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

Options

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

Answer

C

Explanation

The wavelength of spectral line in Balmer series
given by `(1)/(lambda) = R [(1)/(2^(2)) - (1)/(n^(2))]`
For first line of Balmer series, `n = 3`
`rArr (1)/(lambda_(1)) = R [(1)/(2^(2)) - (1)/(3^(2))] = (5R)/(36)`, For second line `n = 4`.
`rArr (1)/(lambda_(2)) = R [(1)/(2^(2)) - (1)/(4^(2))] = (3R)/(16)`
`:. (lambda_(2))/(lambda_(1)) = (20)/(27) rArr lambda_(1) = (20)/(27) xx 6561 = 4860 Å`