The energy of the incident light is given by:

$$E = h \nu = \frac{hc}{\lambda}$$

where $h$ is Planck's constant, $c$ is the speed of light, and $\lambda$ is the wavelength of the light. Substituting the given value of $\lambda$, we get:

$$E = \frac{hc}{hc/e} = e$$

The maximum kinetic energy of the photoelectron is given by:

$KE_{max} = E - \phi$

where $\phi$ is the work function. Substituting the values, we get:

$KE_{max} = e - \phi$

Therefore, the correct answer is Option C.