# Radiation in photoelectric effect.

1. Jul 12, 2013

### khanhhung2512

Textbooks say that by measuring the stopping potential V0, we can determine the maximum kinetic energy with which electrons leave the cathode: eV0 = Kmax
However, as I know, when we apply the stopping potential, the electrons will decelerate and radiate parts of their energies. Thus, the stopping potential measured V will be less than V0 in the equation above.
Could anyone please explain this problem?
Thank you very much.

2. Jul 13, 2013

### Simon Bridge

Welcome to PF;
What makes you think there should be an energy deficit over the stopping potential value?
Are you thinking that accelerated charges must radiate, therefore, electrons must radiate whenever they change speed, for instance, when they are decelerated over an electric potential difference?

3. Jul 13, 2013

### khanhhung2512

Yes, that's what I think. However, textbooks don't seem to incorporate that radiation in photoelectric effect. Is it because that amount of energy (according to Larmor formula) is too small and negligible?

4. Jul 13, 2013

### Simon Bridge

Hmmm ... it is a matter of routine to consider the electric potential energy lost as an electron falls through a potential difference to be entirely exchanged for kinetic energy. This seems to work to a very high precision ... but we tend to treat electrons as quantum particles rather than in terms of classical electrodynamics.

It's generally no taught as part of the photoelectric effect because it is not important to the lesson - which is to gently introduce the student to the concept of quantization.

Certainly nobody seems to think Larmor radiation is important to the physics.

I can see papers concerning Larmor radiation in different applied potentials ...
i.e. http://arxiv.org/abs/1211.2478
... the general upshot is that quantum mechanical effects can suppress the Larmor radiation.
I've been unable to tell if anything like that appears in the photoelectric effect ... the fields in the experimental setups appear non-trivial.

Something is nagging the back of my mind though - consider: in Millikan's experiment, the charges are initially falling in gravity, so they are accelerating, so they must be radiating, ... but when they are stopped by the opposing potential, they no longer radiate and the mass-charge ratio is unaffected.

Anyway - the thought is half formed.
I'll have to sleep on it.

The losses, if any, are certainly very small - they don't affect the construction of CRT TV sets for eg.
You could probably do back-of-envelope calculations to work out an order-of-magnitude correction in the classical theory.

5. Jul 13, 2013

### vanhees71

In principle the OP is right, and there is some radiation loss, whenever a charged particle is accelerated, but for the usual considerations on the photo effect this is negligibly small. The radiation loss is relevant for high-energy electrons in synchrotrons (that's why it's sometimes called synchrotron radiation). That's why one likes to use linear accelerators for electrons.

A more serious problem with the treatment of the photo effect is that it is utterly misleading and wrong. It's as bad as using the Bohr-Sommerfeld model of the atom (although it has its 100th birthday this year). Mostly it's claimed that the photo effect is an empirical evidence for the quantization of the electromagnetic field and then a very wrong "particle picture" of photons is hammered into the minds of young students. It's a hard time to forget this folklore in the real quantum-theory course.

The photo effect, at the level of accuracy in the introductory treatments, can be entirely understood by a standard calculation in first-order time-dependent perturbation theory for the transition rate from a bound state (e.g., electron bound in a hydrogen atom) with the classical (!) radiation field as perturbation. Usually it can be simplified to simply a harmonic external electric field. This gives you that radiation energy to this leading order is absorbed in portions of $\hbar \omega$, where $\omega$ is the frequency of the classical radiation field. This is a very nice calculation (see, e.g., Landau/Lifgarbagez vol. 3) and it gives a much better picture about what's going on than this wrong "photon picture", which unfortunately one still finds often in introductory textbooks on quantum theory.

6. Jul 13, 2013

### ZapperZ

Staff Emeritus
IF, and this is a BIG if, there is any significant effects on such a radiation loss, it would be more apparent in photoemission experiment where the electrons are stopped by a detector more abruptly than the simple photoelectric effect experiments. Considering that this has never been an issue, I'd say it is a non-issue here.

Zz.