Photo electric effect

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Anaranjo
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Photo electric effect

Postby Anaranjo » Wed Jun 29, 2016 8:49 pm

How does assuming that light energy consists of quanta help to explain the plot of kinetic energy against frequency in the study of the photoelectric effect?

Chem_Mod
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Re: Photo electric effect

Postby Chem_Mod » Wed Jun 29, 2016 10:36 pm

We can think of a case where one photoejects only a single electron from the metal surface. Since the photoejection of an electron depends on the frequency, rather than the intensity, it implies that an electromagnetic radiation of a particular frequency must have provided sufficient energy to knock out 1 electron. This in fact is the smallest energy a light source can provide to knock out a single electron, so we can assume light has "quanta" of energy which depends on the frequency or E=hv.

Manpreet Singh 1N
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Photo electric effect

Postby Manpreet Singh 1N » Mon Sep 26, 2016 2:51 pm

I am stuck on this question from the post assessment.
Light hits a sodium metal surface and the velocity of the ejected electron is 6.61 x 105 m.s-1. The work function for sodium is 150.6 kJ.mol-1.
Answer the following three questions.
A. What is the kinetic energy of the ejected electron?

I think I would use the Ek = 1/2 mv^2 equation, but what would I use at the mass of the electron? Is that information provided?

Julia Nakamura 2D
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Joined: Wed Sep 21, 2016 2:57 pm

Re: Photo electric effect

Postby Julia Nakamura 2D » Mon Sep 26, 2016 9:37 pm

Yes, the mass of the electron will be given to you (with the correct number of sig figs that you should use). The mass that I used for that problem was 9.11 x 10^-31. So, multiplying the mass of the electron times the velocity (6.61 x 10^5)^2 and dividing by two should give you 1.99 x 10^-19 J as the kinetic energy of the ejected electron.


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