What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n = 4 to n = 2 of He⁺ spectrum?

For an atom, wave number = 1/λ = R_HZ² [1/n₁²-1/n₂²]

For He⁺ spectrum Z = 2, n₂=4, n₁=2

Therefore [^-v] = 1/λ = R_H× {4× [1/2²-1/4²]}

= R_H× {4× [1/4 – 1/16]}

= R_H× {[4× 3] / 16}

= 3R_H / 4

For hydrogen spectrum, wave number = 3R_H/4, Z = 1

Therefore, wave number = 1/λ = R_H× 1[1/n₁²-1/n₂²]

Or

R_H [1/n₁²-1/n₂²] = 3R_H/4

Or

1/n₁² – 1/n₂² = 3/4

Which can be so for n₁ = 1 & n₂ = 2, i.e. the transition is from n = 2 to n = 1