Biomedical Engineering Reference
In-Depth Information
simply equal to the difference in the masses of the
60
Co and
60
Ni
atoms
.
9)
There-
fore, it follows that one can compute the energy released in beta decay from the
difference in the values
P
and
D
, of the parent and daughter atoms:
Q
β
-
=
P
-
D
.
(3.25)
-
Using the data from Appendix D, we find for the energy released in a
β
transfor-
mation of
60
Co to the ground state of
60
Ni
Q
=
-61.651 - (-64.471)
=
2.820 MeV.
(3.26)
In accordance with (3.23), this energy is shared by the beta particle, antineutrino,
and recoil
60
Ni nucleus. The latter, because of its relatively large mass, receives
negligible energy, and so
E
β
-
+
E
ν
=
Q
,
(3.27)
where
E
β
-
and
E
ν
are the initial kinetic energies of the electron and antineutrino.
Depending on the relative directions of the momenta of the three decay products
(
β
-
,
ν
, and recoil nucleus),
E
β
-
and
E
ν
can each have any value between zero and
Q
, subject to the condition (3.27) on their sum. Thus the spectrum of beta-particle
energies
E
β
-
is continuous, with 0
Q
, in contrast to the discrete spectra
of alpha particles, as required by Eq. (3.18). Alpha particles are emitted in a decay
into two bodies, which must share energy and momentum in a unique way, giving
rise to discrete alpha spectra. Beta particles are emitted in a decay into three bod-
ies, which can share energy and momentum in a continuum of ways, resulting in
continuous beta spectra. The shape of a typical spectrum is shown in Fig. 3.5.
10)
The maximum beta-particle energy is always equal to the
Q
valueforthenuclear
tr
ansition. As a rule of thumb, the average beta energy is about one-third of
Q
:
E
β
-
≤
E
β
-
≤
Q
/3
.
To construct the decay scheme for
60
Co we consult Appendix D. We see that
99 +
%
of the decays occur with
Q
= 0.318
MeV and that both of the gamma pho-
tons occur with almost every disintegration. Therefore, almost every decay must
go through an excited state of the daughter
60
Ni nucleus with an energy at least
1.173 + 1.332 = 2.505
MeV above the ground state. Adding the maximum beta
energy to this gives
2.505 + 0.318 = 2.823
MeV, the value [Eq. (3.26), except for
round-off ] calculated for a transition all the way to the ground state of the
60
Ni nu-
cleus. Therefore, we conclude that the
60
Co nucleus first emits a beta particle, with
Q
=
∼
0.318
MeV, which is followed successively by the two gamma rays. It remains
9 We can think of adding and subtracting 27
electron masses in Eq. (3.24), giving
Q
10 Beta-ray spectra exhibit a variety of shapes.
The spectra of some 100 nuclides of
importance for radiation protection and
biomedical applications are presented by W.
G. Gross, H. Ing, and N. Freedman, “A Short
Atlas of Beta-Ray Spectra,” Phys. Med. Biol.
28
(
M
Co,N
+27
m
)-(
M
Ni,N
+28
m
).
Neglecting the difference in electron binding
energies, then, we have
Q
=
M
Co,
A
-
M
Ni,
A
,
where the subscript
A
denotes the atomic
masses. It follows that
Q
is equal to the
difference in
=
, 1251-1260 (1983).
values for the two atoms.
