Wavelength of high enrgy trabsition fo H-atoms is ` 91 . 2 nm`. Calculate the corresponding wavelength of `He^(+)`.

Explanation

For H atom: `1/(lamda_(H))=R_(H)[1/(n_(1)^(2))-1/(n_(2)^(2))]`…….i
For `He^(+)` ion: `1/(lamdaHe^(+))=R_(H)xxZ^(2)[1/(n_(1)^(2))-1/(n_(2)^(2))]`……….ii
By eqs I and ii `lamda_(He^(+))=lamda_(H)xx1/(Z^(2))`
`:.lamda_(H)=91.2nm, Z` for `He^(+)=2`
`:.lamda_(He^(+))=91.2xx1/(2^(2))=22.8nm`