Chemistry Reference
In-Depth Information
and let
k
x
=
k
y
=
8
π/
L
. Draw out (i) for
t
=
0 and (ii) for
t
=
π/ω
,
the lines along which
1. Calculate the repeat distance of
the wave along the
x
-direction, the
y
-direction, and its direction of
motion.
ψ(
x
,
y
,
t
)
=
3.7
Show that
a
1
)
are a set of primitive (i.e. basis) vectors for the FCC lattice, mean-
ing any lattice site
R
can be generated as a linear combination of an
integer times each of these three vectors:
=
(
0,
a
/
2,
a
/
2
)
,
a
2
=
(
a
/
2, 0,
a
/
2
)
and
a
3
=
(
a
/
2,
a
/
2, 0
R
=
n
1
a
1
+
n
2
a
2
+
n
3
a
3
3.8
Each reciprocal lattice vector
G
for a given crystal lattice defines the
wavevector of a plane wave which has the same periodicity as the
lattice, that is if
R
is a vector joining the same point in two differ-
ent unit cells of the crystal lattice, then we require e
i
G
·
R
=
1, or
G
n
, where
n
is an integer. A particular crystal lattice is
defined by the three primitive vectors,
a
1
,
a
2
and
a
3
. Show that the
reciprocal lattice can then be defined in terms of the three primitive
vectors
b
1
,
b
2
and
b
3
, where
·
R
=
2
π
2
π(
a
2
×
a
3
)
2
π(
a
3
×
a
1
)
2
π(
a
1
×
a
2
)
b
1
=
;
b
2
=
;
b
3
=
a
1
·
(
a
2
×
a
3
)
a
1
·
(
a
2
×
a
3
)
a
1
·
(
a
2
×
a
3
)
3.9 Using the FCC basis vectors of problem 3.7, calculate the basis vec-
tors for the FCC reciprocal lattice, and hence show that the reciprocal
lattice of an FCC lattice is a BCC lattice (and vice versa).
3.10 A 2-D trian
gu
lar lattice is defined by the basis vectors
a
1
=
(
a
,0
)
;
2,
√
3
a
a
2
. Determine the reciprocal lattice basis vectors for
this triangular lattice, and hence show that the reciprocal lattice of
a triangular lattice is itself a triangular lattice.
=
(
a
/
/
2
)
3.11 The triangular lattice of problem 3.10 contains one atom per unit
cell, which has a single orbital with self-energy
E
s
. Show that if this
orbital has an interaction
V
with each of its six neighbours,
then the triangular lattice band structure is given by
(
V
<
0
)
2
V
cos
√
3
k
y
a
E
s
k
=
E
s
+
(
k
x
a
)
+
2 cos
(
k
x
a
/
2
)
cos
(
/
2
)
Show that the lower and upper band edges are then at energies
E
sL
, respectively. Justify why the
most strongly bonding state,
E
sL
, is in this case shifted considerably
further down in energy than
E
sU
is shifted upwards.
=
E
s
−
6
|
V
|
and
E
sU
=
E
s
+
2
|
V
|
