## Photoelectric effect

$c=\lambda v$

Hanniel U 2B
Posts: 76
Joined: Fri Sep 28, 2018 12:16 am

### Photoelectric effect

Does photon have to be greater than work function for photoelectric effect to work?

Manya Bali 4E
Posts: 66
Joined: Fri Sep 28, 2018 12:23 am

### Re: Photoelectric effect

Yup! The photoelectric effect states that electrons are emitted from a metal when an incident light shines on the metal. However, the electron can't be emitted unless the energy of the photon is equal to or greater than the threshold energy/work of the metal. If the photon is able to emit the electron, then depending on how much excess energy the photon contained, the kinetic energy of the emitted electron can be determined.

ran2000
Posts: 66
Joined: Fri Sep 28, 2018 12:15 am

### Re: Photoelectric effect

Yes. The photon of the incident light must have sufficient energy such that it exceeds or equals the work function. Only then, will the electron have sufficient energy to be ionized and get released to be detected and produce the photoelectric effect.

Courtney Quan 1C
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### Re: Photoelectric effect

The energy of an individual photon (calculated by E=hv) must be greater than or equal to the work function (energy required to remove an electron from the metal) in order to displace an electron. Ephoton-Ework function/energy to remove an electron=Ekinetic energy of electron, so a photon with an amount of energy exactly equal to the work function would result in displacing an electron with zero velocity and thus zero kinetic energy. Photons with energy greater than that of the work function will result in removing an electron with some velocity, and therefore some kinetic energy.

Katelyn Phan 2A
Posts: 61
Joined: Fri Sep 28, 2018 12:23 am

### Re: Photoelectric effect

The energy of each photon does not necessarily have to be greater. It needs to reach a certain level of energy to remove an electron. That fence it needs to climb over would be the threshold or work energy.