Sapling HW 2 Question 4

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Marisa Gaitan 2D
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Sapling HW 2 Question 4

Postby Marisa Gaitan 2D » Thu Oct 22, 2020 11:08 am

Hi, I'm having some trouble with the second part of this question. I found the work function, but how would I go about finding the maximum number of electrons?

When a metal was exposed to photons at a frequency of 1.13×1015 s−1, electrons were emitted with a maximum kinetic energy of 3.70×10−19 J.
Calculate the work function, Φ, of this metal. 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 6.84×10−7 J?

Madilyn Schindler 3E
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Re: Sapling HW 2 Question 4

Postby Madilyn Schindler 3E » Thu Oct 22, 2020 11:16 am

The maximum number of electrons would just be the total energy divided by the work function. To explain further, the work function is the amount of energy needed to remove one electron. Therefore, knowing the total energy, you can figure out how many electrons could be ejected. Hope that helps!

Giselle Granda 3F
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Re: Sapling HW 2 Question 4

Postby Giselle Granda 3F » Thu Oct 22, 2020 11:22 am

Hi! To find the number of electrons ejected all you have to do is divide the total energy by the work function (also known as the energy needed to remove and electron). This is because the total energy unit is J, and the work function units are J/electron, so Joules will cancel and you will be left with the number of electrons.

Katie Lam 2J
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Re: Sapling HW 2 Question 4

Postby Katie Lam 2J » Thu Oct 22, 2020 12:32 pm

Hi! The find the maximum number of electrons, you need to divide the total energy (6.84x10^-7 J) by the work function (which is in J/photon). When dividing, Joules should cancel out, so you should be left with photons as your units. Since the ratio of photons to electrons is 1:1, the number of photons you get in your answer is the same number of electrons. Hope this helps!

Sophia Kalanski 1A
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Re: Sapling HW 2 Question 4

Postby Sophia Kalanski 1A » Fri Oct 23, 2020 12:01 am

Hi! So in order to find the amount of electrons, you would need to divide the total energy (using the frequency and plank's constant) and then divide it by the work function because the work function is Jules per electron aka the energy needed to remove one electron. By doing this, you can find exactly how many electrons there are.

RitaThomas_3G
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Re: Sapling HW 2 Question 4

Postby RitaThomas_3G » Sat Oct 24, 2020 8:16 pm

Hey! So for this part of the problem, I like to look at it like dimensional analysis when solving it. This is because you solve it by dividing the energy from part a (units: Joules) by the total energy given in part b (units: Joules/Photon). You can notice that when dividing the two values, the Joules cancel out and you are left with photons, which is the units of the answer!

Renny_kim_2G
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Re: Sapling HW 2 Question 4

Postby Renny_kim_2G » Sun Oct 25, 2020 12:24 pm

It seems like you have the numbers that you can plug in to get the work function. Once you find that, divide the total energy by the work function to get the answer.


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