Excited atoms emit radiations consisting of only certain discrete frequencies or wavelengths. In spectroscopy it is often more convenient to use frequencies or wave numbers than wavelength because frequencies and wave numbers are proportional to energy and spectroscopy involves transitions between different energy levels. The line spectrum shown by a mono electronic excited atom (a finger print of an atom) is called atomic spectrum. `1/(lambda) = Z^2 R [1/(n_1^2 - 1/(n_2^2)]`
The ratio of wavelength for II line of Balmer series and I line of Lyman series is

Options

A. 1
B. 2
C. 3
D. 4

Answer

D

Explanation

For II line of Balmer : `n_1 = 2 , n_2= 4` ,
`1/(lambda_B) = R_H xx [1/(2^2) - 1/(4^2) ] = R_H xx 3/(16)`
For I line of Balmer : `n_1= 1 , n_2 = 2 `,
`1/(lambda_L) = R_H xx [1/(1^2) - 1/(2^2)] = R_H xx 3/4 therefore (lambda_B)/(lambda_L) = 4`