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The wave length of the Balmer transition n = 4 to n = 2 of He⁺ spectrum is same as n = 2 to n = 1 transition in the hydrogen spectrum, then its region is

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Detailed Solution

$\frac{1}{λ_{(H e^{+})}} = R_{H} \times 4 (\frac{1}{2^{2}} - \frac{1}{4^{2}}) = R_{H} (\frac{1}{1^{2}} - \frac{1}{2^{2}})$
As the wave length ( $λ$ ) is equal to that of lyman series of hydrogen spectrum, it is in u.v. region.

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