# The photoelectric effect: What frequency does υₛ refer to in the work function?

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JACKR
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What frequency does υₛ refer to in the work function?
The equation KE=hυ- hυₛ gives the maximum energy an electron can acquire when a photon with a frequency υ is incident on the cathode. I am wondering what does υₛ refer to? Does it mean the frequency of the atom to which the electron is attached? According to Planck, the smallest energy an atom can have is hυ; υ is the frequency in which the atoms oscillate. Is my reasoning right?

I think ##hv_s## is the amount of energy required for an electron to be emitted from its crystal lattice. ##hv## is the energy of the photon, which eventually transfers to the electron; if this energy is greater than ##hv_s## and there is no conversion into thermal energy (or other types), the maximum kinetic energy produced after the electron is emitted is equal to the excess amount of energy: ##K=hv-hv_s##.

Well, here is an explanation from wikipedia

AmanWithoutAscarf said:
I think ##hv_s## is the amount of energy required for an electron to be emitted from its crystal lattice. ##hv## is the energy of the photon, which eventually transfers to the electron; if this energy is greater than ##hv_s## and there is no conversion into thermal energy (or other types), the maximum kinetic energy produced after the electron is emitted is equal to the excess amount of energy: ##K=hv-hv_s##.

Well, here is an explanation from wikipedia View attachment 346216
Yep I know this. What I am asking about is, does υₛ refer to anything about the atom to which the electron is attached? I mean, removing an electron from an atom result in forming a positive ion with an energy greater than the energy of the original atom by an amount equal to hυₛ. My question is, doesn't that mean that the frequency of the absorbing atom is υₛ? According to planck, the energy of an atom is at least hυ̅ where υ̅ is the frequency of the atomic oscillation, and it only increases in quanta, multiple number of hυ̅. That being said, then I've concluded that since the atom absorbed an energy equal to hυₛ, then its energy before absorption must have been nhυₛ and absorbing that quantum must raise the energy of the atom to (n+1)hυₛ, where υₛ is the frequency of the atom?!

If you're looking for the quantum nature of that energy, I'm afriad that I'm not able to answer yet. However, from the semiclassical perspective, I believe this amount of energy is equal the total sum of all types of potentials exerted on the emitting electron by the crystal lattice, the atomic nucleus, and the electron cloud. So, different electron has the different value of ##hv_s##, and it's hard to determine all properties of the atom since it isn't the only factor contributing to ##hv_s##.

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JACKR said:
The equation KE=hυ- hυₛ gives the maximum energy an electron can acquire when a photon with a frequency υ is incident on the cathode. I am wondering what does υₛ refer to? Does it mean the frequency of the atom to which the electron is attached? According to Planck, the smallest energy an atom can have is hυ; υ is the frequency in which the atoms oscillate. Is my reasoning right?
What do you mean by "the frequency in which the atoms oscillate"? Can you cite a textbook or other reference that defines and discusses "the frequency of the atom to which the electron is attached"?

Planck, M. On the Law of Distribution of Energy in the Normal Spectrum.

renormalize said:
Are you sure? Did you even read Planck's paper? According to the version here: https://upload.wikimedia.org/wikipe...nergy_Distribution_in_the_Normal_Spectrum.pdf
the words "atom", "frequency" and "electron" appear nowhere!
I did study it yes. The word atoms isn't mentioned anywhere there because he called them 'resonators' instead. You know, the idea that matter is composed of atoms wasn't something popular back then. However, these resonators turned out to be what we call today atoms! So, as you can see from the paper, the solution to the blackbody problem was achieved when Planck proposed that the blackbody resonators (the blackbody atoms) each has an energy proportional to its frequency of oscillation, i.e., E=nhυ; where n is a positive integer.

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Ok, but we have 2024 now, and we don't use this language for quite a long time now

JACKR said:
What I am asking about is, does υₛ refer to anything about the atom to which the electron is attached?

No. It's just the smallest frequency for which photoelectric effect occurs.

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JACKR said:
these resonators turned out to be what we call today atoms!
No, they're not. The resonators Planck is talking about are part of the radiation field, not matter that emits radiation. What Planck was modeling was a radiation field confined in a cavity with perfectly reflecting walls, except for a small hole through which the cavity could emit radiation to the outside. The "resonators" were what the radiation field inside the cavity was made of: individual wave modes with a specific frequency and wave number.

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