I came across this question in my text book, it says:

What will happen to the x-rays if we change the element of the target in Coolidge tube by another one with a higher atomic number?

I believe that the characteristic spectrum will have a shorter wavelength, but I don't know whether the bremsstrahlung will be affected or not? And if the bremmstrahlung will undergo a change how would be this? (Note: my book says the bremsstrahlung is affected by the voltage between the cathode and the anode and didn't mention anything about the element of the target material).

  • $\begingroup$ The wikipedia article covers this, perhaps in more depth than you might want. $\endgroup$
    – Jon Custer
    May 30, 2017 at 16:27
  • $\begingroup$ You mean that article with the Köhn-Ebert formulas @JonCuster ? Basic answer is that the short-wave cutoff does not change. Intensity is a bit complicated, because self-absorption is also higher in high-Z materials. $\endgroup$
    – user137289
    May 31, 2017 at 12:13

1 Answer 1


As an example for X-ray crystallography, where targets of Molybdenum ($^{42}$Mo) are frequently used as "general workhorse", and Copper ($^{29}$Cu) is more suitable for organic matter, especially proteins, the superposition below may be helpful:

enter image description here


The then selected and used $K_\alpha$ are the at wavelengths of 0.709 (Mo) and 1.54 Å (Cu), respectively.

According to Pieter's comments, however, if the electrical potential is kept fixed, and only the target is exchanged, their short-wavelength cut-off will be the same:

enter image description here

(same source)

  • $\begingroup$ But in that figure, a higher voltage was used for the molybdenum anode (the short wavelength cut off is lower). $\endgroup$
    – user137289
    May 31, 2017 at 12:02
  • $\begingroup$ So there will be a change, right? Could you please explain it simply for me as I am just a high school student? And is there a change in the voltage for molybdenum as Pieter have already mentioned? $\endgroup$
    – Asmaa
    May 31, 2017 at 20:35
  • $\begingroup$ @Asmaa I still think the characteristic K\alpha lines will stay at their wavelength, altering electrical potential and current applied on the X-ray tube may influence if the process takes place and at what intensity (eventually how many X-ray photons are obtained). But from Peter's two comments I extract that I was insofar wrong that 1) changing potential and current influences in any case the shape of spectral distribution of background / bremsstrahlung. And 2) keeping the two fixed while changing the target will yield a common cut-off at short wavelength' s side, regardless the target. $\endgroup$
    – Buttonwood
    May 31, 2017 at 21:04

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