Biomedical Engineering Reference
In-Depth Information
The first process represents a beta decay in which 12 B decays to 12 C*, where the asterisk is
used to indicate that the carbon nucleus is left in an excited state. The excited carbon
nucleus then decays to a ground state by emitting an x-ray. Note that an x-ray emission
does not result in any change in Z or A.
Positron (b 1) b 1 , g) Decay
With the introduction of the neutrino, the beta decay process in its correct form can be
written as
14
6 C
14
7 N
0
!
þ
1 e
þ v
where the bar in the symbol
indicates that this is an antineutrino. To explain what an
antineutrino is, consider the following decay process:
12
7 N
v
12
6 C
0
1 e
v
When 12 N decays into 12 C, a particle is produced that is identical to the electron except that
it has a positive charge of
!
þ
þ
. This particle is called a positron. Because it is like the electron
in all respects except charge, the positron is said to be antiparticle to the electron.
It should be noted that positrons (
þ e
) are created in the nucleus, as if a proton was con-
verted to a neutron and a positron. Positron decay produces a different element by decreas-
ing Z by 1, with A being the same. For example, carbon-11 decays to the predominant stable
isotope of boron (i.e., 11 C transforms to 11 B). Positrons (like negatrons) share their energy
with neutrinos. Positron decay is also associated with
-ray emission. The positron, once
emitted, however, is often annihilated as a result of a collision with an electron within
one nanosecond. This produces a pair of photons of 0.511 MeV that move in opposite direc-
tions. A minimum transition energy of 1.022 MeV is required for any positron decay.
g
EXAMPLE PROBLEM 15.5
12 N beta decays to an excited state of 12 C, which subsequently decays to the ground state with
an emission of 4.43 MeV gamma ray. What is the maximum energy of the emitted beta particle?
Solution
The decay process for positive beta emission is
12
7 N
12
6 C *
þ þ e
12
6 C
g
Conservation of energy tells us that a nucleus X will decay into a lighter nucleus X 1 with an
emission of one or more particles (collectively designated as
!
þ v !
þ
x
) only if the mass of X is greater than
the total mass X 1
þ x
. The excess mass energy is known as the Q value of the decay. In this process
Q
c 2
to determine Q for this decay, we first need to find the mass of the product nucleus 12 C in its excited
state. In the ground state 12 C has a mass of 12.000000 amu, so its mass in the excited state is
12
7 N
12
6 C *
¼½
m
ð
Þ
m
ð
Þ
2m e
4
:
43 MeV
12
:
000000 amu
þ
amu ¼
12
:
004756 amu
931
:
5 MeV
=
Q
¼½
12
:
018613
12
:
004756 amu
2
0
:
000549 amu
931
:
5 MeV
=
amu
¼
11
:
89 MeV
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