Ok, background - studying for the physics part of my radiology exams, and came across a question that went something like this

An electron fired through a tungsten target loses energy by: a) Bremsstrauhlung b) collisions with bound target electrons c) giving off characteristic radiation d) all of the above

Now, by the process of elimination d must be right, because both a and b are

My question regards c - characteristic radiation

My understanding of this process is the excitation of the electron to a higher energy state, usually a more peripheral valence shell, which then releases that extra energy when it drops back (or another drops in to fill the hole)

So, what I don't get is how a free electron (like one fired from the filament to the target) has energy states. By definition, it is unbound, and therefore there are no binding energies to create quantised energy states.

Is there a property I am overlooking, is the question wrong, or is it a fudge?

I apologise if this has been asked. I looked at the suggested answers, and did a google search of stack exchange, but my phone can't use the search bar here directly, so I could have missed something


The (c) option is badly worded, in my opinion. The characteristic radiation is emitted by the atom and this indirectly contributes to the original electron (say, A) losing energy.

This is how it happens, roughly: An electron A fired into the target can ionize one of the atoms which ejects an electron B. In doing so, the electron A loses energy. But the effect is seen when the ionized atom emits characteristic radiation when one of the higher shell electrons falls into the shell from which the atom ejected electron B.

So there's no question of free particles in bound states. Free particles by definition are not in bound states.

  • $\begingroup$ I agree; option c is not stated clearly at all. Your explanation seems right. "Characteristic radiation" is simply ionisation/deionisation of the atom in the collision. $\endgroup$ – Noldorin Jan 21 '11 at 17:36
  • $\begingroup$ Thanks. So answer c is the same as answer b. That's what I thought. It is gonna be a bit of a tricky exam because maybe every tenth question will have a similar pseudo error like this, because the exam is aimed at physics layman. Unfortunately, I have no ability to challenge my result, or even see my paper afterwards $\endgroup$ – SoulmanZ Jan 21 '11 at 23:27

Your Answer

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