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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?

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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.

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  • $\begingroup$ 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 May 25 '17 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. $\endgroup$ – Pieter May 25 '17 at 22:36

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