For He ^ + , a transition takes place from the orbit of radius 105...

The radius of the n^th orbit can be represented as,

$r = 52 . 9 \times \dfrac{{n}^{2}}{z}pm$

$∴ 105 . 8 = \dfrac{52 . 9 \times {n}^{2}}{2} ∴ {n}_{2} = 2$

and $26 . 45 = 52 . 9 \times \dfrac{{n}^{2}}{2} ∴ {n}_{1} = 1$

$∵ \Delta E = {R}_{H}hC \times {z}^{2}\left[1{n}_{1}^{2} - 1{n}_{2}^{2}\right]$

$\dfrac{hc}{\lambda} = {R}_{H}hC \times {z}^{2}\left[1{n}_{1}^{2} - 1{n}_{2}^{2}\right]$

$\dfrac{6 . 6 \times {10}^{- 34} \times 3 \times {10}^{8}}{\lambda} = 2 . 2 \times {10}^{- 18} \times 4 \times \left[\dfrac{1}{1} - \dfrac{1}{4}\right]$

$\dfrac{6 . 6 \times {10}^{- 34} \times 3 \times {10}^{8}}{\lambda} = 2 . 2 \times {10}^{- 18} \times 4 \times \dfrac{3}{4}$

$∴ \lambda = 300A$

$∴ \lambda = 30nm$