Photoelectric effect formula question

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Homework Help Overview

The discussion revolves around the photoelectric effect, specifically focusing on the relationship between the work function of metals and the wavelengths required to emit photoelectrons. Participants are examining the implications of the formula ɸ=hf0=hc/λ0 in determining threshold wavelengths and exploring concepts related to energy and kinetic energy of emitted electrons.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants are questioning whether the formula provides the largest wavelength for photoelectron emission or just the threshold wavelength. There are discussions about the relationship between photon energy, work function, and the kinetic energy of emitted electrons.

Discussion Status

The conversation includes various interpretations of the photoelectric effect and the implications of energy thresholds. Some participants have offered insights into the nature of kinetic energy at the threshold and the utility of eV in calculations, while others are exploring the nuances of the formula and its application.

Contextual Notes

There is an ongoing examination of the definitions and implications of kinetic energy in relation to the work function, as well as the relevance of energy units (eV vs. joules) in calculations. Participants are also considering the experimental context of measuring threshold wavelengths and stopping potentials.

mss90
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Homework Statement


If I know the work function of various metals in eV and need to find the largest wavelenghts required to emit photoelectrons from the metals would i just use ɸ=hf0=hc/λ0?

Homework Equations


ɸ=hf0=hc/λ0

The Attempt at a Solution


The formula only gives the threshold wavelength and not the largest right?

Thanks
 
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Hello M and welcome to PF :)

Bit of a problem to help without answering (as we aren't allowed to do). Look at it this way: if you find the largest wavelength, the energy with which you kick the electron is just enough to overcome the work function. Work function is a kind of binding energy. Shorter wavelengths mean higher energy. What do you think happens to the excess energy ? Where could it go ? Check it out

3. No !
 
BvU said:
Hello M and welcome to PF :)

Bit of a problem to help without answering (as we aren't allowed to do). Look at it this way: if you find the largest wavelength, the energy with which you kick the electron is just enough to overcome the work function. Work function is a kind of binding energy. Shorter wavelengths mean higher energy. What do you think happens to the excess energy ? Where could it go ? Check it out

3. No !
.."the energy of the photoelectrons emitted is exactly the energy of the incident photon minus the material's work function or binding energy.."
The only thing i can think of is to calculate the initial energy of the photon (or the Ek of the electron + work func) and run it through the formula again with known E.

You said "no" to "3", however it states; "and the second term is the work function (ϕ) of the surface with threshold frequency (ƒ0) or threshold wavelength (λ0)" here; http://physics.info/photoelectric/
 
.."the energy of the photoelectrons emitted is exactly the energy of the incident photon minus the material's work function or binding energy.."
And that energy of the photoelectrons emitted is kinetic energy. The exercise wording points at zero kinetic energy.

3. No ! meaning "... threshold wavelength and not which is the largest.. "
 
If Ek=0 doesn't that mean the electron in the metal is only exited not actually emitted?
 
If it did point to that doesn't that mean the following: Kmax = E − ϕ where Kmax=0 E=ϕ hence λmax= hc/ ɸ?
another question, are eV ever used in calculations or do you always convert to joules?
 
It means it comes out of the metal with zero speed. But it does "come out".
The experiment described in the link (under Stopping potential) shows that the maximum kinetic energy at the threshold frequency is zero.

550px-Photoelectric_effect_diagram.svg.png
So finding the threshold wavelength is a matter of plotting the stopping potential versus frequency (a bit more linear than versus wavelength). Where the line crosses the V axis is the threshold.
Usually it's not as straight as the picture suggests and you have to draw a straight line through some points with higher frequencies to find the intercept on the frequency axis.eV are very useful in particle physics, nuclear physics, and so on. Those folks express (almost) everything in eV (or multiples like keV, MeV, GeV,TeV).

(almost) meaning mostly mass and energy, but also momentum. You never put a proton on a balance to weigh it anyway :)
 
Last edited:
mss90 said:
If it did point to that doesn't that mean the following: Kmax = E − ϕ where Kmax=0 E=ϕ hence λmax= hc/ ɸ?
Correct. I think you understand it.
 

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