Environmental Engineering Reference
In-Depth Information
For electric
eld
E
¼
0, the electron is assumed to have energy
U
¼
E
o
at
x
¼
a
o
¼
0.053 nm, the Bohr radius. When
E
>
0, there is a second location, near
x
2
¼
E
o
/
eE
, where
U
is again
E
o.
The problem is to
find the time
t
to tunnel from
x
¼
a
o
to
x
¼
x
2
¼
E
o
/
eE
through the potential barrier
U
(
x
)
E
o
. For small
eld
E
, the
maximum height of the barrier seen by the electron is
E
o
¼
13.6 eV, and the width
t
of the tunnel barrier is essentially
E
o
/
eE
.
If the barrier potential
U
(
x
) were
V
B
(a square barrier independent of
x
), the
tunneling transmission probability
T
would be
D
x
¼
t
¼
x
2
¼
1
=
2
T
¼
exp
½
2
ð
2
mV
B
Þ
D
x
=
h
ð
4
:
5
Þ
for barrier width
h
/2
p
, with
h
Plancks constant.
A more accurate approach would replace (
V
B
)
1/2
D
x
and
h
¼
x
by
Ð
½
1
=
2
d
x
, this is
known as the WKB approximation, and was also used in the Gamow tunneling
probability, see Equation 2.22.
We will simplify further, and calculate on the basis of the
average value
of the
barrier: approximate the barrier function
U
(
x
)
D
U
ð
x
Þ
E
o
E
o
by half its maximum value,
1
/
2
V
BMAX
, over the width
D
x
.
1/
t
, with
t
the desired ionization lifetime, is
Tf
approach
.
The orbital frequency of the electron will be taken from Bohrs rule for the angular
momentum
L
The escape rate,
f
escape
¼
h
/[(2
p
)
2
ma
o
]
10
15
s
1
. The work-
¼
mvr
¼
h
/2
p
,so
f
approach
¼
¼
6.5
ing formula
f
escape
¼
1/
t
is
1
=
2
f
escape
¼
f
approach
exp
½
2
ð
mV
BMAX
Þ
D
x
=
h
:
ð
4
:
6
Þ
ke
2
/
x
To
nd
D
x
one solves the quadratic
E
o
¼
eEx
(
E
is the electric
field):
2
1
=
2
Dx ¼½ðE
o
=eÞ
4
k
C
eE
=E:
ð
4
:
7
Þ
(This limits the applicability of the approximation to
E
<
32.1 V/nm, where
D
x
0.)
So we have the width of the barrier, and nowwe want to
¼
find the location of its peak
and the peak value.
The equation for
U
is
U
(
x
)
k
C
e
2
x
1
¼
eEx
. Taking the derivative with respect to
U
0
, we set
U
0
¼
x
,d
U
/d
x
¼
0to
find the barrier peak. Thus,
U
0
¼
k
C
e
2
x
2
eE
¼
0at
1
=
2
x
0
¼ð
k
C
e
=
E
Þ
;
ð
4
:
8
Þ
which locates the peak of the barrier. Plugging this value,
x
0
, into the expression for
U
,
we
find the peak value of
U
,
U
max
,tobe
k
C
e
2
1
=
2
1
=
2
1
=
2
U
max
¼
ð
E
=
k
C
e
Þ
eE
ð
k
C
e
=
E
Þ
¼
2
e
ð
k
C
Ee
Þ
:
ð
:
Þ
4
9
