PHY 472 Midterm
1
1. The building block of a
lattice is shown in the Figure. All
angles are 90o, the sides are 0.8nm, 0.6nm and 0.2nm, and the
lattice points are at the corners and at the two face centers, as
indicated. The lattice is obtained by
periodic repetition of this unit.
a./ Indicate on the Figure
three possible choices of reciprocal lattice vectors.
b./ What is the volume of the primitive unit
cell?
c./ What are the lengths of the
three reciprocal lattice vectors?
d./ Powder x-ray diffraction is
performed on a sample that has this lattice.
The wavelength of the X-rays is 0.2nm.
What is the lowest angle 2q for the scattered X rays?
Solution:
a./ The a and b lattice vectors
can be chosen as indicated in the Figure.
The c vector is along the
“vertical” edge of the unit. Many other
choices are possible.
b./ In order to determine the
volume, first we need the magnitude of the shaded area in the Figure. It is quite clear that this are is half of
the area of the rectangle: A=(1/2) (0.8nm)(0.6nm)=0.24nm2. The volume is
therefore V=(0.24nm2)(0.2nm)=0.048nm3.
c./ The magnitude of the cross
product axb also relates to the shaded
area: |axb|=0.24nm2. Since the length of the a vector is 0.5nm, other cross product is |axc|=0.10nm2;
and the third one is |bxc|=0.12nm2. The lengths of the corresponding reciprocal
lattice vectors are: 31.4nm-1, 13.1nm-1and 15.7nm-1.
d./ The scattering condition is DK=G., where |DK|=2ksinq. The lowest angle belongs to
the shortest possible G vector, that is to |G|=
13.1nm-1. With k= 2p/(0.2nm)=6.98nm-1 we get q=12.03o and 2q=24.06o .
2. Consider a two-dimensional square lattice of
lattice spacing a, and linear size L (area=L2). The sound
velocity for longitudinal waves is cL, and for transverse waves it is cT
.Assume the sound velocity is isotropic within the plane. Derive an expression for the density of
states D(w) that works for low frequencies. Make sure that the D(w)
is expressed in terms of a, L, cL, and cT
. What is the condition for this low
frequency approximation to be valid? (In
terms of w
being much less than what?)
Solution:
Consider one of the modes. On the two dimensional kx, ky
plane the number of phonon states within a ring of radius k and width dk
is dn=1/(2pL)2
2pk dk. Since
dk=dw(dw/dk)-1=dw/c, we get D(w)=dn/dw=(L2/2p)k/c=(L2/2p) w/c2.
This expression is valid as long as the wavenumber k is
not close to the Brillouin zone boundary, k<<p/a. In terms of w, we get w<<pc/a.
With two modes we have to add up the number of
states, and we get:
D(w)= (L2/2p) w2
(1/ cL2 +1/cT2). The condition of validity is w<<pc/a, with c being the lower of the two velocities.
3. The
specific heat of metals is dominated by the electronic contribution at low
temperatures, and by phonons at high temperatures. At what temperature are the two contributions
equal in rubidium? Use the textbook as a resource for the
necessary data. Shortly describe your
thinking.
Solution:
The phonon contribution is C=234NkB(T/Q)3. The electron contribution is C=gT. For rubidium Q=56K
(Tabe 1, page 126), and g=2.41 mJ/(mole
234 (6x1023)(1.38x10-23)J/K
T2 /(56K)3=2.41x10-3J/
T=0.47K