Step 1: For a hydrogen-like species, the wavenumber ($\bar{\nu}$) is given by $\bar{\nu} = Z^2 \times 109,677 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \text{ cm}^{-1}$. For He⁺, the atomic number $Z = 2$.

Step 2: For the Balmer transition $n = 4$ to $n = 2$ in He⁺, the wavenumber is: $\bar{\nu}_{\text{He}^+} = 2^2 \times 109,677 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = 4 \times 109,677 \left( \frac{1}{4} - \frac{1}{16} \right) = 4 \times 109,677 \left( \frac{3}{16} \right) = 109,677 \left( \frac{3}{4} \right) \text{ cm}^{-1}$.

Step 3: For the hydrogen atom, $Z = 1$, so the wavenumber is $\bar{\nu}_{\text{H}} = 109,677 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \text{ cm}^{-1}$.

Step 4: Since the wavelengths are the same, their wavenumbers must be equal. Equating the two expressions:

Step 5: Solving for $n_1$ and $n_2$, if we take $n_1 = 1$, then $1 - \frac{1}{n_2^2} = \frac{3}{4}$, which gives $\frac{1}{n_2^2} = \frac{1}{4}$. Therefore, $n_2 = 2$. The transition is $n = 2$ to $n = 1$.