# What happens when $f = f_0$ or $hf = \phi$ in the photoelectric effect?

In this answer: What happens when work function $\phi = hf$

It says when $$hf = \phi$$, "The electron will just "go up" to the top of potential barrier and then it will "go down" back to the bottom of the potential barrier and it wont be able to escape the nucleus."

However, Young & Freedman says

Stopping potential is zero at threshold frequency (electrons emerge with zero kinetic energy).

Where they have used the phrase 'emerge with zero kinetic energy', which still sounds like electrons can escape/are ejected.

Similarly, Halliday & Resnick state

At the cutoff frequency, the kinetic energy $$K_{max}$$ is zero. Thus, all the energy $$hf$$ that is transferred from a photon to an electron goes into the electron’s escape, which requires an energy of $$\phi$$.

Again, to me that sounds like if $$hf = \phi$$, the electron can escape.

Who is right?

• @josephh Thank you for providing an answer. I get it now - I missed the part where the Halliday book says "If $hf=\phi$, electrons barely escape but have no kinetic energy." Being an introductory physics text, it does not mention the probability that electrons are recaptured after being ejected from the metal surface from a quantum mechanical point of view. By the way, please see my comment to your answer on physics.stackexchange.com/questions/652543/…. But thank you always for helping out ! Really appreciate it. Oct 31 '21 at 7:09
• no worries and good luck with your studies. Oct 31 '21 at 8:20

If $$h\nu=\Phi$$ then the electron has the minimum amount of energy required for it to escape from the atom (the final state of the electron is one with zero kinetic energy). But since the electron does have zero kinetic energy at this point, from a quantum mechanical point of view, there may always be a probability that the electron will be captured again.

The point is that the work function defines the minimum energy transferred to the electron in order to be ejected, and if the incident photon has an energy $$h\nu\gt\Phi$$ then the electron will be ejected from the metal. Perhaps the higher the energy the incident photon has above the work function, the smaller the probability that the electron will be captured.