## Sapling Week 2,3,4 HW #4

905596651
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### Sapling Week 2,3,4 HW #4

What is the maximum number of electrons that could be ejected from this metal by a burst of photons (at some other frequency) with a total energy of 1.11×10−7 J?

Jessica Katz 1G
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Joined: Wed Sep 30, 2020 10:09 pm
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### Re: Sapling Week 2,3,4 HW #4

since your solution for problem above, when it was asking about calculating the work function, is in the units joules per photon, divide the total energy (just joules) by that value. This will allow you to cancel out joules and be left with the number of photons. The number of photons is then equal to the number of electrons.

Josh Chou 1I
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### Re: Sapling Week 2,3,4 HW #4

Hi! So you would divide (1.11 x 10-7) by the work function you found in the first portion of the problem to get the number of electrons that can be ejected at this new frequency. Hope this helps

Aydin Karatas 1A
Posts: 55
Joined: Wed Sep 30, 2020 9:37 pm

### Re: Sapling Week 2,3,4 HW #4

In the first part of the question, you calculated for $\phi$, which is measured in $J/photon$. Divide $\phi$ by the total energy of the burst of photons to get the number of photons. Since this question is dealing with the photoelectric effect, 1 photon energizes 1 electron. Therefore, the number of photons in the burst = the number of electrons ejected.

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Austin Aldujaili 2E
Posts: 42
Joined: Wed Sep 30, 2020 9:46 pm

### Re: Sapling Week 2,3,4 HW #4

Just as they described above, you want to solve for the number of photons that the metal absorbed because, given the photoelectric effect and the particle model of light, one photon will only be able to eject one electron. Therefore, solving for the number of photons will result in the same value as solving for the number of e-.