Advanced Solid State Physics

14.4.2 Example: Tin

A popular example for structural phase transitions is tin (Chemical symbol: Sn). This element has
two forms:

• α-Sn: Gray tin, appears as a semiconductor, diamond crystal structure, stable below 13. 2 ◦C


• β-Sn: White tin, appears as a metal, tetragonal crystal structure, stable above 13. 2 ◦C

If tin is needed in applications where the temperature is colder than the transition temperature, it
looses its good metal properties. To avoid the phase transition, some impurities have to be added to
the pure metal. To understand the process of these transitions, we have to look at the free energy:

F=U−TS (250)

The system tries to reduce its free energyFand so we can see, that the crystal structures must have
different entropies. An easy model to explain the different entropies of the crystal structures is to
think about the speed of sound, which has to be different, too. A lower speed of sound means softer
phonon modes, which leads to more possible configurations, which means a higher entropy.

In order to calculate the entropy, we have to do some steps. First we need to measure or calculate the
dispersion relationships of the two crystal structures. They are shown in fig. 130 for tin. When we
have the dispersion relationships, we can calculate the density of states, which is plotted in fig. 131.
With this we can write down the equation for the internal energy

U(ω) =

∫∞

0

︸︷︷︸~ω

Energy

D(ω)

︸︷︷︸

DoS

1

exp(k~BωT)− 1

︸ ︷︷ ︸

Probability

dω (251)

and the integral for the entropy:

S=

∫

dU|V=const

T

=

∫

cV

T

dT (252)

If this is done for both crystal structures, we should be able to predict where the structural phase
transition will happen.

In the prior calculations, we said that the crystal tries to minimize the entropy, but at low temperatures
often the enthalpyHis important for the choice of the crystal structure.

H=U+pV (253)

At normal conditions the volume of a solid doesn’t change very much, if the pressure is changed.
So usually the enthalpy is almost the same as the free energy. In many experiments the controlled
parameters are the temperature and the pressure. So we do two Legendre transformations and get
the Gibbs free energy, which we have to minimize:

G=U+pV−TS (254)