Biomedical Engineering Reference
In-Depth Information
atomic electron, producing two 0.511-MeV photons. In small detectors one or both
annihilation photons can escape without interacting, leading to escape peaks at the
energies
h
ν
0
- 0.511
and
h
ν
0
- 1.022
MeV.
An example of a portable NaI gamma analyzer for laboratory and field use is
shown in Fig. 10.32. It has 512 channels and can store 30 spectra in memory. The
instrument displays spectra and dose rates, as seen in the figure. Software includes
peak analysis, nuclide identification search, and capability for data transfer. Ad-
justable alarm thresholds with an audible signal can be set.
Example
Monoenergetic 450-keV gamma rays are absorbed in a NaI(Tl) crystal having an
efficiency of 12%. Seventy-five percent of the scintillation photons, which have an
average energy of 2.8 eV, reach the cathode of a photomultiplier tube, which con-
verts 20% of the incident photons into photoelectrons. Assume that variations in the
pulse heights from different gamma photons are due entirely to statistical fluctua-
tions in the number of visible photons per pulse that reach the cathode. (a) Calculate
the average number of scintillation photons produced per absorbed gamma photon.
(b) How many photoelectrons are produced, on the average, per gamma photon?
(c) What is the average energy expended by the incident photon to produce a pho-
toelectron from the cathode of the photomultiplier tube (the “
W
value”)? (d) Com-
pare this value with the average energy needed to produce an ion pair in a gas or a
semiconductor.
Solution
(a) The total energy of the visible light produced with 12% efficiency is 450 keV
×
0.12 = 54.0 keV. The average number of scintillation photons is therefore 54,000/
2.8
19,300. (b) The average number of photons that reach the photomultiplier cath-
ode is 0.75 × 19,300 = 14,500, and so the average number of photoelectrons that
produce a pulse is 0.20
=
2900. (c) Since one 450-keV incident gamma
photon produces an average of 2900 photoelectrons that initiate the signal, the
“
W
value” for the scintillator is 450,000/2900
×
14,500
=
=
155 eV/photoelectron. (d) For gases,
30 eV ip
-1
; and so the average number of electrons produced by absorption of
a photon would be about 450,000/30
W
∼
3eVip
-1
and the corresponding number of electrons would be 150,000. A “
W
value” of sev-
eral hundred eV per electron produced at the photocathode is typical for scintillation
detectors. Energy resolution is discussed in Section 11.11.
The energy resolution of a spectrometer depends on several factors, such as the
efficiency of light or charge collection and electronic noise. Resolution is also inher-
ently limited by random fluctuations in the number of charge carriers collected when
a given amount of energy is absorbed in the detector. As the last example shows, the
number of charge carriers produced in germanium is substantially larger than that
in a scintillator. Therefore, as discussed in the next chapter, the
relative
fluctuation
about the mean—the inherent resolution—is much better for germanium. The up-
per panel in the example presented in Fig. 10.33 compares pulse-height spectra mea-
sured with a NaI detector for pure
133
Ba and for a mixture of
133
Ba and
239
Pu. The
soft Pu gamma photons, at energies marked by the arrows, are not distinguished.
=
15,000. For a semiconductor,
W
∼
