# 2.11: Questions

- Page ID
- 9395

1. Which of these properties of a crystal may be anisotropic?

a. Density

b. Young's modulus

c. Surface energy

d. Refractive index

e. Electrical conductivity

f. Thermal conductivity

g. Heat Capacity

h. Melting point

i. Coefficient of thermal expansion

**Answer**-
a. No. Density is a

*scalar*property so cannot be anisotropic.b. Yes. Young's modulus relates

*vector*quantities so can be anisotropic.c. Yes. Surface energy relates

*vector*quantities so can be anisotropic.d. Yes. Refractive index relates

*vector*quantities so can be anisotropic.e. Yes. Electrical conductivity relates

*vector*quantities so can be anisotropic.f. Yes. Thermal conductivity relates

*vector*quantities so can be anisotropic.g. No. Thermal conductivity relates

*vector*quantities so can be anisotropic.h. No. Melting point is a

*scalar*property so cannot be anisotropic.i. Yes. Coefficient of thermal expansion relates

*vector*quantities so can be anisotropic.

2. Two similar transparent uniaxial crystals show the same (principal) extraordinary refractive index, *n*_{e}. However one is optically positive, and the other is optically negative. In which will the light travel faster along the optic axis?

**Answer**-
In the optically positive crystal. In an optically negative material, the extraordinary refractive index is smaller than the ordinary refractive index. The light will therefore experience a larger (ordinary) refractive index than in the optically positive material. A large refractive index implies a slower propagation of light through the medium, so when travelling parallel to the optic axis, the light will travel slower than if the crystal was optically positive.

3. Which of these could not induce anisotropy in an initially isotropic material?

a. Application of a stress

b. Application of an electric field

c. Application of a magnetic field

d. Application of a high temperature

**Answer**-
A is incorrect. For example, in polymers, a stress can align the chains along a common direction, creating anisotropy. This may lead to optical birefringence .

B is incorrect. An electric field can create an overall electric polarisation direction in a material. This may lead to the electro-optic effect.

C is incorrect. A magnetic field can create an overall magnetic dipole direction in a material.

D is correct. Isotropy can be induced in a number of different ways. Most obviously, if the sample is heated through the melting temperature, it will form an isotropic liquid, but isotropy can also be induced in the solid state. For example, in polymers, heating can allow chains to resort to a coiled state, destroying any common alignment in the bulk polymer, and in perovskites such as barium titanate, a phase transition to an isotropic cubic form occurs at high temperature (phase transitions due to an increase in temperature commonly involve an increase in symmetry).

4. Below 0°C a particular material has a crystal structure that gives rise to anisotropic thermal conductivity. At room temperature the thermal conductivity of a sample of this material is found to be isotropic. In what circumstances would the following hypotheses explain this observation?

a. The sample is polycrystalline

b. The sample has undergone a phase transition when brought up to room temperature

c. The principle values of thermal conductivity have changed with temperature

**Answer**-
A. In this case the thermal conductivity is not truly isotropic. If the sample is polycrystalline and the crystallites (grains) are randomly oriented the anisotropy will be averaged out. (Note: in many polycrystalline samples the grains are not randomly oriented but have preferred orientations or "texture".)

B. If the transition is to a cubic crystal structure the material will then display true isotropic thermal conductivity.

C. In this case the thermal conductivity is not truly isotropic. If the changes of the principal values with temperature are such as make them equal within the accuracy of the experimental technique at room temperature the thermal conductivity will appear to be isotropic.

5. Which of these materials will show isotropy in its mechanical properties?

a. Wood

b. Carbon fibre reinforced polymer

c. Window glass

d. Extruded polyethylene

**Answer**- C

6. An olivine crystal has the following diffusion constants for Ni at a certain temperature:

\[\begin{aligned}

&D_{x}=6 \times 10^{-8} \mathrm{m}^{2} \mathrm{s}^{-1}\\

&D_{y}=4 \times 10^{-18} \mathrm{m}^{2} \mathrm{s}^{-1}\\

&D_{z}=120 \times 10^{-18} \mathrm{m}^{2} \mathrm{s}^{-1}

\end{aligned}\]

The unit cell dimensions are as follows:

\[\begin{array}{l}

a=0.5 \mathrm{nm} \\

b=1.0 \mathrm{nm} \\

c=0.6 \mathrm{nm}

\end{array}\]

By considering the representation surface, what will the diffusion constant be when the concentration gradient lies along the [101] direction?

**Answer**-
Using simple geometry, the [101] direction lies 50.19° from the x-axis (to 2 decimal places). Since [101] lies in the

*x*-*z*plane, a simplified version of the equation of the ellipsoid representation surface can be used:\[\frac{x^{2}}{A^{2}}+\frac{z^{2}}{C^{2}}=1\]

where A and C are equal to \(\frac{1}{\sqrt{D}}\) in the

*x*and*z*directions respectively.Hence

*x*^{2}*D*+_{x}*z*^{2}*D*=1_{z}Letting the length of the radius in the [101] direction be

*r*, the equation becomes:\[r^{2} \cos ^{2} 50.19 D_{x}+r^{2} \sin ^{2} 50.19 D_{z}=1\]

Since

*r*will be equal to \(\frac{1}{\sqrt{D}}\) in the [101] direction, substituting in the given values gives a diffusion constant of 73.3 x 10^{-18}m^{2}s^{-1}.

7. A certain orthorhombic crystal has the following principal values of thermal conductivity:

\[\begin{array}{l}

k_{x}=6.25 \mathrm{Wm}^{-1} \mathrm{K}^{-1} \\

k_{\mathrm{y}}=1.00 \mathrm{Wm}^{-1} \mathrm{K}^{-1} \\

k_{z}=1.75 \mathrm{Wm}^{-1} \mathrm{K}^{-1}

\end{array}\]

**Answer**-
By simple geometry, the lengths of the [110] and the [111] vectors can be calculated. These are 1 nm and \(\sqrt{2}\) nm respectively, and using these values, the direction cosines can be obtained. Using equations and notation from the TLP, the direction cosines are found to be:

\[I=\frac{x}{r}=\frac{0.8}{\sqrt{2}}, m=\frac{y}{r}=\frac{0.6}{\sqrt{2}}, n=\frac{z}{r}=\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}}\]

The thermal conductivity can be calculated using:

\[\begin{aligned}

&k=k_{\mathrm{x}} R+k_{\mathrm{y}} m^{2}+k_{\mathrm{z}} n 2\\

&k=6.25\left(\frac{4}{5 \sqrt{2}}\right)^{2}+1.00\left(\frac{3}{5 \sqrt{2}}\right)^{2}+1.75\left(\frac{1}{\sqrt{2}}\right)^{2}\\

&k=2+0.18+0.875=3.055\\

&k=3.06 \mathrm{Wm}^{-1} \mathrm{K}^{-1}(3 \mathrm{s} . \mathrm{f})

\end{aligned}\]

8. Explain how anisotropy is involved in the operation of the following devices:

- a liquid crystal display
- a fuel cell that uses a sold state electrolyte
- a pyroelectric intruder alarm

9. Suggest ways of distinguishing between the answers to Question 4.