
88 Layered nanostructures
greater than the potential barrier, U
0
, the transmission coefficient D 1, and
the reflection coefficient is close to unity. As the ratio E/U
0
increases, the
transmission coefficient, D, rapidly increases and at E/U
0
= 2 its maximal
value is D ≈ 0.9. The electron’s reflection from the potential well at E > U
0
is a purely quantum effect, which is caused by the wave properties of the
electron.
With the increase of the electron energy, E, both coefficients, D and R,
oscillate. This can be explained by invoking the interference of the de Broglie
waves reflected from the potential barrier at the boundaries of the well at x = 0
and x = L. Since, at kL = nπ , the electron does not undergo reflections, R = 0
and D = 1. The energy of such an electron has to be equal to
E
n
=
π
2
h
-
2
2m
e
L
2
n
2
, (3.132)
where n is an integer number from n = n
min
to n →∞, where n
min
is defined
from the condition E
n
≥ U
0
with E
n
found from Eq. (3.132). The absence
of electron reflection from the potential well is analogous to the effect of an
antireflecting coating of a lens. This effect is connected with the destructive
interference of waves reflected from the two boundaries of the potential well,
which takes place at
L = n
λ
Br
2
, (3.133)
where n = 1, 2, 3,... and λ
Br
is the de Broglie wavelength that is defined by
Eq. (2.88): λ
Br
= 2π/k. The opposite situation, i.e., the minima of the transmis-
sion coefficient and, correspondingly, the maxima of the reflection coefficient,
occurs if the following condition is satisfied:
kL = (2n − 1)
π
2
, (3.134)
which can be rewritten as
L =
2n − 1
4
λ
Br
. (3.135)
Example 3.6. An electron travels over a symmetric rectangular well of width
L and barrier height U
0
. Find the energy of the electron, E, that corresponds
to equal values for the coefficients of transmission, D, and reflection, R.The
well should be considered sufficiently narrow, i.e., the condition kL 1must
be satisfied.
Reasoning. From the given condition we have that R = D = 1/2 and, taking
into account Eqs. (3.129) and (3.131), we obtain
sin
2
(kL) = 4
E
U
0
E
U
0
− 1
. (3.136)