3
$\begingroup$

When hf ≥ work function, Then the electron still comes out. So, if I say kinetic energy of ejected electron = 0, it should still come out. Right ?

Then, how does the electron even move out or gets ejected if its $v=0$?

$\endgroup$
1
  • 1
    $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – andselisk
    Mar 6, 2021 at 17:11

2 Answers 2

2
$\begingroup$

Rather than debating whether an electron with no kinetic energy is ejected consider how an experiment to measure the photoelectron effect might be conducted. A photosensitive electron emitter is irradiated and the kinetic energy of ejected electrons is measured. One way to do this experiment is described in MIT Physics Department lab course notes. The setup is described in the following schematic:

enter image description here

A photodiode is irradiated with the filtered output of a mercury lamp. Only one of the available narrow bands of the lamp's spectrum is selected by the filter. The light strikes the K coated surface of the cathode of the photodiode. Emitted electrons are collected at a ring-shaped anode separated from the cathode by a gap, setting up a small current. An adjustable retarding voltage can be applied between the cathode and the anode to suppress the current, and the minimal cutoff voltage V which completely suppresses the current is measured. This voltage is theoretically related to the work function $\phi$ of the material and to the energy $h\nu$ of the incident light as $$V = \frac{h}{e}\nu - \frac{1}{e} \phi$$

Now note perhaps the most important point with regard to your question: only a limited number of light frequencies are available in the experiment, corresponding to the bands generated by the mercury lamp. In addition the motion of a collection of electrons (a current) is measured, not the velocity of a single emitted electron. While it is possible to perform single particle experiments, at no point in the MIT experiment is the kinetic energy of individual electrons directly measured.

The lab notes also provide a short history of the discovery of the photoelectric effect:

Crude though the early data were, the qualitative fact of the dependence of the critical cutoff voltage on the wavelength of light emerged with sufficient clarity to induce the young Albert Einstein, working as a patent examiner in the Swiss Patent Office in 1905, to link the effect with the recent idea, introduced by Planck in 1900,that matter radiates its energy in quanta of energy hν.

and

It was not until 1912 that the technical problems of making precision measurements of the photoelectric effect were overcome [...]

In other words, 7 years passed between the time that Einstein postulated his theoretical description and a confirmation based on "clean" data.

Note also from the lab notes how inherently "messy" such an experiment can be:

It is not possible to precisely determine the work function for removing an electron, because the cathode surface interacts with the remaining gases in the photocell as a getter, so that the surface characteristics change a little from the ideal case. It is important to note that the electronic work function φ is a material constant which incorporates the different emission potentials of the cathode and anode. The former is a difficult quantity to estimate due to the manufacturing process which makes the cathode surface inhomogeneous. It is composed of a mixture of potassium, potassium oxide and oxidized silver. For this reason, you need to take care that the same area is always illuminated. The emission work function of the photoelectrons can vary locally!

$\endgroup$
1
$\begingroup$

There is the direct analogy to the Solar system.

Imagine an asteroid ejected from the Solar system with zero kinetic energy. It makes sense only if the asteroid is at infinite distance from Sun, otherwise it is not ejected yet. It would start long fall on the Sun.

Or, if other stellar systems are considered, than gravity of the Sun must be counteracted by gravity of the other stars, so there is zero potential gradient.

Now, replace stars by atoms and the asteroid by an electron. An electron with zero kinetic energy would be truly ejected only if there is mutually cancelled attraction toward neighbour atoms and ions, or if it is at infinite distance to them.

So the zero kinetic energy of ejected electron has sense and meaning rather merely mathematical as the limit point on the chart where electron energy as function of wavelength crosses the zero point.

Zero kinetic energy of electron has also quantum obstacles. Zero kinetic energy means zero speed and that would mean certainty of the electron position. Following Heisenberg uncertainty principle , momentum of electron could have any value.

Electron wave function, determining electron energy in an atom, is not function of time, so precision of predicted energy is not limited.

$\endgroup$
5
  • $\begingroup$ You might have a location between various star systems where the graviational attraction from the systems cancels. That could be used as a reference. I think the photoelectric effect experiment is useful as an introduction to quantum phenomena and it might be good to avoid invoking classical analogies. $\endgroup$
    – Buck Thorn
    Mar 8, 2021 at 11:00
  • 1
    $\begingroup$ @buckthorn You might have a location between various star systems where the graviational attraction from the systems cancels. You may have missed I have mentioned that. About avoiding classical analogies - it is case dependent. It applies here very well, while elsewhere is better to avoid it, if the essential is a quantum effect where there is no counterpart. But the limit case of zero kinetic energy at zero potential energy of the central force is where both worlds meet quite well. $\endgroup$
    – Poutnik
    Mar 8, 2021 at 11:03
  • $\begingroup$ That's true. I guess what the QM explanation does is reveal why the photon frequency has to be shorter than a given threshold for a current to be measured. The classical analogy would be that a planet can be freed from a given orbit by only one collision with an extrasolar obhect of sufficiently high energy (which is not entirely true but is the key takeaway for an introduction: light interacts as discrete wavepackets we call photons). $\endgroup$
    – Buck Thorn
    Mar 8, 2021 at 12:04
  • 1
    $\begingroup$ @BuckThorn Sure, but I have not been addressing the nature of the photoelectric effect, that has no classical solution. I was addressing ejected electron state with zero kinetic energy, from the electrostatic point of view, related to gravitational analogy with quadratic central force in both cases. $\endgroup$
    – Poutnik
    Mar 8, 2021 at 12:11
  • $\begingroup$ Yes, the field is treated classically. $\endgroup$
    – Buck Thorn
    Mar 8, 2021 at 14:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.