Photoelectric Effect

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NayeliPlayas_1L
Posts: 130
Joined: Wed Sep 30, 2020 9:59 pm

Photoelectric Effect

Postby NayeliPlayas_1L » Mon Oct 12, 2020 1:37 pm

Can anyone define/explain the photoelectric effect in simple terms?
Also can anyone explain the equation E(photon)- E(energy remove e-) = E(excess) = Ek (e-)

Darlene Lien 3E
Posts: 117
Joined: Wed Sep 30, 2020 9:37 pm

Re: Photoelectric Effect

Postby Darlene Lien 3E » Mon Oct 12, 2020 1:40 pm

In simple terms, the energy of a photon + the energy required to remove an electron = the energy of the removed electron (the excess/what remains). Hope this helps!

Samantha Pedersen 2K
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Re: Photoelectric Effect

Postby Samantha Pedersen 2K » Mon Oct 12, 2020 1:50 pm

The photoelectric effect is the phenomenon where light of certain frequencies projected onto a metal causes electrons to be ejected from the metal.

As for the equation, every metal has a certain energy threshold that each photon must meet in order to eject an electron. This energy threshold is denoted by E (energy remove e-). Any additional energy of the photon above that threshold will be transferred to the electron in the form of kinetic energy, which is denoted by Ek (e-). You can determine the amount of additional energy available to be transferred to the electron in the form of kinetic energy, Ek (e-), by subtracting the energy threshold per photon required to remove an electron, E (energy remove e-), from the actual energy of the photon, E (photon). I hope this helps!

DMaya_2G
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Re: Photoelectric Effect

Postby DMaya_2G » Mon Oct 12, 2020 3:00 pm

What exactly are the "unexpected outcomes" that make the photoelectric effect?

Saumya Tawakley 1E
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Re: Photoelectric Effect

Postby Saumya Tawakley 1E » Mon Oct 12, 2020 3:25 pm

I'm not quite sure if this is what you were referring to, but the unexpected outcome that was observed during the photoelectric experiment was that no electrons were ejected even when a higher intensity light was used. That led scientists to decrease the wavelength of light to the UV region and that's when they observed electrons ejected (even when they were using a lower intensity light). That led to the eventual conclusion that the energy of the photon must be greater than or equal to the energy needed to eject the electron from the metal, and that can be accomplished using a shorter wavelength light as opposed to higher intensity light (which merely increased the number of photons hitting the metal surface rather than changing their actual energy).

Wil Chai 3D
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Re: Photoelectric Effect

Postby Wil Chai 3D » Mon Oct 12, 2020 3:31 pm

Regarding the equation, you could also think about it in terms of the Law of Conservation of Energy, ie, the total energy of the incoming photon (E photon) = the energy that is used to remove the electron from the metal (E energy to remove e-) plus the energy that is left over (E excess), which is transferred to the electron (Ek e-). You can represent this by rewriting the equation:

E(photon) = E(energy to remove e-) + E(excess). Keep in mind E(excess) is the same as Ek (e-) since the excess energy is transferred to the electron.

Hope this helps.


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