PHY.K03 Molecular and Solid State Physics Exercsie Exam
27.06.2023

1. Diffraction
A tetragonal crystal has lattice constants $a=b=14.1$ Å, $c=6.8$ Å.

(a) Calculate the primitive reciprocal lattice vectors.

(b) In an x-ray diffraction experiment, x-rays with a wavelength of 6.28 Å are used. What is the Bragg angle for the (110) reflection?

(c) The drawing below shows the Ewald sphere drawn in the (001) plane. Include the primitive reciprocal lattice vectors, the reciprocal lattice vector $\vec{G}_{110}$ and the Bragg angle into this drawing.

Solution

(a) The reciprocal lattice vectors are $\vec{b}_1=\frac{2\pi}{a}\hat{x} = 0.45$ Å-1, $\vec{b}_2=\frac{2\pi}{b}\hat{y} = 0.45$ Å-1, $\vec{b}_3=\frac{2\pi}{c}\hat{z} = 0.92$ Å-1.

(b) Many students wrote down the Bragg formula in the form $n\lambda = 2d\sin\theta$. This comes from a derivation that avoids the discussion of reciprocal lattice vectors. Here $d$ is the distance between the atomic planes and $n$ is the order of the reflection. Parallel reciprocal lattice vectors are distinguished by the order of the reflection. The reflection $\vec{G}_{100}$ corresponds to $n=1$ and $\vec{G}_{200}$ to $n=2$. Generally $\vec{G}_{nh\,nk\,nl} = n\vec{G}_{hkl}$.

A more sophisticated version of the Bragg condition is $\lambda = 2d_{hkl}\sin\theta$. Here $d_{hkl}=2\pi/|\vec{G}_{hkl}|$ is the distance between the net planes. The net planes are parallel to the atomic planes but they can be more closely spaced than the atoms. Now the order $n$ is built into $d_{hkl}$.

A third version of the same concept is called the Laue condition or the diffraction condition, $\vec{k}'-\vec{k} = \vec{G}$. This is a statement of the conservation of momentum. The conservation of energy requires that $|\vec{k}'| =|\vec{k}|$. The combination of the conservation of energy and the conservation of momentum yields, $\frac{|\vec{G}|}{2} = |\vec{k}|\sin\theta$ so

$$\frac{|\vec{G}_{hkl}|}{2|\vec{k}|}= \frac{\lambda}{2d_{hkl}}= \sin\theta.$$

In this case the Bragg angle is $\theta = 18.4$°. A drawing of the vectors and the angle in the Ewald sphere would be:

You can imagine leaving the circle where it is and rotating the blue dots around the origin at Oreziprok. $\vec{G}_{110}$ must be one step in the $\vec{b}_1$ direction and one step in the $\vec{b}_2$ from Oreziprok and it must lie on the circle. There is only one point satisfying these conditions.

See also: <a href="http://lampz.tugraz.at/~hadley/ss1/crystaldiffraction/ewald2.php">Ewald sphere</a>.

2. Density of states
The density of a material is 8.96 g/cm³. The molar mass of the particles that make up the material is 63.5 g/mol. Every particle contributes one electron to the free electron density. 1 mole = $6.023 \times 10^{23}$ particles.

(a) Calculate the electron density of this material.

(b) Use the density of states given below to determine the Fermi energy of this material.

(d) Is the assumption that the material can be described by the free electron model reasonable? Explain why or why not.

Solution

(a)

$$8.96\frac{\text{g}}{\text{cm}^3}/63.5 \frac{\text{g}}{\text{mol}} =  0.1411 \frac{\text{ mol}}{\text{cm}^3}$$

$$ 8.49 \times 10^{22} \,\frac{\text{ particles}}{\text{cm}^3}\quad\rightarrow\quad n= 8.49 \times 10^{28} \,\frac{\text{ electrons}}{\text{m}^3}$$

(b) The calculated electron density would be enough to fill about 8.5 squares of this density of states. This makes the Fermi energy $E_f = 11 \times 10^{-19}$ J. Since the Fermi energy is in the middle of a band, this material is a metal.

(c) In the free electron model, the states in reciprocal space are occupied up to the Fermi wave vector $k_F$. The volume of reciprocal space that is occupied is $4\pi k_F^3/3$. The volume of a single $k$ state is $(2\pi)^{-3}$ per unit volume and each $k$ state can hold two electrons (spin) so,

$$ n = \frac{2\cdot4\pi k_F^3}{3(2\pi)^3}\quad \rightarrow \quad k_F = (3\pi^2 n)^{1/3}.$$
$$E_F = \frac{\hbar^2k_F^2}{2m} = 11.3 \times 10^{-19}\text{ J.}$$

(d) From the density of states we have determined that this material is a metal and the Fermi energies agree quite nicely so we conclude that this material can be described by a free electron model.

However, the properties of a metal depend mostly on the states near the Fermi energy and the value of the Fermi energy is irrelevant. What is relevant is the density of states at the Fermi energy $D(E_F)$ and the derivative of the Fermi energy at the Fermi energy $D'(E_F)$. In the free electron model,

$$D(E_F) = \frac{3n}{2E_F}= 1.13\times 10^{47} \,\frac{1}{\text{J m}^3},\qquad D'(E_F) = \frac{3n}{4E_F^2}= 5.0\times 10^{64}\,\frac{1}{\text{J}^2\text{ m}^3}$$

From the give density of states,

$$D(E_F) = 1\times 10^{47} \,\frac{1}{\text{J m}^3},\qquad D'(E_F) = 0\,\frac{1}{\text{J}^2\text{ m}^3}$$

The density of states at the Fermi energy matches but the derivative does not. This means that most of the thermodynamic properties will be prediced correctly but the temperature dependence of the chemical potential will not follow the behavior predicted by the free electron model.

3. Thermal properties
The Fermi energy of a material is $E_F= 7.0$ eV. At low tempertures the specific heat of this material is linear with the temperature.

$$c_v= \frac{\pi^2}{2}\frac{k_BT}{T_F}\qquad \qquad T_F = \text{Fermi temperature}$$

(a) At 30 K the heat capacity is measured to be 15 J mol^-1 K^-1. What is the heat capacity at 15 K?

(b) Is the heat capacity dominated by the electron contribution to the heat capacity or the phonon contribution to the heat capacity? Explain your reasoning.

Constants

$$\hbar = 1.05\times 10^{-34}\text{ Js}, \quad m_e = 9.11\times 10^{-31}\text{ kg}, \quad\ e = 1.60\times 10^{-19}\text{ C}, \quad\ k_B = 1.38\times 10^{-23}\text{ J/K}$$

PHY.K03 Molecular and Solid State Physics Exercsie Exam27.06.2023

PHY.K03 Molecular and Solid State Physics Exercsie Exam
27.06.2023