Confusion on Photoelectric Effect

[hazelyang2E]

Hi! I understand the general concept of photoelectric effect, but started to get confused once more specific concepts and examples were discussed during lecture. I was most confused by the diagram of the metal and the uv radiation that was pictured in the lecture slides and how the different formulas (E=hv , Ek, Work Function, etc.) factor into that diagram. Would anyone be able to explain this concept to me in a different way? Thank you!

[Aaron Ang 4H]

The photoelectric experiment shines light on a metal surface to measure energy needed to remove electrons from different metals. As the light is shined on the metal, the energy generated is conserved thus the threshold energy is equal to the energy of the electron that is deflected. Therefore, E=hv minus the work function is equal to 1/2MeVe^2

[klarratt2]

In the photoelectric effect experiment, UV light is shone on a piece of metal. Electrons are sometime ejected from the metal because of the UV light, which is picked up by a detector.

Not all frequencies of light can cause electrons to be ejected. The equation E=hv tells us the energy of the incoming light. In order for an electron to be ejected, this energy has to be greater than the work function of the metal (ϕ).

Each metal has a different work function. If a photon of light is exactly the same as the value of the work function, an electron will be ejected with 0 kinetic energy. This can be seen with this equation: E(incoming photon) - ϕ=E(kinetic). When E(incoming photon) = ϕ, kinetic energy of the electron is 0. If E(incoming photon) is greater than the work function, the ejected electron will have a positive kinetic energy. If E(incoming photon) is less than the work function, no electrons will be ejected.

The results of this experiment were surprising because until it was performed, it was generally accepted than light behaved as a wave. They thought that if an electron was not ejected, increasing the intensity of the light would cause them to eject. However, increasing the intensity of the light had no effect on whether or not electrons were ejected. The only thing that played a role in electron ejection was the frequency of light, which is directly proportional to the energy of the light (E=hv).

Hope this helps!