Question

For a hydrogen atom, the ratio of the largest wavelength of Lyman series to that of the Balmer series is.

(1) \(5 : 36\)

(2) \(5 : 27\)

(3) \(3: 4\)

(4) \(27 : 5\)

Answer 1

Correct option is : (2) \(5 : 27\) 

Lyman

\(\frac{1}{\lambda_1} = R \left[\frac{1}{1}- \frac{1}{4}\right] = \frac{3R}{4}\)

\(\lambda _1 = \frac{4}{3R}\)  ...(1)

and Balmer

\(\frac{1}{\lambda_2} = R \left[\frac{1}{4}- \frac{1}{9}\right] = \frac{5R}{36}\)

\(\lambda _2 = \frac{36}{5R}\) 

Then, \(\frac{\lambda _1}{\lambda_2} = \frac{5}{27}\)