Question

Photons of what minimum frequency are required to remove electrons from gold? note: the work function for gold is 4.8 ev.

Answer 1

In the photoelectric effect, the energy given by the incoming photon is used partially to extract the electron from the metal (work function) and the rest is converted into kinetic energy of the electron:
$hf= \phi + K$
where
hf is the energy of the photon, with h being the Planck constant and f the frequency of the photon
$\phi$ is the work function
K is the kinetic energy of the electron

When K=0, we have the minimum energy required to extract the electron from the metal, so the equation becomes
$hf= \phi$ (1)

If we convert the work function of gold into Joules:
$\phi=4.8 eV = 7.69 \cdot 10^{-19}J$
We can re-arrange eq.(1) to find the minimum energy of the photon:
$f= \frac{\phi}{h}= \frac{7.69 \cdot 10^{-19}J}{6.6 \cdot 10^{-34} Js} =1.17 \cdot 10^{15} Hz$