Let's break down the problem step by step.

The photoelectric equation is given by:

$$K_{\text{max}} = \frac{hc}{\lambda} - \phi$$

where:

$$K_{\text{max}}$$ is the maximum kinetic energy of the electrons.

$$\frac{hc}{\lambda}$$ is the energy of the incident photon.

$$\phi$$ is the work function of the metal.

For the initial light source of wavelength $$\lambda$$, we know:

$$2 \text{ eV} = \frac{hc}{\lambda} - 1 \text{ eV}$$

Solving for $$\frac{hc}{\lambda}$$ gives:

$$\frac{hc}{\lambda} = 2 \text{ eV} + 1 \text{ eV} = 3 \text{ eV}$$

Now, if the wavelength is halved to $$\frac{\lambda}{2}$$, the photon energy becomes:

$$\frac{hc}{\lambda/2} = \frac{2hc}{\lambda} = 2 \times 3 \text{ eV} = 6 \text{ eV}$$

The maximum kinetic energy for the new wavelength is then:

$$K'_{\text{max}} = 6 \text{ eV} - 1 \text{ eV} = 5 \text{ eV}$$

Therefore, the maximum kinetic energy of the ejected electrons when the surface is illuminated by a light source of wavelength $$\frac{\lambda}{2}$$ is 5 eV.

The correct answer is Option A.