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

Explanation

For H- atom : ` 1/( lambda_H) = R_H [ 1/n_1^2 - 1/n_2^2]` ..(1)
From `He^+ ion ,`
` 1/( lambda_(He^+)) = R_H xx Z^2 [ 1/n_(2)] `
` = 91. 2 xx 1/2^2 = 22 . 8 nm`.