(f) During each collision, the energy of the neutron is reduced by the factor 1 –
0.89 =
0.11. If
E
i
is the initial energy, then the energy after
n
collisions is given by
E
= (0.11)
n
E
i
.
We take the natural logarithm of both sides and solve for
n
. The result is
ln( /
)
ln(0.025 eV/1.00 eV)
7.9
8.
ln 0.11
ln 0.11
i
EE
n
==
=
≈
The energy first falls below 0.025 eV on the eighth collision.
23. (a) Let
v
ni
be the initial velocity of the neutron,
v
nf
be its final velocity, and
v
f
be the
final velocity of the target nucleus. Then, since the target nucleus is initially at rest,
conservation of momentum yields
m
n
v
ni
= m
n
v
nf
+ mv
f
and conservation of energy yields
1
2
2
1
2
2
1
2
2
mv
nn
i
f
f
=+
.
We solve these two equations simultaneously for
v
f
. This can
be done, for example, by using the conservation of momentum equation to obtain an
expression for
v
nf
in terms of
v
f
and substituting the expression into the conservation of
energy equation. We solve the resulting equation for
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 05/19/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Momentum, Neutron

Click to edit the document details