PHYS 6210: Condensed Matter Phyiscs I - HMW #5
Jiang Xiao
October 27, 2004
Problem 1:
the basis for diamond structure is (0, 0, 0), a (1, 1, 1), and the lattice structure is fcc, let the
4
potential created by each atom is Ua (r), thus
2
U (r)ei a (2,0,0)
PHYS 6210: Condensed Matter Phyiscs I - HMW #5
Jiang Xiao
October 27, 2004
Problem 1:
the basis for diamond structure is (0, 0, 0), a (1, 1, 1), and the lattice structure is fcc, let the
4
potential created by each atom is Ua (r), thus
2
U (r)ei a (2,0,0)
PHYS 6210: Condensed Matter Phyiscs I - HMW #3
Jiang Xiao
September 18, 2004
Problem 1(A&M 6.1):
(a). from lecture notes, we have for cubic
Table 1: sin2 /2
A
B
C
0.129598 0.0618467 0.133135
0.17329 0.122645 0.355484
0.345492 0.183985 0.491274
0.476447 0.
PHYS 6210: Condensed Matter Phyiscs I - HMW #3
Jiang Xiao
September 18, 2004
Problem 1(A&M 6.1):
(a). from lecture notes, we have for cubic
Table 1: sin2 /2
A
B
C
0.129598 0.0618467 0.133135
0.17329 0.122645 0.355484
0.345492 0.183985 0.491274
0.476447 0.
PH672 Problem Set #4
WINTER 2005
1.
Marder, Chapter 17, problem #6. Hints: Part (b): Assume that the Joule heat generated in a short length of an element is conducted away along the wire. Part (d): The quantity R in Eq. (17.182) and in (17.81) as well sho
PH672 Problem Set #4
WINTER 2005
1.
Marder, Chapter 17, problem #6. Hints: Part (b): Assume that the Joule heat generated in a short length of an element is conducted away along the wire. Part (d): The quantity R in Eq. (17.182) and in (17.81) as well sho
Solutions for Homework Set 1
1. The face-centered cubic is the most dense and the simple cubic is the least dense of the three cubic Bravais
lattices. The diamond structure is less dense than any of these. One measure of this is the packing fraction for
t
Solutions for Homework Set 1
1. The face-centered cubic is the most dense and the simple cubic is the least dense of the three cubic Bravais
lattices. The diamond structure is less dense than any of these. One measure of this is the packing fraction for
t
PHYS 6210: Condensed Matter Phyiscs I - HMW #4
Jiang Xiao
October 9, 2004
Problem 1:
(a). use i = 1, 2 to denote the region [a, 0] and [0, a], and assume the state in these two
regions are(k0 = 2m/ 2 )
i = Ai ek0 x + Bi ek0 x
then we have
i = k0 (Ai ek0 x
PHYS 6210: Condensed Matter Phyiscs I - HMW #4
Jiang Xiao
October 9, 2004
Problem 1:
(a). use i = 1, 2 to denote the region [a, 0] and [0, a], and assume the state in these two
regions are(k0 = 2m/ 2 )
i = Ai ek0 x + Bi ek0 x
then we have
i = k0 (Ai ek0 x
PHYS 6210: Condensed Matter Phyiscs I - HMW #5
Jiang Xiao
October 27, 2004
Problem 1:
the basis for diamond structure is (0, 0, 0), a (1, 1, 1), and the lattice structure is fcc, let the
4
potential created by each atom is Ua (r), thus
2
U (r)ei a (2,0,0)
PHYS 6210: Condensed Matter Phyiscs I - HMW #5
Jiang Xiao
October 27, 2004
Problem 1:
the basis for diamond structure is (0, 0, 0), a (1, 1, 1), and the lattice structure is fcc, let the
4
potential created by each atom is Ua (r), thus
2
U (r)ei a (2,0,0)
PHYS 6210: Condensed Matter Phyiscs I - HMW #3
Jiang Xiao
September 18, 2004
Problem 1(A&M 6.1):
(a). from lecture notes, we have for cubic
Table 1: sin2 /2
A
B
C
0.129598 0.0618467 0.133135
0.17329 0.122645 0.355484
0.345492 0.183985 0.491274
0.476447 0.
PHYS 6210: Condensed Matter Phyiscs I - HMW #3
Jiang Xiao
September 18, 2004
Problem 1(A&M 6.1):
(a). from lecture notes, we have for cubic
Table 1: sin2 /2
A
B
C
0.129598 0.0618467 0.133135
0.17329 0.122645 0.355484
0.345492 0.183985 0.491274
0.476447 0.
PHYS 6210: Condensed Matter Phyiscs I - HMW #7
Jiang Xiao
November 14, 2004
Problem 1:
(a). when r = 0:
Q k 0 r
2Q
e
+ k0 ek0 r
r
r
1d
d Q k 0 r
2Q
= 2
r2
+ k0 ek0 r
e
r dr
dr r
r
1d
2Q
Q(1 + k0 r )ek0 r + k0 ek0 r
=2
r dr
r
Q k 0 r
Q k 0 r
2
2
= k0 e
+ k
PHYS 6210: Condensed Matter Phyiscs I - HMW #7
Jiang Xiao
November 14, 2004
Problem 1:
(a). when r = 0:
Q k 0 r
2Q
e
+ k0 ek0 r
r
r
1d
d Q k 0 r
2Q
= 2
r2
+ k0 ek0 r
e
r dr
dr r
r
1d
2Q
Q(1 + k0 r )ek0 r + k0 ek0 r
=2
r dr
r
Q k 0 r
Q k 0 r
2
2
= k0 e
+ k
2. A&M Problem 13.5. For a metal subject simultaneously to a nonzero electric eld E
thermal gradient T 0, the heat production per unit volume dq/dt is given by
0 and nonzero
dn
dq du
=
,
dt
dt
dt
(2.1)
where u is the energy per unit volume, the chemical
2. A&M Problem 13.5. For a metal subject simultaneously to a nonzero electric eld E
thermal gradient T 0, the heat production per unit volume dq/dt is given by
0 and nonzero
dn
dq du
=
,
dt
dt
dt
(2.1)
where u is the energy per unit volume, the chemical
PHYSICS 880.06 (Fall 2004)
Problem Set 2 Solution
2.1
2.2
(a) Carrier concentration n, sign of the carrier (positive for holes or negative for electrons), and carrier collision
relaxation time .
The results on the high Tc material are anomalous in that Ha
PHYSICS 880.06 (Fall 2004)
Problem Set 2 Solution
2.1
2.2
(a) Carrier concentration n, sign of the carrier (positive for holes or negative for electrons), and carrier collision
relaxation time .
The results on the high Tc material are anomalous in that Ha