Solutions to problem set 10 (141A Sp07)
1
1. (ISSP 8.1).
(a) ISSP, page 210, equation (51):
E
d
=
13
.
6
2
m
e
m
eV
=
13
.
6
18
2
0
.
015
eV
=
6
.
3
×
10

4
eV
(b) ISSP, page 210, equation (52):
a
=
0
.
53
×
m
e
/
m
Å
=
(0
.
53)18
0
.
015
Å
=
636Å
(c) The wavefunctions of the impurity electrons need to overlap if they are to form a band:
then the density needs to be at least one electron per (636Å)
3
, ie
n
=
1
(636Å)
3
=
1
.
6
×
10
7
m

3
2. (ISSP 8.2).
(a) ISSP, page 213, equation (53),
n
=
√
n
o
n
d
e

E
g
/
(2
k
B
T
)
with
n
0
=
2
mk
B
T
2
π
2
3
/
2
=
2
(0
.
01)(9
.
1
×
10

31
)(1
.
38
×
10

23
)4
2
π
(1
.
05
×
10

34
)
2
3
/
2
=
3
.
9
×
10
19
m

3
=
3
.
9
×
10
13
cm

3
and so
√
n
0
n
d
=
√
3
.
9
×
10
26
=
1
.
97
×
10
13
cm

3
.
Also, since 4
K
corresponds to 0
.
345
×
10

3
eV
, we have
n
=
(1
.
97
×
10
13
cm

3
)
e

1
/
(2(0
.
345))
=
4
.
6
×
10
12
cm

3
(b) In SI units,
R
H
=
1
/
ne
=
1
(4
.
6
×
10
18
)(1
.
6
×
10

19
)
=
1
.
3
m
3
/
C
3. (ISSP 8.3).
Recall that
j
x
=
σ
xx
E
x
+
σ
xy
E
y
j
y
=
σ
yx
E
x
+
σ
yy
E
y
Since the transverse current must be zero,
E
x
=

σ
yy
E
y
/σ
yx
, and so
j
x
=
σ
xy

σ
xx
σ
yy
σ
yx
E
y
giving
R
H
=
E
y
j
x
B
=
1
B
σ
xy

σ
xx
σ
yy
σ
yx
=
σ
yx
B
(
σ
xy
σ
yx

σ
xx
σ
yy
)
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Solutions to problem set 10 (141A Sp07)
2
Note that
σ
xy
and
σ
yx
are of order
B
, and so
σ
xy
σ
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 Spring '08
 SOUZA
 Solid State Physics, Trigraph, 4k, 2 m, 2KB, xx yy yx

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