Biomedical Engineering Reference
In-Depth Information
where X is the unknown daughter element. Requiring that the mass numbers and atomic numbers
balance on the two sides of the arrow, we find that the daughter nucleus must have a mass
number of 222 and an atomic number of 86:
4
2
He
The periodic table shows that the nucleus with an atomic number of 86 is radon, Rn.
226
88
Ra
222
86
X
!
þ
EXAMPLE PROBLEM 15.3
In Example Problem 15.2, we showed that the
226
88
Ra nucleus undergoes alpha decay to
222
86
Rn.
226
88
Ra to be
Calculate the amount of energy liberated in this decay. Take the mass of
226.025406 amu, that of
222
86
Rn to be 222.017574 amu, and that of
2
He to be 4.002603 amu.
Solution
After decay, the mass of the daughter,
m
d
, plus the mass of the alpha particle,
m
a
,is
m
d
þ
m
a
¼
222
:
017574 amu
þ
4
:
002603 amu
¼
222
:
020177 amu
Thus, calling the mass of the parent nucleus
M
p
, we find that the mass lost during decay is
D
m
¼
M
p
ð
m
d
þ
m
a
Þ¼
226
:
025406 amu
226
:
020177 amu
¼
0
:
005229 amu
Using the relationship 1 amu
¼
931.5 MeV, we find that the energy liberated is
E
¼ð
0
:
005229 amu
Þð
931
:
50 MeV
=
amu
Þ¼
4
:
87 MeV
Negatron (b-) or (b-, g) Decay
When a radioactive nucleus undergoes beta decay, the daughter nucleus contains the
same number of nucleons as the parent nucleus, but the atomic number (Z) is increased
by 1. A typical beta decay event is
14
6
C
14
7
N
0
1
e
The superscripts and subscripts on the carbon and nitrogen nuclei follow our usual
conventions, but those on the electron may need some explanation. The -1 indicates that
the electron has a charge whose magnitude is equal to that of the proton but is negative.
The 0 used for the electron's mass number indicates that the mass of the electron is almost
zero relative to that of carbon and nitrogen nuclei.
The emission of electrons from a nucleus is surprising because the nucleus is usually
thought to be composed of protons and neutrons only. This apparent discrepancy can be
explained by noting that the electron that is emitted is created in the nucleus by a process
in which a neutron is transformed into a proton. This can be represented by the following
equation:
!
þ
1
1
0
0
n
!
1
p
þ
1
e
