If the wavelength of the first line of the Balmer series of hydrogen atom is `6561Å`, the wavelength of the second line of the series should be

Options

A. `13122Å`
B. `3280Å`
C. 4860Å
D. `2187Å`

Answer

C

Explanation

Wavlength of spectral line in Blamer series is given by
`" " (1)/(lambda)=R[(1)/(2^(2))-(1)/(n^(2))]`
For first line of Blamer series,n=3
`:." " (1)/(lambda)=R[(1)/(2^(2))-(1)/(3^(2))]=(5R)/(36)`
For second line of Balmer series n=4
`" " (1)/(lambda)=R((1)/(2^(2))-(1)/(4^(2)))=(3R)/(16)`
`:. " " (lambda_(2))/(lambda_(1))=(20)/(27)`
`:." " lambda_(1)=(20)/(27)xx6561=4860Å`