Biomedical Engineering Reference
In-Depth Information
The units are those of
k
0
e
4
n
/
mc
2
:
(Nm
2
C
-2
)
2
C
4
m
-3
J
N
2
m
Nm
=
Jm
-1
.
(5.30)
=
Converting to the more common units, MeV cm
-1
,wehave
10
-42
z
2
n
β
8.12
×
J
m
×
1
1.60 × 10
-13
MeV
J
1
100
m
cm
×
2
10
-31
z
2
n
β
5.08
×
MeV cm
-1
.
=
(5.31)
2
In the dimensionless logarithmic term in (5.23) we express the energies conve-
niently in eV. The stopping power is then
ln
1.02
2
MeV cm
-1
.
10
-31
z
2
n
β
10
6
2
-
d
E
5.08
×
×
β
d
x
=
-
β
(5.32)
2
I
eV
(1 -
β
2
)
This general formula for any heavy charged particle in any medium can be written
10
-31
z
2
n
β
-
d
E
5.08
×
)-
ln
I
eV
]MeVcm
-1
,
(5.33)
d
x
=
[
F
(
β
2
where
10
6
2
ln
1.02
×
β
2
.
F
(
β
)
=
-
β
(5.34)
2
1-
β
Example
Compute
F
(
β
) for a proton with kinetic energy
T
=
10 MeV.
Solution
In the second example in Section 5.2 we found that
2
β
=
0.02099. Substitution of this
value into Eq. (5.34) gives
F
(
β
)
=
9.972.
2
and
F
(
) are given for protons of various energies in Table 5.2.
Since, for a given value of
β
, the kinetic energy of a particle is proportional to its rest
mass, the table can also be used for other heavy particles as well. For example, the
ratio of the kinetic energies
T
d
and
T
p
of a deuteron and a proton traveling at the
same speed is
The quantities
β
β
T
d
T
p
=
M
d
M
p
=
2.
(5.35)
The value of
F
(
9.973 that we just computed for a 10-MeV proton applies, there-
fore, to a 20-MeV deuteron. Linear interpolation can be used where needed in the
table.
β
)
=
