2
$\begingroup$

Iv been looking at gamma ray spectroscopy of late for a project that I am currently researching and going to be doing in the lab, but what I cannot seem to figure is how the barium x-rays are being shown in the spectrum?

enter image description here

enter image description here

Looking at the decay scheme and comparing it to a c137 spectrograph I cant see how the x-ray photons are being generated. My only reasoning is that there is some sort of characteristic x-ray being formed from during the transition gamma photo being ejected during the decay.

$\endgroup$
  • 1
    $\begingroup$ Look at your energy level diagram - 94.6% of the decays go to the excited level of the Barium nucleus. This excited level then decays by emission of the 662keV gamma. That gamma has more than enough energy to knock out an electron from a Ba atom, leading to characteristic x-rays. Why do you not think that is possible? $\endgroup$ – Jon Custer Oct 9 '18 at 21:27
  • $\begingroup$ @BenCrowell I did not realize that internal conversion was the same as what Jon described, so I learned something and removed my comments. $\endgroup$ – Pieter Oct 10 '18 at 7:12
3
$\begingroup$

Beta decay changes $Z$ which may produce a hole in the K shell ("shake-up"). That hole then decays. But that is then the K-shell of the progeny nucleus: barium.

Edit: probably more important: the metastable state of $^{137}$Ba decays by internal conversion in which a K electron was emitted.

https://www.ld-didactic.de/software/524221en/Content/Appendix/Cs137.htm

Indeed, when one washes out the barium product, the spectrum of the eluate looks the same.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ @BenCrowell Skake-up: a sudden change in $Z$ projects the $1s$ orbitals on a slightly different potential. Overlap is not perfect. $\endgroup$ – Pieter Oct 10 '18 at 0:21
  • 2
    $\begingroup$ I see. Sorry, in earlier comments I was thinking "electron conversion" when you wrote "internal conversion." This seems like the most likely of the reasons we've discussed, since it explains why K is preferred. The order of magnitude of the probability of x-ray emission seems about right for this process as well; internal conversion is most probable for low-energy gamma decay, and this one is a fairly high energy, so it makes sense that the probability appears to be $<\sim 0.1$. $\endgroup$ – user4552 Oct 10 '18 at 0:31
  • $\begingroup$ @BenCrowell Yes, there is also shake-up for other shells. But L emission is at much lower energy, not detected in this experiment. $\endgroup$ – Pieter Oct 10 '18 at 0:33
  • 2
    $\begingroup$ In the comment thread under the question, we discussed Jon Custer's idea that the gamma was interacting with the K shell electrons on the way out. There are two ways we could imagine this happening: (A) absorption of the gamma by a K-shell electron, or (B) scattering of the gamma by a K-shell electron. I think B is probably what Jon had in mind, but A is in fact a description of internal conversion. If B happens, then we should detect gammas with less than the full energy of the photopeak. This may very well happen, but would be hard to distinguish from Compton scattering in the detector. $\endgroup$ – user4552 Oct 10 '18 at 0:57

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.