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_(H_(2)))RZ_(H)^(2)[(1)/(4)-(1)/(9)]=R(1)^(2)[(5)/(36)]`
`(1)/(lambda_(He))=RZ_(He)^(2)[(1)/(4)-(1)/(16)]=R(4)[(3)/(16)]`
`(lambda_(He))/(lambda_(He))=(1)/(4)[(16)/(3)xx(5)/(36)]=(5)/(27)`
`lambda_(He)=(5)/(27)xx6561=1215+A214Å`