In hydrogen spectrum L_α line arises due to transition of electron from the orbit n=3 to the orbit n=2. In the spectrum of singly ionized helium there is a line having the same wavelength as that of the L_α line.
This is due to the transition of electron from the orbit:

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n = 5 to n = 3

n = 6 to n = 4

n = 3 to n = 2

n = 4 to n = 2

answer is D.

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Detailed Solution

For H–atom , z = 1, n₁ = 2, n₂ = 3
$\begin{array}{l} ∴ \frac{1}{λ} = {Rz}^{2} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}) \\ ∴ \frac{1}{λ_{L_{α}}} = R (1)^{2} (\frac{1}{2^{2}} - \frac{1}{3^{2}}); λ_{L_{α}} = \frac{36}{5 R} \to 1 \end{array}$
For He–atom, z = 2, n₁ = 4, n₂ = 6
$∴ \frac{1}{λ_{L_{α}}^{'}} = R (2)^{2} (\frac{1}{4^{2}} - \frac{1}{6^{2}}); λ_{L_{α}}^{'} = \frac{36}{5 R} \to 2$
i.e.., ${(λ_{L_{α}})}_{H - atom} = {(λ_{L_{α}}^{'})}_{He - atom}$
at n = 6 to n = 4