Monochromatic light of wavelength λ ejects photoelectrons from a...

For hydrogen atoms to emit visible light, they must be excited to at least the $n=3$ state, as visible light corresponds to the Balmer series (transitions to $n=2$).

The minimum energy required to excite a hydrogen atom from the ground state ($n=1$) to $n=3$ is:

$\Delta E = 13.6 \left( \dfrac{1}{1^2} - \dfrac{1}{3^2} \right) = 13.6 \times \dfrac{8}{9} = 12.09 \text{ eV}$

The photoelectrons must have a maximum kinetic energy K_{\text\max} of at least $12.09 \text{ eV}$ to cause this excitation.

Using the photoelectric equation:

K_{\text\max} = \dfrac{hc}{\lambda} - \Phi

Substituting the known values and $hc \approx 1242 \text{ eV}\cdot\text{nm}$:

$12.09 = \dfrac{1242}{\lambda} - 1.9$

$\dfrac{1242}{\lambda} = 13.99$

$\lambda = \dfrac{1242}{13.99} \approx 88.8 \text{ nm}$

The closest maximum wavelength from the given options is $89 \text{ nm}$.