442 Nuclear Medicine Physics
8.2.7 The Photoelectric Effect
Theoretical analysis of the photoelectric effect is difficult, because the wave
functions of the final states, which are solutions of the Dirac relativistic
equation of the scattered electron, can only be obtained in the exact form
as an infinite sum of partial waves. In addition, for high energies, a large
number of terms are needed.
This is more complex than the Compton effect due to the existence of the
bonding energy of the electron, and there is no simple equation, such as that
of Klein–Nishina discussed earlier.
The wave function of the initial state for the bond electron is, in general,
assumed to be an electron state in a central potential of the nucleus, and the
effect of all the remainder electrons in the atom are taken in account as if the
field of the nucleus has an electric charge Z −S
i
. The constant S
i
is named the
shielding constant and represents the decrease of the nuclear electric charge
arising from the presence of all the electrons of the atom. For further analysis,
see, for example, Davisson [7].
In the photoelectric effect, a photon disappears and an electron is ejected
from the atom. Instead of looking at this interaction as an effect between a
photon and an electron, it must be seen as a process between a photon and an
atom. In fact, complete absorption between only a photon and a free electron
cannot occur, because the linear momentum will not be conserved [8].
The nucleus absorbs the momentum but acquires relatively little kinetic
energy due to its large mass. It is clear that the photoelectric effect can only
occur if the incident photon has energy greater than the bonding energy of
the electron that will be removed. The hole created by the ejected electron is
filled by electrons from any outer orbital, which can simultaneously produce
fluorescence radiation, emission of Auger electrons, or both.
The competition between the emissions of a fluorescent K photon and the
emission of Auger electrons is described by Y
K
, termed the fluorescence yield
of the K shell, which is defined by the number of K photons emitted per hole
in the shell K. The probability that a photon K will be emitted is nearly 1 for
elements with high Z and almost zero for materials with low Z.
Inthe photoelectric effect, the photoelectronacquiresenergyw −E
s
,where
E
s
is the bonding energy of the shell from where the photon was ejected. In
the filling of the holes created, on average, part of the energy, E
s
, is emitted
as characteristic radiation, whereas part is deposited through Auger elec-
trons. For this reason, the mean energy transferred (to electrons),
¯
E
tr
,inthe
photoelectric process is given by
w −E
s
<
¯
E
tr
< w. (8.21)
For materials with high Z, this becomes more complicated [9]. For tissue
equivalent materials of interest in radiological applications, it is much sim-
pler, because in these materials the bonding energies of the K shell are very
small (approximately 500 eV); therefore, the photoelectron acquires almost