Hexagonal emit-packed structure Consider first Brillouin zone of a crystal with a simple hexagonal lattice in three dimensions with lattice constants a and c. Let Gc denote the shortest reciprocal lattice vector parallel to the c axis of the crystal lattice.
(a) Show that for a hexagonal-close-packed crystal structure the Fourier component U(Gc) of the crystal potential U(r) is zero.
(b) Is U (2Gc) also zero?
(c) Why is it possible in principle to obtain an insulator made up of divalent atoms at tile lattice points of a simple hexagonal lattice?
(d) Why is it nut possible to obtain an insulator made up of monovalent atoms in a hexagonal-close-packed structure?
The Correct Answer and Explanation is :
(a) Fourier Component ( U(G_c) ) of the Crystal Potential ( U(r) ):
The Fourier transform of the crystal potential ( U(r) ) is given by:
[
U(\mathbf{G}) = \int e^{-i\mathbf{G} \cdot \mathbf{r}} U(\mathbf{r}) d\mathbf{r}
]
where ( \mathbf{G} ) is a reciprocal lattice vector. For a hexagonal-close-packed (hcp) structure, the reciprocal lattice vectors along the c-axis, denoted ( G_c ), are given by:
[
G_c = \left( 0, 0, \frac{2\pi}{c} \right)
]
The potential ( U(r) ) in a crystal is periodic, so if the potential does not vary in the direction of the c-axis, the Fourier component ( U(G_c) ) will be zero. In particular, for the hcp structure, the potential has a symmetry where the lattice points are regularly spaced along the c-axis, but there is no variation of the potential in this direction. Hence:
[
U(G_c) = 0
]
This result is because the integral of ( U(r) ) multiplied by the complex exponential ( e^{-i G_c \cdot r} ), which corresponds to the periodic lattice structure along the c-axis, vanishes due to the symmetry and periodicity of the lattice potential.
(b) Is ( U(2G_c) ) also zero?
The Fourier component at ( 2G_c ) involves a reciprocal lattice vector that corresponds to twice the frequency along the c-axis. Since the crystal potential ( U(r) ) is periodic, this higher-order Fourier component will also be zero. This can be explained by the fact that higher harmonics of the reciprocal lattice vectors for a periodic potential also lead to vanishing Fourier components due to the periodicity of the lattice in the c-direction.
Thus, ( U(2G_c) = 0 ) for the same reason as ( U(G_c) = 0 ), namely that the potential is constant along the c-axis in the hexagonal-close-packed structure.
(c) Insulator Made of Divalent Atoms at the Lattice Points of a Simple Hexagonal Lattice:
In principle, it is possible to obtain an insulator made of divalent atoms in a simple hexagonal lattice because divalent atoms have two valence electrons, which can form a filled bonding band, leading to a completely occupied valence band. This results in an insulating electronic structure because all the electrons are paired and there are no available states for conduction. A simple hexagonal lattice can have an electronic band structure that supports such insulating behavior, especially if the band gap is large enough to prevent electron movement.
(d) Insulator Made of Monovalent Atoms in a Hexagonal-Close-Packed Structure:
For monovalent atoms, each atom contributes only one electron. In a hexagonal-close-packed (hcp) structure, the electronic bands formed by monovalent atoms are typically only partially filled. In this case, the conduction band will have partially occupied states, allowing for electrical conductivity. This makes it impossible for an hcp structure made of monovalent atoms to become an insulator under normal conditions, as the partially filled bands provide free electrons that can conduct electricity.
Therefore, the primary difference lies in the number of electrons contributed by the atoms in the lattice. Divalent atoms can form a fully filled valence band that prevents electrical conductivity, while monovalent atoms cannot achieve such a configuration, leading to metallic or conducting behavior.