# More points vs. precision

I have a dilemma.

In my lab exercise, I was measuring spectra with HPGe detector of several sources (gamma spectroscopy). To determine the energy of the unknown spectrum I first needed to calibrate the detector by making measurements of known elements: Co 60, Cs 137, and Ba 133. The spectra of Co and Cs have nice discernible peaks, and from table I know their energies. For Co ~1332 and 1173 keV peaks, and Cs ~661 keV peak. But I also had Ba 133, and there were 4 peaks, and even tho I have the energies of those peaks (I don't know with 100% certainty that they are correct), when I draw the calibration graph (energy/channel plot), I have some scattering, and my calibration fit is E=a*n+b, where a=($0.78\pm 0.01$) keV, and b=($80\pm 10$) keV.

But if I only take the three distinct peaks (Co and Cs), the calibration parameters are a=($0.8200\pm 0.002$) keV and b=(26.1\pm 0.3)keV. The relative error in this case is minute.

The picture is:

the dark cyan one is fit with all 7 points, and the red dashed is only 3 points.

Now, my dilemma is: do I take more points (I remember from my stat class, that the more the better, since you never know how your data will behave in different regimes), or do I take these three which are totally distinct and I know with 100% the energy of those peaks. And then use that as my calibration?

Is the ~40keV (around 200th channel) difference between these two, a big difference in determining the unknown spectra?

Even tho my relative uncertainty is higher with Barium spectra, maybe it's good to include them, especially if my unknown spectra have peaks on lower energies (channel).

What would you do?

EDIT: Barium spectra:

And my detector has 2048 channels.

-
A statistician would tell you that the minimum number of data points for fitting to $n$ parameters is 2n+1 if you are to report uncertainties. You certainly cant find all of $a$, $\Delta a$, $b$, and $\Delta b$ with only three points with any confidence. What tool are you using that isn't letting you know about these problems? –  dmckee Mar 10 '13 at 13:53
Next question. You say you get 4 distinct peaks from the Ba source. The table of the isotopes entry for $^{133}\mathrm{Ba}$ suggests to me that this is a strange number to get. I see 356, 81, 302, 384, 276 keV in the gamma table and 31, 31, 34, 4.3, 36 keV from the x-ray table. using a cutoff around 5% ranching ratio. I always found Ba sources to be difficult, and I seem to recall a minor degree of non-linearity at the lowest end of the energy range for Ge detectors. –  dmckee Mar 10 '13 at 14:10
I'm not sure I understood this: "What tool are you using that isn't letting you know about these problems?" :\ You mean, what program I'm using to record this or with which I'm processing my data? Oh I'll show Ba spectrum, which is kinda... difficult to interpret for me, that's the point why I asked the question :D I'm not 100% sure to which peak I need to attribute the energies :\ –  dingo_d Mar 10 '13 at 14:19
Random thoughts. I used to use Co-60, Cs-137, Na-22, and K-40 and maybe Co-57, Mn-54 and/or Ba-133. If you have a thorium source available there is a 2.6 MeV gamma line. If you have a low background device (i.e. deep underground and/or a cosmic veto) you can use a banana as a K-40 source. Or any other high potassium fruit or vegetable. If the crystal itself is small you may have a significant Compton edge associated with each full absorption line. The quantum efficiency of these devices is rather harder to nail down than the energy calibration (which is very linear over most of their range). –  dmckee Mar 10 '13 at 14:26
–  dmckee Mar 10 '13 at 20:15

Not knowing your situation in detail it is hard to offer detailed advice, but here is one course of action that should be available to you and will give a sufficient energy calibration:

• Use the Co-60 and Cs-137 source as you have been (3 points)
• Use only the 356 keV line from the Ba-133 source (1 point)
• Use a banana (suitably wrapped in shrink wrap and ziplock bags to avoid contaminating the detector) as a K-40 source source (1 point)

That gives you five points which is sufficient for a full linear fit with uncertainties.

Now that I see the OP's data with it's very good signal to noise ratios (I was using a rather aged and low activity barium source), I would modify this as several of the barium lines are usable, and there appears to be a K-40 background in the detector visible in the Co-60 data which means that the banana is not actually needed.

-
I've done my measurements so the banana one is out, but I'll use your advice on those 4 points :) Thank you. I must say the banana one is really interesting :D –  dingo_d Mar 10 '13 at 15:08