Absorption and Spectroscopy: Why increased counts at lower energies?

lamadwp
Messages
3
Reaction score
0
Hi!
In my lab class, we are using a photomultiplier to examine gamma ray emission lines and determine absorption coefficients of different radioactive materials.

Homework Statement



As we test for absorption (by layering on metals, incrementally increasing the thickness), we consider spectral graphs. We compare each subsequent graph (one from each thickness increment) to that of the nucleus without any absorption material and notice that, while the peaks we are following (they are the decay energies, eg. for Barium, we followed energy 0.356 MeV) are decreasing (the height of the peak is determined by the number of counts recorded by the photomultiplier) with increased metal thickness, other peaks are increasing in count.

We want to know what is causing the increase in lower-energy readings.

We also want to understand all the spectral features of the graphs as we are very unsure of any peaks outside the given decay energies..

Homework Equations



{see attachment for graph} Each Series in the legend denotes an increase in absorber thickness of about 6mm.

The Attempt at a Solution


Not indicated on the graph is that we followed the peak at about 100 on the x-axis (it's not actually 100 MeV or anything--the program we used filtered energies into channel bins, which we calibrated. The indicated energy here is 0.356 MeV). We attribute its decrease to absorption of the gamma rays as we added more material between our source and the detector. However, the peaks before channel 50 increase with material thickness. This is what we are trying to figure out.

We are considering Compton scattering, elastic scattering, and photoelectric absorption, but are unsure as to where these are applicable.
 

Attachments

  • balead.JPG
    balead.JPG
    22.7 KB · Views: 391
Last edited:
Physics news on Phys.org
6mm should be sufficent to block any secondary electrons which do not come from the last absorber plate, Compton electrons should decrease with increasing absorber thickness. Series 2 is with one absorber plate?

Photons, emitted from secondary electrons?
Radioactivity from the plates? Did you check the results without the source?
 
Yes, Series 2 is with the first plate of lead (~6mm), Series 3 is with about 12mm of lead shielding, etc. Series 1 is the source without any shielding.

We did not check the plates without the source :T
Thanks for the information about Compton electrons--I'll look it up some more, but might you be able to explain why those counts decrease with thickness, too?
 
More absorbing plates -> more photons get absorbed.
That is an exponential decay.
 
Maybe the peaks below channel 50 are the K-alpha and K-beta characteristic x-rays of lead. If you know the energies corresponding to the channels, you could compare them to the known x-ray energies for lead.

Also see here for some sample gamma spectra that show x-ray peaks from lead.
 
Thank you!

Thank you for your responses, everyone!

TSny, those resources were incredibly helpful, thank you!
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top