Environmental Engineering Reference
In-Depth Information
The potential barrier has a peak at the closest spacing,
r ¼ r
Z
þ r
a
.Inthis
gure,
we see three regions: an inner region where bound states composed of oppositely
traveling waves describe a particle bouncing around inside a fused nucleus, an
intermediate range where the potential barrier exceeds the kinetic energy (the
wave function will decay exponentially), and an outer region where traveling
waves again exist.
These regions can be described accurately with Schrodinger
s
equation in spherical polar coordinates
. To simplify, the main features are usefully
understood in a one-dimensional model, which we will also use for discussion of
the state of electrons in a metal.
2.2.3
Bound States Inside a One-Dimensional Potential Well, E>0
In the context of nuclear fusion, the potential
U
inside the nucleus will be strongly
negative, on an MeV scale, compared to the potential outside the nucleus. For
the purpose of simple 1D calculation, we will take the zero of energy to be that
inside the nucleus, and for purpose of simple calculation, assume the potential at
the outer radius
L
of the nucleus is in
nite. In one dimension, suppose
U¼
0
for 0
<
x
<
L
,and
U¼¥
elsewhere, where
(
x
)
¼
0. For 0
<
x
<
L
, the equation
becomes
d
2
d
x
2
2
ðxÞ=
þð
2
mE
=
h
Þ ðxÞ¼
0
:
ð
2
:
11
Þ
This has the same form as the classical equation for the motion of a mass on a
spring, the simple harmonic oscillator, so the solutions can be adapted from that
familiar example. (For a mass on a spring,
F¼ma
, for
F¼Kx
, gives the differential
equation
d
2
x
/d
t
2
þ
(
K
/
m
)
x¼
0, with solution
x ¼
sin ((
K
/
m
)
1/2
t
). The spring con-
stant is
K
, in Newtons/
m
. The role of
K
/
m
is taken on by 2
mE
/
h
2
in the potential well.)
Thus, one writes
ðxÞ¼A
sin
kx þB
cos
kx
;
ð
2
:
12
Þ
where
2
1
=
2
k ¼ð
2
mE
=
h
Þ
¼
2
p=l:
ð
2
:
13
Þ
The in
nite potential walls at
x¼
0 and
x¼L
require
(0)
¼
(
L
)
¼
0, whichmeans
that
B¼
0. Again, the boundary condition
(
L
)
¼
0
¼A
sin
kL
means that
kL ¼ np;
with
n ¼
1
;
2
;
...
ð
2
:
14
Þ
This, in turn, gives
2
2
2
m ¼ n
2
h
2
8
mL
2
E
n
¼
h
ðnp=
LÞ
=
=
;
n ¼
1
;
2
;
3
;
...
ð
2
:
15
Þ
and wavefunctions, normalized to one electron per state
1
=
2
sin
ðnpx
n
ðxÞ¼ð
2
=
LÞ
=
LÞ:
ð
2
:
16
Þ
