Biomedical Engineering Reference
In-Depth Information
(b) Applying Eq. (9.5) with
Q
= 0.75 MeV and
E
n
= 2.6 MeV, we obtain
cos
2
θ
=
57.5
◦
for the scattering angle of the proton (Fig. 9.6). Therefore, the
neutron scattering angle is 90.0
◦
- 57.5
◦
= 32.5
◦
.
(c) The average energy loss,
Q
avg
, is approximately one-half the maximum possible.
With
M
= 1and
m
= 12 in Eq. (9.3), we find for neutron collisions with carbon
0.288, giving
θ
=
4
×
12
×
1
Q
max
=
E
n
=
0.284
E
n
.
(9.12)
(1 + 12)
2
Thus,
Q
max
=
0.369 MeV.
(d) As measured by an observer at rest with respect to the center of mass of the
colliding neutron and proton, neither particle loses energy in the collision. The two,
having equal masses, first approach each other with equal speeds. The energy lost by
the neutron in the center-of-mass system is zero.
0.284
×
2.6
=
0.738 MeV; and
Q
avg
=
Example
(a) In the last problem, what are the energies of the neutron and proton in the center-
of-mass system? (b) How much energy is associated with the motion of the center of
mass in the laboratory system?
Solution
(a) Since the speed of both particles in the center-of-mass system is
2
V
and the masses
are equal, the neutron and proton each have the energy
1
2
M
(
2
V
)
2
1
8
MV
2
1
4
E
n
=
=
=
=
0.650 MeV.
(9.13)
(b) To an observer at rest relative to the center of mass, each of the colliding par-
ticles has an energy
. The total energy associated with this relative motion is
1
2
E
n
=
rel
=
1.30 MeV. In the laboratory system, the total energy
E
n
is the
sum of the energy of relative motion,
2
=
rel
, and the energy of motion of the center
of mass,
E
com
. Therefore
E
n
-
2
E
n
=
1
2
E
n
=
E
com
=
E
n
-
rel
=
1.30 MeV,
(9.14)
which is the desired answer. An alternative way of analyzing the problem is the fol-
lowing. As we have seen (Chapter 2, Problem 31), the total energy in the laboratory
system is the sum of the energies of the center-of-mass motion and the relative mo-
tion. The total mass,
M
+
M
= 2
M
, moves with speed
2
V
. The energy of this motion
is
1
2
(2
M
)(
2
V
)
2
1
4
MV
2
1
2
E
n
,
E
com
=
=
=
(9.15)
in agreement with (9.14). The effective mass for the relative motion is given by
Eq. (2.20) for the reduced mass, which with equal masses is
m
r
=
1
2
M
. The relative
velocity of the neutron and proton is
V
. Thus, the energy of relative motion is
1
2
m
r
V
2
1
4
MV
2
1
2
E
n
,
rel
=
=
=
(9.16)
as found earlier.
