Biomedical Engineering Reference
In-Depth Information
3.3
Alpha Decay
Almost all naturally occurring alpha emitters are heavy elements with
Z
83.The
≥
principal features of alpha decay can be learned from the example of
226
Ra:
226
88
Ra
222
86
Rn +
2
He.
(3.11)
→
The energy
Q
released in the decay arises from a net loss in the masses
M
Ra,N
,
M
Rn,N
,and
M
He,N
, of the radium, radon, and helium nuclei:
Q
=
M
Ra,N
-
M
Rn,N
-
M
He,N
.
(3.12)
This nuclear mass difference is very nearly equal the atomic mass difference,
which, in turn, is equal to the difference in
values.
4)
Letting
P
,
D
,and
He
denote the values of the parent, daughter, and helium atoms, we can write a general
equation for obtaining the energy release in alpha decay:
Q
α
=
P
-
D
-
He
.
(3.13)
Using the values in Appendix D for the decay of
22
88
Ra to the ground state of
22
86
Rn,
we obtain
Q
=
23.69 - 16.39 - 2.42
=
4.88 MeV.
(3.14)
The
Q
value (3.14) is shared by the alpha particle and the recoil radon nucleus,
and we can calculate the portion that each acquires. Since the radium nucleus was
at rest, the momenta of the two decay products must be equal and opposite. Letting
m
and
v
represent the mass and initial velocity of the alpha particle and
M
and
V
those of the recoil nucleus, we write
mv
=
MV
.
(3.15)
Since the initial kinetic energies of the products must be equal to the energy re-
leased in the decay, we have
1
2
mv
2
+
2
MV
2
=
Q
.
(3.16)
Substituting
V
=
mv
/
M
from Eq. (3.15) into (3.16) and solving for
v
2
, one finds
2
MQ
m
(
m
+
M
)
.
v
2
=
(3.17)
4
Specifically, the relatively slight difference in
the binding energies of the 88 electrons on
either side of the arrow in (3.11) is neglected
when atomic mass loss is equated to nuclear
mass loss. In principle, nuclear masses are
needed; however, atomic masses are much
better known. These small differences are
negligible for most purposes.
