A beam of monochromatic light of wavelength λ is used to eject...

The energy required to excite a hydrogen atom from the ground state ($n=1$) to the first excited state ($n=2$) is given by:

$\Delta E = 13.6 \left( \dfrac{1}{1^2} - \dfrac{1}{2^2} \right) \text{ eV} = 10.2 \text{ eV}$

For the ejected photoelectrons to excite the hydrogen atoms, their maximum kinetic energy must be at least equal to this excitation energy. To find the maximum wavelength $\lambda$ of the incident light, we consider the case where the maximum kinetic energy of the photoelectrons is exactly $10.2 \text{ eV}$.

Using Einstein's photoelectric equation:

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

Substituting the given values:

$10.2 \text{ eV} = \dfrac{hc}{\lambda} - 1.9 \text{ eV}$

$\dfrac{hc}{\lambda} = 10.2 \text{ eV} + 1.9 \text{ eV} = 12.1 \text{ eV}$

Using the standard approximation $hc \approx 1240 \text{ eV} \cdot \text{nm}$:

$\lambda = \dfrac{1240}{12.1} \text{ nm} \approx 102.4 \text{ nm}$

The closest value among the given options is $102 \text{ nm}$.