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Subatomic Physics

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4.1. Scintillation Counters 55<br />

The efficiency of these inorganic crystals for gamma rays is high, but the decay of<br />

each pulse is slow, about 0.25µs. Moreover, NaI(Tl) is hygroscopic and large crystals<br />

are very expensive. Plastic scintillators, for instance polystyrene with terphenyl<br />

added, are cheap; they can be bought in large sheets and can be machined in nearly<br />

any desired shape. The decay time is only a few ns, but the efficiency for photons<br />

is low. They are therefore mainly used for the detection of charged particles.<br />

A few remarks are in order concerning the mechanism of observation of gamma<br />

rays in NaI(Tl) crystals. For a gamma ray of less than 1 MeV, only photoeffect<br />

and Compton effect have to be considered. Photoeffect results in an electron with<br />

an energy Ee = Eγ − Eb, whereEb is the binding energy of the electron before it<br />

was ejected by the photon. The electron will usually be completely absorbed in the<br />

crystal. The energy deposited in the crystal produces a number of light quanta that<br />

are then detected by the photomultiplier. In turn, these photons result in a pulse<br />

of electric charge proportional to Ee andwithacertainwidth∆E. This photo<br />

or full-energy peak is shown in Fig. 4.3. The energy of the electrons produced by<br />

the Compton effect depends on the angle at which the photons are scattered. The<br />

Compton effect therefore gives rise to a spectrum, as indicated in Fig. 4.3. The<br />

width of the full-energy peak, measured at half-height, depends on the number of<br />

light quanta produced by the incident gamma ray; typically ∆E/Eγ is of the order of<br />

20% at Eγ = 100 keV and 6–8% at 1 MeV. At energies above 1 MeV, the incident<br />

gamma ray can produce an electron–positron pair; the electron is absorbed, and<br />

the positron annihilates into two 0.51 MeV photons. These two photons can escape<br />

from the crystal. The energy deposited is Eγ if no photon escapes, Eγ − mec 2 if<br />

one escapes, and Eγ − 2mec 2 if both annihilation photons escape.<br />

The energy resolution ∆E/E deserves some additional consideration. Is a resolution<br />

of about 10% sufficient to study the gamma rays emitted by nuclei? In<br />

some cases, it is. In many instances, however, gamma rays have energies so close<br />

together that a scintillation counter cannot separate them. Before discussing a<br />

counter with better resolution, it is necessary to understand the sources contributing<br />

to the width. The chain of events in a scintillation counter is as follows: The<br />

incident gamma ray produces a photoelectron with energy Ee ≈ Eγ. The photoelectron,<br />

via excitation and ionization, produces n1q light quanta, each with an energy<br />

of E1q ≈ 3eV(λ ≈ 400nm). (For clarity we call the incident photon the gamma ray<br />

and the optical photon the light quantum.) The number of light quanta is given by<br />

n1q ≈ Eγ<br />

ɛlight,<br />

E1q<br />

where ɛlight is the efficiency for the conversion of the excitation energy into light<br />

quanta. Of the n1q light quanta, only a fraction ɛcoll are collected at the cathode<br />

of the photomultiplier. Each light quantum hitting the cathode has a probability<br />

ɛcathode of ejecting an electron. The number ne of electrons produced at the input

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