- #1
JulienB
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Homework Statement
Hi everybody! My homework this week is to discuss the results we obtained in an experiment last week, which was about determining the gamma-rays absorption coefficient of lead ##\mu## with a Geiger-counter and then read from a graph the gamma photon energy of the radioactive material caesium-137.
For the absorption coefficient, we measured the time taken by the Geiger-counter to make 1000 counts for 5 lead plates with different thicknesses and without plate, then performed a linear regression of ##\ln(I(d))## (##I## being counts/time - background level) which gave us ##\mu## (see attached pics). The background level was measured with nothing inside the compartments.
It is important to note that since we did only one measurement of ##I_0## (counts/s without plate), we could not plot ##ln(\frac{I(d)}{I_0})## because we could then not calculate the covariance in the uncertainty. We were told to do it that way and use ##I_0## as a parameter for the fit.
The thicknesses of the plates were 1, 1.7, 3.3, 6.8 and 11.2, all given in mm.
Homework Equations
The relation between all those values is: ##I(d) = I_0 \cdot exp(-\mu \cdot d)##.
I will also give here our results: the absorption coefficient was found to be ##\mu = (0.097 \pm 0.004) mm^{-1}##. As a consequence, we found that the half-thickness was ##d_{1/2} = (7.1 \pm 0.3)##mm and that the mass attenuation coefficient was ##(0.086 \pm 0.004)##. After reading the graphs we were given (see attached pics), we determined the photon energy of ##^{137}##Cs to be ##E = (0.8 \pm 0.1)##MeV.
Also important about discussing the fit (see below), ##I_{0,parameter} = (1.54 \pm 0.03)## and ##I_{0,measured} = (1.65 \pm 0.06)##.
The Attempt at a Solution
The problem is that the reference value I found for the energy is ##E_{ref} = 661.64##KeV, and I must now explain this non-negligible difference in results. I am in first year, that is I never had any lecture about quantum physics so I am a bit clueless. I figured that the error is most probably located in the fit, and if ##\mu## would be bigger then ##E## would tend towards ##E_{ref}##. Through research and thinking, I have some suggestions but I can't be sure whether they are right or wrong:
- if we did more measurements of ##I_0##, we would be able to calculate the covariance and perform the linear fit with ##ln(\frac{I(d)}{I_0})##. That would most probably result in a bigger ##\mu## since ##I_{0,measured} > I_{0,parameter}##;
- a Geiger-counter is only 1% efficient at detecting gamma-rays. Though it is a limitation of the measuring instrument, I am not sure whether this has any statistical impact on our measurements;
- the Geiger-counter creates a "dead time" of ##\tau = 100 \mu s## after each detection, which could prevent another detection to be made during this time. There is formula to calculate its influence: ##n = \frac{n_{measured}}{1 - n_{measured} \cdot \tau} = 1111## counts. I find this very big, does that make sense?
- 3 other teams performed the same experiment near to us. Could that affect our measurement of the background level?
- the thickness of the lead plates were given without uncertainties, and they could have been inhomogeneous. Our probe was located just under the plate. I've read that this could create a scattering effect increasing the number of counts. Could that be the case? All of the areas were the same, and we didnt pile up the plates (there were 5 different ones);
- there are two types of interaction happening inside the tube: photo-absorption and Compton scattering. If the collision between a photon and an electron is of the type Compton scattering, could it be that the Geiger-counter detects two gamma photons instead of one or is the dead time large enough to prevent it?
As you can see, this is for me very confusing. Do you remarks about what I just wrote or clues about other sources of uncertainty? I hope I didn't forget something important.Thank you very much in advance.Julien.
