Taking the wavelength of first Balmer line in hydrogen spectrum (n = 3 to n = 2) as 660 nm, the wavelength of the `2^(nd)` Balmer line (n = 4 to n = 2) will be :

Options

A. 488.9 nm
B. 642.7 nm
C. 388.9 nm
D. 889.2 nm

Answer

A

Explanation

`(lambda_(2))/(660)=((1)/(2^(2))-(1)/(3^(2)))/((1)/(2^(2))-(1)/(4^(2)))rArr lambda_(2) = 488.88 ~~ 48.9 Å`