Lecture 4: Diffusion: Ficks second law
Todays topics
Learn how to deduce the Ficks second law, and understand the basic meaning, in
comparison to the first law.
Learn how to apply the second law in several practical cases, including homogenization,
interd
1
Physics 7440, Solutions to Problem Set # 8
1. Ashcroft & Mermin 12.2
For both parts of this problem, the constant offset of the energy, and also the location of the minimum at k0 ,
have no effect. Therefore we work with the simpler form for the energy
(
Solution manual - Chapter 4
Jens Zamanian
March 10, 2014
4.2
From Fig. 1 below we clearly see that for (n1 , n2 , n3 )
a) if the ni are either all odd or all even we get a BCC lattice and
b) if the sum of ni is even we get a FCC lattice.
Figure 1: BCC and
Solution manual - Chapters 810
Jens Zamanian
March 24, 2014
1
Chapters 8-10
Kronig-Penney model
We want to solve the Schr
odinger equation for the potential
0 0 < x < a, a + b < x < 2a + b, . . .
V (x) =
V0
b < x < 0, a < x < a + b, . . .
(1)
which can b
Solution manual - Chapter 22-23
Jens Zamanian
May 7, 2014
Chapter 22
Problem 22.2 - Diatomic Linear Chain
(a) To obtain the dispersion relation we must generalize the discussion of a lattice with a basis
to the case where the dierent atoms in the basis ha
Solution manual - Chapter 12
Jens Zamanian
March 27, 2014
Chapter 12
12.3
When the energy is in the form
Ek = E0 +
~2
(k
2
1
k0 ) M
(k
k0 ) ,
(1)
where M is independent of k we get
v(k) =
1
~2 1
rk E k =
M
~
2 ~
1
(k
k0 ) + (k
k0 )M
1
.
(2)
However, Mij1
Solution manual - Chapter 1
Jens Zamanian
March 9, 2014
1.1 Poisson Distribution
a) Divide the time t in small increments dt. For each of these increments the probability of a
collision is dt/ and hence the probability of no collision is 1 dt/ . The numbe
Solution manual - Chapter 6
Jens Zamanian
March 24, 2014
1
Chapter 6
6.2 - The FCC lattice as a simple cubic with a basis
a) The FCC lattice can be represented as a simple cubic lattice with the basis
d1 = 0,
d2 =
a
) ,
(
x+y
2
d3 =
a
) ,
(
y+z
2
d4 =
a
)