Surface Physics - Exercise II Model Solutions

Due: 20.1.2014
Surface Physics - Exercise II
Model Solutions
Problem V: Calculate the nearest neighbour distances and step heights for the
following crystal facets:
a) Ag(111) (bulk lattice constant 409 pm, FCC crystal structure)
b) Fe(111) (bulk lattice constant 287 pm, BCC crystal structure)
c) Co(0001) (bulk lattice constants: a=b=251 pm, c=407 pm, HCP crystal
structure)
Solution:
a) FCC crystal: Any atom in the FCC(111) plane has six nearest neighbours seperated from it by
√
2
a = 289pm
dN N =
2
where a is the lattice constant of the FCC structure. To calculate the step
height, note that the two (111) planes in figure 1 cut the space diagonal
into three equal parts. From there, it is straightforward to see that
√
3
zSH =
a = 236pm
3
.
b) BCC crystal: The nearest neighbour distance here is
√
dN N = 2a = 405pm
The (111) planes now cut the diagonal of the cube into six equal parts,
taking into account the planes formed by the cube centers in the neighbouring unit cells, making the step height:
√
3
zSH =
a = 83pm
6
c) The nearest neigbour distance is simply the length a=b=251 pm, while
the step height is c/2 = 204pm.
1
For a more thorough derivation of the step heights, see the solution to exercise
VII below.
Problem VI: Draw the structure and unit vectors of the following facets of a
primitive cubic lattice:
a) (133)
b) (33¯
1)
c) (113)
Solution: For easier viusalisation, see e.g.:
http://www.doitpoms.ac.uk/tlplib/miller indices/printall.php
The (133) and (33¯
1)) planes have the same structure and spacing, while the
angles for the (113) are different.
Problem VII: Construct the reciprocal lattices of a FCC and a BCC crystal
structure. How do they relate to each other?
Solution: We start with the FCC case and first need to find the primitve unit
vectors of the FCC structure. These are the vectors pointing from the origin,
taken at a corner of the cubic unit cell, to the centers of the three faces that
include it. This gives the following set of primitive unit vectors (coordinates are
given relative to the cubic unit cell):


 
 
1
0
1
1
1
1
~a1 = a  1  ~a2 = a  1  ~a3 = a  0 
2
2
2
0
1
1
2
Where a is the period of the cubic unit cell. The reciprocal lattice vectors are
defined as:
~bl = 2π
~am × ~an
|~al (~am × ~an )|
(1)
Plugging the unit vectors into this equation and computing the cross products
leads to the following vectors in reciprocal space:


1
2π
~b1 =
 1 
a
−1




−1
1
2π
2π
 1  ~b3 =
 −1 
b~2 =
a
a
1
1
Now, let’s look at the BCC case. Here, too, we need to find a primitive unit cell
before we can get the reciprocal lattice of the BCC structure, which in this case
are three vectors pointing from the corner of a cubic unit cell to three space
centers around it, e.g.:






1
1
−1
1
1
1 
1  ~a2 = a  −1  ~a3 = a  1 
~a1 = a
2
2
2
−1
1
1
Plugging this set of vectors into equation (1) yields:






−1
−1
0
2π
2π
2π
~b1 =
 −1  ~b2 =
 0  ~b3 =
 −1 
a
a
a
0
−1
−1
As one can see by comparing the sets of vectors with one another, the reciprocal
lattice of the FCC structure is a BCC lattice and vice-versa (though the lattice
constants in reciprocal space are obviously different).
We can also use this information to calculate the distance between the crystal
planes from the reciprocal lattice vectors. The plane separation for a set of
planes described by the miller indices (hkl) can be shown to be
dhkl =
2π
~
|G|
~ is a vector of the reciprocal lattice given by
where G
3
~ = h~b1 + k~b2 + l~b3
G
Now, try to visualise the FCC and BCC lattices with their primitive unit vectors
as we have defined them above. In both cases, the (111) planes are equivalent
in both of the structures! This is very easy in canse of the FCC structure where
one simply ends up in three corners of the cubic unit cell by walking two steps
along the primitive unit vectors, which are the three points defining the (111)
plane: the (111) planes in both basis sets are crystallographically equivalent!
Now, we can calculate the plane separation as described above:
CC
dF
111
=
=
=
dBCC
111
=
=
=
1
a
2π 2π ~bF CC + ~bF CC + ~bF CC 3
2
1
a
1 1 1 a
a
√
=√
2
2
2
3
1 +1 +1
a
1
2π 2π ~bBCC + ~bBCC + ~bBCC 1
2
3
a
−2 −2 −2 a
a
a
p
=√ = √
2
2
2
12
2 3
(−2) + (−2) + (−2)
Problem VIII: Draw sketches of the following adsorption structures and convert them from Woods notation to matrix notation or vice-versa:
√ √
a) (100)c(2 2x 2)R45
3 −1
b)
on a FCC (111) surface
−1 1
c)
1
0
√0
2
on a BCC (110) surface
Solution:
Sketches of the adsorption structures can be found in the figure below. The
converted structure notations are:
4
a)
3
2
−1
1
2
1
or
3
2
1
2
1
2
3
2
b) c(2x2) or (2x1) depending on the choice of unit vectors
c) This structure
√ is not commensurate with the surface due to the irrational
spacing of 2. Since in this case, the unit cell vectors of the adsorbate
structure are colinear with the
√ ones of the substrate, the structure can
still be written down as (1x 2). Should this not be the case, i.e. if
the unit vectors of the adstructure have irrational values AND are rotate
with respect to the substrate, the repeat distance becomes infinite and the
structure can no longer be described by the Wood’s notation, though a
matrix notation description does still exist. In this sense, the matrix notation is more powerful than Wood’s notation as it can describe arbitrary
periodic surface structures, while the advantage of Wood’s notatio lies in
the more intuitive visualisation of the structures.
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