The wavelength of the first spectral line in the Balmer series of hydrogen atom is `6561 A^(@)` . The wavelength of the second spectral line in the Balmer series of singly - ionized helium atom is

Options

A. `1215 Å`
B. `1640 Å`
C. `2430 Å`
D. `4687 Å`

Answer

A

Explanation

`(1)/(lambda_(Hz)) = RZ_(H)^(2) [(1)/(4) - (1)/(9)] = R (1)^(2) [(5)/(36)]`
`(1)/(lambda_(He)) = RZ_(He)^(2) [(1)/(4) - (1)/(16)] = R (4)^(2) [(3)/(16)]`
`(lambda_(Hz))/(lambda_(He)) = (1)/(4) [(16)/(3) xx (5)/(36)] = (5)/(27)`
`lambda_(He) = (5)/(27) xx 6561 = 1215 Å`