Sapling HW #4 Part 2 Photoelectric

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Inderpal Singh 2L
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Sapling HW #4 Part 2 Photoelectric

Postby Inderpal Singh 2L » Fri Oct 16, 2020 1:43 pm

When a metal was exposed to photons at a frequency of 1.40×1015 s−1, electrons were emitted with a maximum kinetic energy of 3.30×10−19 J.
Calculate the work function, Φ, of this metal.
Φ =5.97×10^−19 J/photon
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 7.47×10−7 J?

Can someone help me with the second part of this problem? I got the first part but I have no idea how to even start finding the maximum number of electrons.

Sana Nagori 2H
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Re: Sapling HW #4 Part 2 Photoelectric

Postby Sana Nagori 2H » Fri Oct 16, 2020 1:50 pm

I haven't done it yet but my guess is that first, you would calculate the E(photon) which is the energy for one photon. Then you would divide the total energy given in the question to find out how many photons there are. Since we know one photon interacts with one electron the value you get for photons will be the # of electrons. Someone else who has done it can correct me if I'm wrong though.

Mikayla James 2A
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Re: Sapling HW #4 Part 2 Photoelectric

Postby Mikayla James 2A » Fri Oct 16, 2020 1:50 pm

Hi! In the lecture on the photoelectric effect, it is mentioned that one photon interacts with one electron. Therefore, if each proton has sufficient energy to remove each electron, the maximum number of electrons ejected would be equal to the number of photons. So, you would divide the total energy mentioned in part two by the work function to find the maximum number of electrons ejected. Hope this helps!

Javier Perez M 1H
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Re: Sapling HW #4 Part 2 Photoelectric

Postby Javier Perez M 1H » Fri Oct 16, 2020 5:10 pm

Well for this question you will have to keep in mind the following two equations E(photon energy)=hv and c=wavelength*frequency. You are only given K.E and Frequency. You basically plug in frequency into your "C" equation to get the wavelength then plug in your wavelength into your E(photon) equation to get energy of photon. Once you have energy of photon you subtract the K.E (given) to arrive to your value of energy necessary to remove the electron or should I say work function. = 6.60x10^-19
Work function means energy needed to remove just one electron. However, it asks that how many can be removed with that given beam hitting the solid from which you are removing electrons.
so....
you divide the beam by your work function to get the literal number of electrons you can possibly remove with that amount of energy:
1.21x10^12

hope this helps.


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