34B Photoelectric effect module

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melissa_dis4K
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34B Photoelectric effect module

Postby melissa_dis4K » Tue Oct 23, 2018 1:32 pm

34. B. If molybdenum is irradiated with 194 nm light, what is the maximum possible kinetic energy of the emitted electrons?
For the previous question part a I got E=7.22 x 10^-19. However, I am not sure how to solve this problem. I think I should use Kinetic Energy=1/2Me*Ve^2. But, I don't know how to get the velocity from the given 194 nm of light.

Chem_Mod
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Re: 34B Photoelectric effect module

Postby Chem_Mod » Tue Oct 23, 2018 2:16 pm

Energy of photon - threshold energy = Kinetic energy. From wavelength, you can find frequency and thus energy of the photon using Einstein's equation E=hv where v is frequency.

janeane Kim4G
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Re: 34B Photoelectric effect module

Postby janeane Kim4G » Tue Oct 23, 2018 2:44 pm

Hi,
From the given wavelength, you can calculate the energy of the light with the equation E= hc/lambda
(6.626*10^-34 Js)(3*10^8ms)/(194*10^-9m) = 1.023*10-18 J

Recall that you found the work function ( E needed to eject an electron) in part A and that
E(kinetic)= E(photon)-work function. Thus subtracting the A value of 7.22*10^-19 from 1.023*10^-18 to get 3.03*10^-19 J.


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