{5}{36}$

Shortest wavelength $\frac{1}{{\lambda }_S}=k(\frac{1}{2^2}-\frac{1}{\alpha })=R.\frac{1}{4}$

$\therefore \frac{{\lambda }_{\alpha }}{{\lambda }_S}=\frac{\frac{R}{4}}{R.\frac{5}{36}}=\frac{9}{5}$