Part 1: "When a metal was exposed to photons at a frequency of 1.19×1015 s−1, electrons were emitted with a maximum kinetic energy of 3.60×10−19 J. Calculate the work function, Φ, of this metal."
Part 2: "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 6.78×10−7 J?"
I was able to solve for the first part, but I'm confused for part 2 as to what's asking. The work function is 4.28494×10−19 J/photon. What equation or process would I need to use to solve for the maximum number of ejected electrons?
Sapling #4 Homework Part 2
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Re: Sapling #4 Homework Part 2
The second part of the question is basically asking how many electrons will be ejected when the energy of each photon is equal to the work function. This occurs when the kinetic energy of the electrons is zero. You would divide the given 7.21 x 10^7 J by the found work function to find the number of photons with energy equal to the threshold energy. I believe the answer is 1.70 x10^12.

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Re: Sapling #4 Homework Part 2
Divide the total energy by the work function to see how many electrons the energy can displace.
6.78×10−7 J/4.28494×10−19 J/photon
Important concepts:
One photon with enough energy to displace an electron will only displace one electron. In other words, there is a onephoton to one electron ratio if the threshold energy is met.
6.78×10−7 J/4.28494×10−19 J/photon
Important concepts:
One photon with enough energy to displace an electron will only displace one electron. In other words, there is a onephoton to one electron ratio if the threshold energy is met.

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Re: Sapling #4 Homework Part 2
Neel Sharma 1G wrote:Divide the total energy by the work function to see how many electrons the energy can displace.
6.78×10−7 J/4.28494×10−19 J/photon
Important concepts:
One photon with enough energy to displace an electron will only displace one electron. In other words, there is a onephoton to one electron ratio if the threshold energy is met.
Thank you, this explanation makes sense. I was just wondering for future problems though, do you know if it would be possible to find the # of electrons ejected if we aren’t given the work function? If there are other concepts or information that could be used?
Re: Sapling #4 Homework Part 2
Neel Sharma 1G wrote:Divide the total energy by the work function to see how many electrons the energy can displace.
6.78×10−7 J/4.28494×10−19 J/photon
Important concepts:
One photon with enough energy to displace an electron will only displace one electron. In other words, there is a onephoton to one electron ratio if the threshold energy is met.
Is it possible to find the electrons that were displaced if you aren't given the work function?

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Re: Sapling #4 Homework Part 2
haley f wrote:Neel Sharma 1G wrote:Divide the total energy by the work function to see how many electrons the energy can displace.
6.78×10−7 J/4.28494×10−19 J/photon
Important concepts:
One photon with enough energy to displace an electron will only displace one electron. In other words, there is a onephoton to one electron ratio if the threshold energy is met.
Is it possible to find the electrons that were displaced if you aren't given the work function?
For this problem you had to find the work function in part one and you used it in part two. I think in general you need the work function to find out how many electrons were emitted and you may just have to solve for it yourself or it will be given.

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Sapling Week 2 Question 3
How do you determine wavelength, frequency, and energy from the amount of photons a wave of light produces?

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Re: Sapling #4 Homework Part 2
I'm still a little confused on why we divide the Energy per photon by the Energy of the work function. How do we know the kinetic energy is zero if all we're given is that we need to find the maximum number of electrons ejected? Thank u!

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Re: Sapling #4 Homework Part 2
You can think of the number of electrons ejected by a burst of photons that have a net energy of 6.78x10^7 j as number of electrons = 6.78x10^7 j / n where n is the energy per photon which must be equal to or greater than the work function. Photons with energy larger than the work function would cause n to be larger thus making 6.78x10^7/n smaller. As such you want the smallest n possible, which is in this case the value of the work function.
In other words, since a photon with an energy greater than or equal to the work function will only at max eject 1 electron, higher energy photons would result in less electrons ejected since the max net energy cannot exceed 6.78x10^7. The kinetic energy being 0 is just a consequence of that since KE = E(photon)  Work Function.
In other words, since a photon with an energy greater than or equal to the work function will only at max eject 1 electron, higher energy photons would result in less electrons ejected since the max net energy cannot exceed 6.78x10^7. The kinetic energy being 0 is just a consequence of that since KE = E(photon)  Work Function.

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Re: Sapling #4 Homework Part 2
Essentially what I've gotten from this post is that after finding the work function, I would need to break up the energy needed to release each electron between photons. So you cannot actually find how many electrons would be ejected with the given information without thw work function as each proton will eject a single electron.
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