The wavelength corresponding to maximum energy for hydrogen is 91....

Accordinq to Rydberg's equation

$\overset{¯}{v} = \dfrac{1}{\lambda} = {RZ}^{2}\left[1{n}_{L}^{2} - 1{n}_{H}^{2}\right]$

where,

$\overset{¯}{v} =$ wave number

$\lambda =$ wavelength

$R =$ Rydberg constant \left(= 109677{cm}^{- 1}\right)$</p><p>$Z =$atomic number</p><p>${n}_{L} =$lower energy state</p><p>${n}_{H} =$higher energy state.</p><p>i.e.$\dfrac{1}{\lambda} ∝ {Z}^{2}$</p><p>Thus,$\dfrac{\lambda\left({He}^{+}\right)}{\lambda\left(H\right)} = \dfrac{{Z}^{2}\left(H\right)}{{Z}^{2}\left({He}^{+}\right)}$</p><p>To find$\lambda\left({He}^{+}\right) = ?$</p><p>Given,$\lambda\left(H\right) = 91 . 2nm$</p><p>$Z\left(H\right) = 1$</p><p>$Z\left(He\right) = 2$</p><p>$∴ \lambda\left({He}^{+}\right) = \dfrac{1 \times 91 . 2}{4} = 22 . 8nm