Biomedical Engineering Reference
In-Depth Information
the element is found and on the state of condensation of the material, solid, liquid,
or gas (Bragg additivity rule).
When the material is a compound or mixture, the stopping power can be calcu-
lated by simply adding the separate contributions from the individual constituent
elements. If there are
N
i
atoms cm
-3
of an element with atomic number
Z
i
and
mean excitation energy
I
i
, then in formula (5.23) one makes the replacement
(5.27)
n
ln
I
=
N
i
Z
i
ln
I
i
,
i
where
n
is the total number of electrons cm
-3
in the material (
n
=
i
N
i
Z
i
). In
this way the composite
ln
I
value for the material is obtained from the individual
elemental
ln
I
i
values weighted by the electron densities
N
i
Z
i
of the various ele-
ments. When the material is a pure compound, the electron
densities n
and
N
i
Z
i
in
Eq. (5.27) can be replaced by the electron
numbers
in a single molecule, as shown
in the next example.
Example
Calculate the mean excitation energy of H
2
O.
Solution
We obtain the
I
values for H and O from Eqs. (5.24) and (5.25), and then apply (5.27).
For H,
I
H
=
105 eV. The electronic densities
N
i
Z
i
and
n
can be computed in a straightforward way. However, only the ratios
N
i
Z
i
/
n
are needed to find
I
, and these are much simpler to use. Since the H
2
O molecule has
10 electrons, 2 of which belong to H (
Z
=
19.0 eV, and for O,
I
O
=
11.2+11.7
×
8
=
1) and 8 to O (
Z
=
8), we may write from
Eq. (5.27)
2
×
1
10
ln
19.0 +
1
×
8
10
ln
I
=
ln
105
=
4.312,
(5.28)
giving
I
=
74.6 eV.
5.8
Table for Computation of Stopping Powers
In this section we develop a numerical table to facilitate the computation of stop-
ping power for a heavy charged particle in any material. In the next section we use
the table to calculate the proton stopping power of H
2
O as a function of energy.
The multiplicative factor in Eq. (5.23) can be written with the help of the con-
stants in Appendix A and Appendix C as
k
0
z
2
e
4
n
mc
2
10
-19
)
4
n
9.11 × 10
-31
(3.00 × 10
8
)
2
10
9
)
2
z
2
(1.60
4
π
4
π
(8.99
×
×
=
β
2
β
2
= 8.12 × 10
-42
z
2
n
β
Jm
-1
.
(5.29)
2
