7
$\begingroup$

In compton scattering, a photon may deliver only some of its energy to an electron.

But when dealing with photon electron interaction in an atom, it's all or nothing.

Why the difference?

Also, within compton scattering, given the initial wavelength of the photon, we get a relationship between scattering angle, and wavelength of the scattered photon. But we can't know the scattering angle without knowing the wavelength of the scattered photon. So there's a free variable. What determines the wavelength of the scattered photon? Is it some type of random process?

Thanks in advance.

Improve this question
$\endgroup$

2 Answers 2

Reset to default
4
$\begingroup$

A free electron does not have any potential to which it is constraint so it can have any energy possible. i.e. it's energy states are continuum much like a free particle in classical mechanics. So scattering of such a particle is possible.

On the other hand, when in an atom, the electron is subjected to coulomb potential and this potential dictates the energy levels accessible to the electron. It's no more continuous. This means that the electron can only absorb or emit energies in steps. Thus it's an all-or-nothing phenomena. What are the energy states accessible is dictated by solving Schrodinger's equation. Thus any photon cannot pluck out the electron only a specific protons can do that(for whom the energy matches the energy difference of states).

For the second part of your question, In Compton effect the change of wavelength is determined by the scattering angle. The more the angle the more is the change. And you can treat the initial wavelength as the free parameter in this case.

Improve this answer
$\endgroup$
2
  • $\begingroup$ What determines the scattering angle? $\endgroup$
    – Ameet Sharma
    Commented Mar 14, 2016 at 15:09
  • 1
    $\begingroup$ That's interesting really. After the collision the photon can be scattered to any angle. It's state is in quantum-superposition of all the scattered states. Only the experiment collapses that state to a given scattering angle. But the probability of collapse is not same for all angles. $\endgroup$
    – Ari
    Commented Mar 14, 2016 at 16:19
2
$\begingroup$

In Compton scattering the energy of the incident photon is very much greater than the "binding energy" of the target electron so the target electron can be considered as an essentially stationary "free" electron and its final state is not being bound to the nucleus that it was originally bound to. So there is no well defined energy level that it must jump to.
In the "all or nothing" bit of you question I assume you are writing about the promotion of an electron between two well defined energy levels which requires a photon of a well defined energy.

In the Compton experiment the detector is placed at a particular angle and the intensity of the radiation detected is plotted against the wavelength of the radiation detected.
Two peaks are found. One of them which occurs at a wavelength very close to the incident photon wavelength as a result of the incident photons colliding with tightly bound inner shell electrons and the other due to Compton scattering.

The wavelength of the scattered photon is determined by the laws of conservation of energy and momentum.

More here

Improve this answer
$\endgroup$
2
  • $\begingroup$ Given a single collision, what determines the scattering angle? Conservation of energy and momentum only give a relationship between scattering angle and final wavelength. $\endgroup$
    – Ameet Sharma
    Commented Mar 14, 2016 at 15:11
  • 1
    $\begingroup$ The best way I can explain it is if you consider the photon and the electron as billiards balls and the photon does not not the electron full on but at a glancing angle. Changing where the photon hits the electron changes the angle of deflection. I can only explain this in terms of what is called "billiard ball dynamics" and not in terms of quantum mechanics. $\endgroup$
    – Farcher
    Commented Mar 14, 2016 at 16:10

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.