Imagine a hypothetical world where angular momentum is quantized as...

In Bohr's model, the angular momentum is given by $L = \dfrac{nh}{2\pi}$.

Given that angular momentum is quantized as even integral multiples of $\dfrac{h}{2\pi}$, the allowed principal quantum numbers are $n = 2, 4, 6, 8, \dots$

The energy of an electron in the $n$-th orbit of a hydrogen atom is $E_n = -\dfrac{13.6}{n^2} \text{ eV}$.

For a transition to emit a photon in the visible range (approximately $400 \text{ nm}$ to $700 \text{ nm}$, corresponding to energies between $1.8 \text{ eV}$ and $3.1 \text{ eV}$), the electron must transition to the $n = 2$ state. The allowed transitions to $n = 2$ in this hypothetical world are from $n = 4, 6, 8, \dots$

The longest wavelength corresponds to the minimum energy difference, which occurs for the transition from $n = 4$ to $n = 2$.

The energy difference is:

$\Delta E = E_4 - E_2 = -13.6 \left( \dfrac{1}{4^2} - \dfrac{1}{2^2} \right) = 13.6 \left( \dfrac{1}{4} - \dfrac{1}{16} \right) = 13.6 \times \dfrac{3}{16} = 2.55 \text{ eV}$

The corresponding wavelength is:

$\lambda = \dfrac{hc}{\Delta E} \approx \dfrac{1242 \text{ eV nm}}{2.55 \text{ eV}} \approx 487 \text{ nm}$

Answer: $487 \text{ nm}$