2
$\begingroup$

I had this statement in my physics textbook

Photoelectric effect is seen only when electrons are bound, because free electrons cannot absorb the whole energy of the photon while conserving momentum and energy.

I found a simple proof for this statement here.

However, the proof states in the beginning itself, that

If the electron is bound to an atom then the atom itself is able to act as a third body repository of energy and momentum

Here's my issue: I've also read that "free electrons" are the loosely bound electrons of an atom, also called the conduction electrons.

If that is the case, then the statement in by book should be wrong, because if the free electrons were to be defined as loosely bound electrons, then it would be possible for them to absorb the whole energy of the photons. In fact, photoelectric effect is generally defined as the emission of the loosely bound electrons from a metal surface when light is incident on it.

On the other hand, though, if we consider that "free electrons" refers to isolated electrons, i.e. when there is no repository in the form of an atom, then the statement seems perfectly fine.

So my question is:

  1. Does the definition of "free electron" depend on the context?
  2. If yes, then how do we determine which of the definitions is being referred to?

P.S. Wikipedia does say that the definition depends on the context. So the question now boils down to (2) only.

$\endgroup$
  • $\begingroup$ A conduction electron is not a free electron. $\endgroup$ – my2cts May 10 at 23:10
  • $\begingroup$ @my2cts Well, wiki says otherwise . Also, Compton scattering is said to occur in free electrons, but actually the electrons are the loosely bound ones of graphite crystal. So that further lends weight to the fact that conduction electrons can be defined as free electrons. $\endgroup$ – SDFG May 11 at 9:32
  • $\begingroup$ Conduction electrons in many aspects behave like free electrons. However in the context of the photo electric effect, they don't. $\endgroup$ – my2cts May 11 at 9:50
1
$\begingroup$

You are asking about free electrons.

Now there are two main cases here:

  1. a truly free electron in vacuum

  2. an electron in a conductor, that is able to move and create electricity

Now in 1., this truly free electron in a vacuum cannot create the photoelectric effect for you, this electron is only able to emit photons in case it is accelerated by a external magnetic (EM) field.

Now in the case of 2., an electron in a conductor, there are two main opinions on this site:

  1. a free electron in a conductor is loosely bound to the valence shells of the atoms

  2. a free electron in a conductor is delocalized

This is from wiki:

The photoelectric effect is the emission of electrons or other free carriers when light hits a material. Electrons emitted in this manner can be called photoelectrons.

Now you are asking about the photoelectric effect. In this case, the photons that are interacting with the conductor's surface atoms, need to be of the energy level around the work function of these loosely bound or delocalized electrons. When the photons' energies are about (they need to be close to the work function, it is all about probabilities) the work function of the electrons, this means that they reach the level of energy needed to knock the electron off the atoms' valence shells that they are loosely bound to (we can call them delocalized too).

$\endgroup$
  • $\begingroup$ Is there any difference between "a loosely bound electron" and a "delocalized" electron? I thought that they were pretty much the same. Also, please help me how to determine which definition works in which context. For eg. if I were to see this statement : "Free electrons do not show photoelectric effect" I would immediately say it was false. But it isn't, if we were to define free electrons as isolated electrons. Please sort out this confusion. Thanks! $\endgroup$ – SDFG May 11 at 9:28
  • 1
    $\begingroup$ @SDFG In reality, there is no difference between a loosely bound and a delocalized electron. Loosely bound is QM, and delocalized is more classical. So it is more of a theoretical difference between describing the same phenomena. Your statement "Free electrons do not show photoelectric effect" depends on how we define the electron. If it is in vacuum, truly free, then the statement is true. But if it is in a conductor, these electrons are not free, we just call them that. In reality, in a conductor these electrons are loosely bound and do show photoelectric effect. $\endgroup$ – Árpád Szendrei May 11 at 16:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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