Kinetic Energy of Electrons  [ENDORSED]

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Kinetic Energy of Electrons

Postby Brandon_Wu_2L » Thu Sep 29, 2016 8:58 pm

When we are using the equation:
E(photon)(hv) - threshold energy = excess kinetic energy of electron (1/2mv2)
We know increasing E(photon) will result in more excess electron kinetic energy.
But how do we know whether that excess KE is either more electrons (more mass) being ejected from a surface or the same number of electrons just ejected at a higher velocity? Or is it a combination of both increasing electron mass and average ejected electron velocity?

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Re: Kinetic Energy of Electrons

Postby Armo_Derbarsegian_3K » Thu Sep 29, 2016 9:40 pm

I believe the formula finds the (minimum) energy needed to remove a single electron. Therefore, I think it is an increase in kinetic energy of a single electron.

Hyun Young Lee 2H
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Re: Kinetic Energy of Electrons

Postby Hyun Young Lee 2H » Fri Sep 30, 2016 8:19 am

when doing photoelectric experiments, it is important to keep in mind the kinetic E- formula.

Photon- Energy moved = E(k)
so whatever the excess energy is, will be kinetic energy ( 1/2 mv2 ) of a single atom
i hope this helps!

Jonathan Sarquiz 3F
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Re: Kinetic Energy of Electrons  [ENDORSED]

Postby Jonathan Sarquiz 3F » Fri Sep 30, 2016 9:47 am

When dealing with the photoelectric effect, it is a 1:1 ratio of photon:electron. For each photon of light that comes into contact with the metal, one electron gets removed from the metal surface. Therefore, we calculate the kinetic energy using the equation E(photon) = E(threshold) + KE. Then, we can use the equation for kinetic energy (KE = 1/2 mv2) and substituting in the mass of an electron (9.11 x 10-31 kg), we can calculate the velocity for the single electron.

Jinghui Song 2J
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Re: Kinetic Energy of Electrons

Postby Jinghui Song 2J » Fri Sep 30, 2016 10:19 am

First of all, I think we have to understand that, in photoelectric effect, the light shows its particle property, which means light is composed of photons. When the light strikes the metal surface, each electron on the metal surface can only absorb one single photon. The frequency of photon can decide the energy of that photon. As long as the frequency of that photon is higher than threshold frequency, one electron that absorbs that photon will have enough energy to overcome the attraction by other electrons, and then that electron can be emitted. The excess energy after overcoming the attraction force becomes kinetic energy of that electron. Thus, excess energy can only be applied to that particular electron, which absorbed the photon. As a result, if the E(photon) increases, the number of electrons being emitted will not increase, but the kinetic energy of those electrons will increase.

Hope this will help!

Ah, i got something else to introduce. In the sense of particle property of light, the intensity of light refers to the number of photons. The larger the intensity, the more photons will be. Don't confuse the number of photons with the energy of photons. In other words, don't confuse intensity with frequency of light in the sense of particle property.

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