RESOLUTION
Every detector has a specific resolution, or ability to resolve two peaks that are fairly close together in Energy. The resolution of a photopeak is defined as:
R = FWHM/H0 (H0 = mean pulse height of the peak)
There are a lot of contributions to the resolution; usually, in gamma-ray spectroscopy, the most significant is related to the statistical fluctuations of the charge collection. If we assume that this process is described by the Poisson statistics, the standard deviation of the number of photoelectrons is the square root of the mean number, so we can write:
R = FWHM/H0 = (2.35 sigma) / H0 = B sqrt(E) /E = B / sqrt(E)
Another significant contribution to the resolution is due to the fluctuations in PM tube gain. This is a non statistical source of broadening so we treat it as a constant term:
R proportional to A
We can also (have also to) consider contributions of electric noise (although, as we will see, they’re usually negligible) and we have
R = C * (1/E) (ma qualcuno sa perchè va come 1/E?) (sullo Knoll non c'è?
There could be other contributes to the peak broadening: the intrinsic crystal resolution (due to the local fluctuations in the efficiency of the crystal), the “transfer variance” (linked to the probability that a photon produces a photoelectron collected by the dynodes) , the variation in the detector response over its volume and so on. However, this contributes are less important than the previous one, so we neglect them in our study.
As it can be seen in image 006 (dal mio intervento NaI resolution), we calculate the resolution for each peak of 232Th and 22Na’s spectra and then fit these values with the following function:
R = sqrt (A2+B2/E + C2/E2)
(scriverei poi 2 righe sui valori trovati da A B e C mostrano che l'ultimo è trascurabile)
EFFICIENCY
In the definition of relative efficiency for HpGe we divide the counts of each peak for the line’s intensity (%) and fit the relation between efficiency and Energy with the following function (vedere quella del germanio). Obviously with the NaI this is not possible (because of the poor resolution) (io scriverei: this is not possibly for the NaI detector because of...). Since each peak is formed by several lines, we divided the counts of each gaussian for the sum of intensity of all lines that form the peak . The method is basically wrong because it doesn’t considerer the presence of Compton scattering in the peak, but it may be interesting for a qualitative study , as we can see in image007 (dal mio intervento NaI efficiency):

































