Question

A photon of wavelength 350nm and intensity 1.00W/m2 is directed at a Potassium surface (Work Function = 2.2 eV)

Show if the incident energy is greater than the work function.

Answer 1

The energy of a photon is given by:
\[E = hf = \frac{hc}{\lambda}\]
where h is Planck's constant, f is frequency, lambda is wavelength, and c is the speed of light.

The intensity just describes how many of these photons there are, but doesn't effect their individual energy.

As stated, the question seems to be asking "Is the energy of a single photon greater than the work function?" So, just find the energy of a photon of 350 nm light, and see if it's 2.2eV or more.