Photoelectric Effect and energy

[MHD93]

Hello

Before a photon hits an electron, the electron originally had a kinetic energy, if f = f0 we know that
Kmax = hf -hf0 = zero

shouldn't the electron's kinetic energy increase the photon hits it? not the converse?

 

[Bill_K]

Kmax = hf -hf0 = zero

Sorry Mohammad, have you misquoted this? Photoelectrons have a range of energies from zero up to E_max = hν - φ, where φ is the work function of the metal. In other words it's just the opposite of what has been said - the maximum energy is determined by the energy of the photon, and not by the original energy of the electron.

 

[MHD93]

Sorry Mohammad, have you misquoted this? Photoelectrons have a range of energies from zero up to E_max = hν - φ, where φ is the work function of the metal. In other words it's just the opposite of what has been said - the maximum energy is determined by the energy of the photon, and not by the original energy of the electron.

It's what I ask..
Why isn't the maximum energy is the energy of the photon added to the original electron's energy

 

[Bill_K]

Oh OK, that's because this is the maximum energy. Electrons in the metal have a maximum energy, E_F the Fermi energy, which is an amount φ below the vacuum outside. So hν - φ is the maximum energy the emitted photoelectron can have. If the photoelectron came from below the Fermi level, then of course its energy will be less.

 

[Gokul43201]

It's what I ask..
Why isn't the maximum energy is the energy of the photon added to the original electron's energy

It is, but that's not the whole picture. As explained by Bill, there's an additional potential energy barrier that needs to be overcome. The maximum KE of the Photoelectron is therefore equal to its original energy (E_F) + the photon's energy (hv) - the final potential energy (E_vac).

KE(max) = hv - (E_vac - E_F) = hv - φ

φ is called the Work Function, and if you define, φ = hv₀, that gives you: KE(max) = hv - hv₀

 

[MHD93]

THanks.. that's given me the whole picture, helpful.

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[MHD93]

THanks.. that's given me the whole picture, helpful.

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