Assume that under specific conditions, hydrogen atoms are permitted...

The visible range of the hydrogen emission spectrum corresponds to the Balmer series, where the final state is $n_f = 2$.

Given the condition that the principal quantum number changes by $2$ ($\Delta n = 2$), the initial state must be $n_i = 2 + 2 = 4$.

The energy difference for the transition from $n_i = 4$ to $n_f = 2$ is given by:

$\Delta E = 13.6 \left( \dfrac{1}{n_f^2} - \dfrac{1}{n_i^2} \right) \text{ eV}$

$\Delta E = 13.6 \left( \dfrac{1}{2^2} - \dfrac{1}{4^2} \right) \text{ eV} = 13.6 \left( \dfrac{1}{4} - \dfrac{1}{16} \right) \text{ eV}$

$\Delta E = 13.6 \times \dfrac{3}{16} \text{ eV} = 2.55 \text{ eV}$

The corresponding wavelength $\lambda$ can be calculated using the relation $\lambda = \dfrac{hc}{\Delta E}$:

$\lambda \approx \dfrac{1242 \text{ eV nm}}{2.55 \text{ eV}} \approx 487 \text{ nm}$

This wavelength falls within the visible range ($380 \text{ nm}$ to $780 \text{ nm}$). Other transitions with $\Delta n = 2$ (such as $3 \rightarrow 1$ or $5 \rightarrow 3$) fall in the ultraviolet or infrared regions, respectively.