Photoelectric effect formula question

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SUMMARY

The discussion centers on the photoelectric effect formula, specifically the relationship between the work function (ϕ) of metals and the threshold wavelength (λ0) for emitting photoelectrons. Participants clarify that the formula ϕ = hf0 = hc/λ0 determines the threshold wavelength, where the energy of the incident photon must equal the work function for electrons to be emitted. If the kinetic energy (Kmax) of the emitted electrons is zero, it indicates that the photon energy is just sufficient to overcome the work function, resulting in the largest wavelength for emission. The conversation emphasizes the importance of understanding energy conservation in this context.

PREREQUISITES
  • Understanding of the photoelectric effect and its significance in quantum physics
  • Familiarity with the concepts of work function (ϕ) and threshold frequency (f0)
  • Knowledge of energy units such as electronvolts (eV) and their conversion to joules
  • Basic grasp of photon energy calculations using the formula E = hf
NEXT STEPS
  • Study the derivation and implications of the photoelectric effect formula ϕ = hf0 = hc/λ0
  • Explore the concept of stopping potential and its relationship to the photoelectric effect
  • Learn about the applications of electronvolts (eV) in particle and nuclear physics
  • Investigate experimental methods for measuring the photoelectric effect and determining threshold wavelengths
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Students of physics, educators teaching quantum mechanics, and researchers interested in the applications of the photoelectric effect in modern technology.

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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