E(photon)>or=E(energy remove e-)  [ENDORSED]

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paulacamara1E
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Joined: Fri Apr 06, 2018 11:05 am

E(photon)>or=E(energy remove e-)

Postby paulacamara1E » Sun Apr 15, 2018 11:31 pm

Can someone further explain the statement presented in lecture that says, "unless E(photon)>or=E(energy remove e-) then e- not emitted even for high intensity light"?

Bianca Nguyen 1B
Posts: 36
Joined: Fri Apr 06, 2018 11:04 am

Re: E(photon)>or=E(energy remove e-)

Postby Bianca Nguyen 1B » Sun Apr 15, 2018 11:44 pm

Basically, in the photoelectric effect, light is not acting like a classical wave, where the bigger or more intense the wave is, the more energy it has. So increasing the intensity of light does not add to the energy of photons. Rather the energy of a photon is proportional to its frequency, so even a low intensity light can eject electrons if it has high enough frequency, meaning that the energy of the photons are high enough to remove these electrons from their surface. Therefore, if the energy of the photon is not greater than or equal to the energy required to move an electron, then electrons will not be emitted even for high intensity light, as the key to removing electrons is energy here, not intensity.

Alexandra Wade 1L
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Re: E(photon)>or=E(energy remove e-)  [ENDORSED]

Postby Alexandra Wade 1L » Sun Apr 15, 2018 11:56 pm

Basically this means that if the energy to remove the electron is greater than the initial energy harnessed in the photons then electron cannot be emitted.


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