## Photoelectric Effect post module

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Joined: Wed Sep 30, 2020 9:49 pm

### Photoelectric Effect post module

Hi! I'm a bit lost on how to solve problem 31) on the post assessment. The problem is:

A. If 3.607 x 10-19 J is required to remove an electron with zero kinetic energy from a metal surface, what would be the longest wavelength light that could do this?

Darlene Lien 1B
Posts: 54
Joined: Wed Sep 30, 2020 9:37 pm

### Re: Photoelectric Effect post module

Hi, for this question the work function is given and since kinetic energy equals to 0, the energy per photon is equal to the work function. Therefore, you could solve for the wavelength by using the formula lambda=hc/E, where E=3.607E-19 J. Hope this helps!

Bronson Mathos 3J
Posts: 40
Joined: Wed Sep 30, 2020 9:36 pm

### Re: Photoelectric Effect post module

For this problem, since the question says the kinetic energy is zero, we can reason that the energy of a photon hitting the metal surface is equal to the threshold energy. And with this idea, you simply use the given energy 3.607x10^-19J and the equation E=hc/wavelength and solve for wavelength. I hope this helps!

Brianna Martilla 2C
Posts: 45
Joined: Wed Sep 30, 2020 9:58 pm

### Re: Photoelectric Effect post module

This looks like the same example Professor Lavelle did in his lecture on October 14. You rearrange E=hv and c=lambda*v to get lambda=(hc)/E. E is given and h and c are constants.

BrittneyMyint2C
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Joined: Wed Sep 30, 2020 9:59 pm
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### Re: Photoelectric Effect post module

Hello!

3.607 x 10-19 J will be the threshold energy, since it is this energy is required to remove an electron. Since there's 0 kinetic energy, this means that E(photon) = threshold energy, based on the E(photon) - threshold energy = kinetic energy equation.
So if we want to see the longest wavelength that can eject an electron from this metal, we need to use the E = (hc)/λ equation. If we isolate λ, the equation will be λ = (hc)/E.
*Note that the E that we're using in the λ = (hc)/E equation is the energy of the photon because we want the wavelength of light that ejects an electron; since the threshold and E(photon) is the same we substitute in 3.607 x 10-19 J.

If you substitute in the values, you get λ = (6.626 x 10^-34 Js)(2.99792 x 10^8 m/s)/(3.607 x 10^-19 J) = 551 nm.

Extra concept stuff: this is the longest wavelength that can eject an electron from this metal because a wavelength longer than this would have insufficient energy, so no electron would be ejected if the wavelength was longer.

I also think the professor solved this question in his video module, but I hope this helps!