Sapling Week 2,3,4 HW #4

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905596651
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Joined: Wed Sep 30, 2020 10:01 pm

Sapling Week 2,3,4 HW #4

Postby 905596651 » Tue Oct 27, 2020 9:50 am

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

Postby Jessica Katz 1G » Tue Oct 27, 2020 11:10 am

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

Postby Josh Chou 1I » Tue Oct 27, 2020 8:36 pm

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
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Joined: Wed Sep 30, 2020 9:37 pm

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

Postby Aydin Karatas 1A » Tue Oct 27, 2020 8:44 pm

In the first part of the question, you calculated for , which is measured in . Divide 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
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Joined: Wed Sep 30, 2020 9:46 pm

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

Postby Austin Aldujaili 2E » Tue Oct 27, 2020 8:46 pm

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-.


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