0
$\begingroup$

I was looking at the excitation spectrum of a few elements in the first page of this document: http://www.amptek.com/pdf/Choosing%20the%20anode%20material%20in%20an%20x-ray%20tube.pdf In particular I was looking at Rhodium. I was wondering where does the continuum for Rhodium above k-alpha and k-beta peaks come from(k-alpha for Rh is 20.2kev and k-beta is 22.7kev)?

Is the reason that the two peaks for k-alpha and k-beta highest because the probability of an electron from the L-shell and M-shell filling the ejected k-shell electron is higher? If so, does that mean the continuum past the two peaks is the result of electrons from the outer shells(N,O,etc...) filling the k-shell, but it's just a lower probability of that happening hence there are no peaks, just a continuum? Or is the continuum a result of electrons from N-shell and O-shell filling ejected L-shell electrons? or a combination?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

Classically, the continuum is caused by deflection of the incoming electron when it passes very close to the nuclear charge. There is no reason why that should be limited by the K lines.

At the Duane-Hunt limit, the total energy of the electron is converted to the photon energy. This is inverse photoemission, a quantum process, where the incoming electron makes a transition to an unoccupied bound state. In a metal, these are the states just above the Fermi level.

Improve this answer
$\endgroup$
2
  • $\begingroup$ I see, I think that makes sense. So then why is the 'count' for the continuum so much lower than the count at the k-alpha and k-beta? $\endgroup$
    – tinker102
    Commented May 25, 2017 at 22:08
  • $\begingroup$ @tinker102 Bremsstrahlung is a continuum, so the integrated intensity over all energies is usually larger than that of the characteristic lines. But yeah, I think it is less then one percent of the power is converted into x-rays. Competition with low-energy excitations that slow down the incoming electron. And with Auger processes for the atomic lines. $\endgroup$
    – user137289
    Commented May 25, 2017 at 22:36

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.