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

Answer

b

Explanation

`(1)/(lambda_(1))= R((1)/(2^(2))- (1)/(3^(2)))= (5R)/(36)`
`(1)/(lambda_(2))= R((1)/(2^(2))- (1)/(4^(2)))= (3R)/(16)`
`(lambda_(2))/(lambda_(1))= (80)/(108)`
`lambda_(2)= (80)/( 108)lambda_(1)= (80)/( 108) xx 660 = 488.9nm`