SOLID STATE PHYSICS (3C25) Solutions to problem paper 1 1 ...

SOLID STATE PHYSICS (3C25) Solutions to problem paper 1
1. Define the terms Bravais lattice, basis and Miller indices. [3]
Bravais lattice: A set of points generated by a set of 3 non-coplanar translation
vectors.
Basis: The set of atoms associated with each lattice point.
Miller indices: The coefficients of the primitive reciprocal lattice vectors for
the shortest reciprocal lattice vector that is normal to the planes of atoms.
Take the diagram at the end of this question sheet, and draw the following
on it
(a) The primitive translation vectors [2]
(b) A primitive unit cell [2]
(c) A Wigner cell [2]
See diagram below.
How many atoms are there in the primitive unit cell? [1]
There are two atoms per unit cell: one light and one dark.
2. A body centered cubic cell has a conventional lattice constant a.
(a) How many atoms are there in the conventional cubic unit cell? [1]
There are two atoms per conventional cubic unit cell. One corner atom
plus one body centred atom.
(b) Calculate the volume per atom. [2]
The volume per unit cell is a 3 . Thus the volume per atom is a 3 /2
(c) Calculate the packing fraction. [2]
The spheres touch along the diagonal of the cube. Therefore the radius
of the sphere is a √ 3/4. The volume of the sphere is then (4π/3)(a √ 3/4) 3 .
The packing fraction is the ratio of the volume of the sphere to the volume
of the unit cell per atom, which is [(4π/3)(a √ 3/4) 3 ]/[a 3 /2] = √ 3/8π ≈
0.68
(d) Consider planes of atoms parallel to the (100) plane. What is the spacing
between planes? [2]
One type of (100) plane contains the corner atoms, while the other
contains the body centred atoms. Thus the spacing between planes is
a/2.
(e) Using the formula d = a/ √ h 2 + k 2 + l 2 , find the corresponding value of h.
[1]
(h,k,l) are the Miller indices. Since the system is cubic k=l=0 for (100)
planes. Thus d = a/h. But since d = a/2 we have that h=2.
1

(f) Why is this surprising? [2]
The body centred cubic crystal has a cubic cell for which h=1. However
the basis introduces extra planes.
3. When electrons are scattered from a crystal surface
(a) Why does the angle of incidence equal the angle of reflection? [2]
To ensure that the waves scattered off neighbouring surface atoms are
in phase.
(b) Why do you only see reflections at some angles of incidence? [2]
Because only for certain angles of incidence are the waves scattered off
neighbouring planes in phase.
(c) Why can you study only atoms near the crystal surface? [2]
Electrons cannot penetrate deep into a crystal because they are scattered
by the surface layers of atoms.
4. Consider the scattering of x-rays of wavelength λ from a face centred cubic
crystal with cubic lattice constant a.
(a) What are the allowed scattering angles when the scattering is from the
(001) planes of the crystal? (Use Bragg’s law.) [3]
2d sin θ = nλ
d = a/2
⇒ sin θ = nλ
a
(b) Using the simple cubic representation of a face centred cubic crystal,
write down the primitive translation vectors and calculate the reciprocal
lattice vectors. [3]
a = a(100) A = 2π
a (100)
b = a(010) B = 2π
a (010)
c = a(001) C = 2π
a (001)
(c) Using the Laue construction, determine the allowed scattering vectors for
scattering from (001) planes. [4]
In general, a reflection from a (001) surface implies that ∆ k is parallel
to C (the primitive reciprocal lattice vector with indices (001)). From the
Laue construction we know that scattering vectors must equal reciprocal
lattice vectors for strong reflection:
2

Thus ∆k = n C = 2πn(001),
where n is an integer.
a
(d) What is the structure factor for the face centred cubic crystal? [2]
The basis vectors di are (0, 0, 0), (a/2, a/2, 0), (a/2, 0, a/2), and (0, a/2, a/2).
Let ∆k = h A + k B + l C = 2π
a (h, k, l). Thus ∆k · di is
∆ k · (0, 0, 0) = 0
∆ k · (a/2, a/2, 0) = (h + k)π
∆ k · (a/2, 0, a/2) = (h + l)π
∆ k · (0, a/2, a/2) = (k + l)π
and the structure factor is therefore
S = f
1 + e i(h+k)π + e i(h+l)π + e i(k+l)π
(e) Which reflections are forbidden? Explain why they are forbidden. [2]
If h, k and l are all even or all odd then we have S = 4f, otherwise
we have S = 0. Thus the forbidden reflections are those which either
one or two of h, k, l is odd. The basis atoms introduce extra planes and
scattering from these planes can create destructive interference, even
though there would be constructive intereference without them.
3

Figure for question 1.
4