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Photoelectric effect, calculating work function

  • #1
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Homework Statement


(Given a voltage against frequency graph)
Calculate the work function of Sodium and state any assumptions you have made.
My question is, what assumptions have I made?

Homework Equations


hf = work function
V = IR

The Attempt at a Solution


I know on a Voltage against Frequency graph, the x intercept is the threshold frequency as it's the point where there is no kinetic energy given to the electrons to produce a current as V = IR, and so I can use
hf = work function, to calculate work function.

However I'm not sure what I'm assuming in this process, and I'm sure it's not something like 'Assuming the data is correct'
 

Answers and Replies

  • #2
kuruman
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To understand what you are assuming, you have to understand the equations that you are using, how you are using them and why they are relevant. For example, you quote hf = work function. What physical process does that describe? You mention kinetic energy, yet there is no equation in the list that contains KE or ½mv2. What happened to it? You mention V = IR. What exactly is R? Think of what might be going physically in the photoelectric effect, think of your analysis if this is an experiment you did, and then think of how you would describe that mathematically and what simplifications you might have to make in order to bring about the mathematical description.
 
  • #3
52
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To understand what you are assuming, you have to understand the equations that you are using, how you are using them and why they are relevant. For example, you quote hf = work function. What physical process does that describe? You mention kinetic energy, yet there is no equation in the list that contains KE or ½mv2. What happened to it? You mention V = IR. What exactly is R? Think of what might be going physically in the photoelectric effect, think of your analysis if this is an experiment you did, and then think of how you would describe that mathematically and what simplifications you might have to make in order to bring about the mathematical description.
Ok, so I know that R is the resistance in the circuit and when there is no current as a result of the opposing voltage that is provided to negate the energy of the electrons. Which means at this point there is no Ek and therefor using Ek = hf - work function I know that if Ek is 0, the hf = work function

The energy provided by the incident light is transferred to the electrons on the surface of the metal and loses energy equal to the work function which is the minimum amount needed to release an electron from the metal. and Ek = hf - workfunction represent the maximum Ek an electron will have as some lose a bit more.

So I'm assuming that the energy lost is exactly the work function? Have I missed anything else?
 
  • #4
kuruman
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Ok, so I know that R is the resistance in the circuit
What part of the circuit? What is the material that has resistance R?
The energy provided by the incident light is transferred to the electrons on the surface of the metal and loses energy equal to the work function which is the minimum amount needed to release an electron from the metal.
What assumptions are hiding in this statement? Take it apart and justify every assertion you have made. For example, how is the energy transferred to the electrons? Can you get two or three electrons from one photon?
You also need to consider if processes other than the photoelectric effect might kick electrons from the surface. How might these affect your results? Do some research.
 
  • #5
ZapperZ
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There's a bunch of confusing information here. The problem is that the description is rather vague.

Assuming that you performed the standard photoelectric effect experiment that is done in general physics labs, then the "voltage" that you have is the stopping voltage that corresponds to the KE of the most energetic electrons.

ALWAYS start with the general equation first. In this case, the relevant photoelectric effect equation is

eVs = hf - φ

where Vs is the stopping potential, f is the frequency of the incoming light, and φ is the work function.

Now, you said that you have a graph of Vs versus f (V plotted on the vertical axis, while f plotted on the horizontal axis). Then, the first thing you have to do is rearrange the equation so that it matches a straight-line equation (we know a priori that this is a straight line function). In this case, the equation above becomes:

Vs = (h/e)f - φ/e

This is an identical form to y = mx + b.

So if you want to find the work function, you find the y-intercept, and then divide by e. As a bonus, the graph also gives you the ability to find "h", Planck constant, by finding the slope and dividing that by e as well.

I know that this was not what you were asking, but upon reading your description, I had to make sure that we are all on the same page here.

As far as the "assumptions" are concerned, this can be as complicated as you want to make it. I'll list a few based on what I think might be relevant, but without seeing your setup, there could easily be more:

1. If your task is to compare with the book-value of the Na work function, then you're making an explicit assumption that you are doing an experiment on a pure, pristine Na surface, which is often not true since Na reacts and forms an oxide very easily, even in partial vacuum. So you might be finding the work function of a Na oxide surface.

2. You are assuming that your light source is truly monocromatic. Depending on what you used, your "f" may have a spread in frequency, and the higher edge of that spread might affect the value of Vs.

3. Now, since this is an experiment where you vary the frequency ONLY, and then recording the stopping potential, you are assuming that the Einstein model of the photoelectric effect experiment is valid, i.e. you are assuming that changing the INTENSITY of the light at a particular frequency does not affect Vs. The experiment to verify this is not easy if you do not have the necessary setup. If your light source is a discharge lamp, and you select the frequency using narrow-band filters or a monochrometer, then your ability to change intensity of that light is limited to using a set of neutral density filters. I assume that this was not available to you, or that this study was not done. So if that's the case, you have to accept and assume that changing the intensity does not change the energy of the most energetic KE of the photoelectrons.

Zz.
 

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