Photoelectric Effect...Watch this thread
I answered, 'When photons of energy greater than the work function of the metal are incident, they will cause electrons to be emitted.'
In the mark scheme it talked about the threshold frequency. I'm a bit confused when I'd use the threshold frequency or the work function in my answer?
"Different materials require different levels of energy for the photoelectric effect to occur. Some, like alkali metals, experience photoelectric effect at a very low energy threshold (even visible light). Some non-metals require extreme ultraviolet light to get it started. We consider the amount of energy it requires to launch an electron W, or the work function. A photon must contain energy at the level of the work function or higher, or the photoelectric effect will not be activated.
It is the energy per photon, not the overall energy from a light source which matters in the photoelectric effect. This is why a high-frequency, low-intensity light can cause it while a low-frequency, high-intensity light may not."
I think there's a formula for what the work function is for different materials.
The energy of the photon is equal to hf and h is a constant so the amount of energy it carries is proportional to the frequency. The light has to be of a certain frequency (threshold) in order to reach work function and have some energy left over for the KE upon release. So threshold frequency is required to cause the photoelectric effect, but it happens when the threshold allows work function to be reached with some energy for KE when the photo-electron is released
This site may be of use: http://physicsnet.co.uk/a-level-phys...ectric-effect/
particularly this graph at the bottom: http://physicsnet.co.uk/wp-content/u...tric-graph.jpg