The wavelength of radiation emitted, when He⁺ makes a transition from the state n = 3 to the state n = 2 will be:

(Take Rydberg constant R = 1.097 × 10⁷ m^-1)

164.1 nm

$ \frac{1}{\lambda} = R Z^2( \frac{1}{2^2} - \frac{1}{3^2}) = 4R (\frac{1}{4} - \frac{1}{9} ) = 4R(\frac{5}{36}) = \frac{5R}{9}$

$ \lambda = \frac{9}{5R} = \frac{9}{5} \times 91.2nm = 164.1 nm$