9.3 Solutions 517
9.55 LabelAis due to photopeak of 1.275 MeVγ-rays. Some of the primary pho-
tons will be absorbed within the crystal after undergoing Compton scattering.
Such events merely enhance the photopeak. In other cases, the Compton-
scattered photon will escape from the crystal, the light output will now be
proportional to the energy of the recoil electron which will be absorbed in a
large crystal. There will be a energy continuum of the recoil electrons with
energy ranging from zero to maximum. The label B represents the Compton
shoulder. The strong peak labeled C marks the photopeak at 0.511 MeV due
to electron-positron annihilation leading to absorption of one of the photons.
The annihilation may take place from the positron emitted by the source or
by the pair production caused by the primary photon. Now the total kinetic
energy available by the electron-positron pair will be (hν− 1 .02) MeV. When
all of the energy is dissipated, the positron will be annihilated. If both the
annihilation photons escape from the crystal, a peak (called Escape peak) will
occur at an energy (hν− 1 .02) MeV= 1. 275 − 1. 02 = 0 .255 MeV, represented
by label D. If one of the photons escapes, another peak will occur at an energy
(hν− 0 .511) MeV= 1. 275 − 0. 511 = 0 .764 MeV (not shown in the Fig. 9.6).
If both the photons are absorbed, full output will be realized and this will be
added up to the photo peak.
Fig. 9.6
If the annihilation peak at 0.511 MeV and the photo peak due to primary
photon occur simultaneously within the resolving time of the instrument then
their energies are added and the events are recorded as a single event, known
as sum peak at energy 1. 275 + 0. 511 = 1 .786 MeV (label E), usually with
small amplitude.9.56 The normal distribution is
p(x)=(
σ√
2 π)− 1
exp[−(x ̄−x)/ 2 σ^2 ](1)