Chemistry Reference
In-Depth Information
5.5 Because the Hall effect involves two voltage and current components,
Ohm's Lawneeds to be generalised to a tensor formwhen considering
the Hall effect:
V
x
V
y
R
xx
R
xy
R
xy
R
yy
I
x
I
y
=
By calculating the inverse of the above resistancematrix, showthat the
elements of the conductance matrix, describing how current depends
on voltage, are given by
R
yy
R
xx
R
yy
R
xy
R
xx
R
yy
−
G
xx
=
;
G
xy
=
R
xy
R
xy
−
−
From symmetry, we expect
R
xx
R
yy
, leading to the apparent para-
dox that the longitudinal conductance,
G
xx
=
=
0, at the same time as
0.
5.6 Consider a two-dimensional structure for which the conduction
(valence) band density of states per unit area,
g
c
(
v
)
(
the longitudinal resistance,
R
xx
=
m
c
(
v
)
/π
2
,
E
)
=
with the band edges at
E
E
v
, respectively. Use eqs
(5.37) and (5.38) for the quasi-Fermi levels to show that the conduction
electron density,
n
, depends on the quasi-Fermi level energy
F
c
as
=
E
c
and
E
=
m
c
kT
π
e
(
F
c
−
E
c
)/
kT
n
=
ln
[
+
1
]
2
with an equivalent expression for the valence hole density,
p
. Invert
this expression to show that the Fermi energy varies with carrier
density
n
as
kT
ln
exp
n
1
2
m
c
kT
π
F
c
−
E
c
=
−
5.7 Consider an ideal quantum well laser for which the conduction
(valence) band density of states per unit area,
g
c
(
v
)
(
m
c
(
v
)
/π
2
,
E
)
=
with the band edges at
E
E
v
, respectively. It can be
shown that the peak gain,
g
max
, occurs at the band edge energy, and
is given by
=
E
c
and
E
=
g
max
=
G
0
(
f
c
−
f
v
)
where
f
c
and
f
v
are the values of the quasi-Fermi functions at the
conduction and valence band edges respectively and
G
0
is a constant.
