Thermal Properties of Solids Questions and Answers

Answer: A

The low-temperature specific heat of a metal is \(\alpha T + \beta T^{3}\).

Two contributions add. The conduction electrons give a term linear in \(T\), because only the electrons within about \(k_{B}T\) of the Fermi level can be excited. The lattice vibrations give the Debye \(T^{3}\) law at temperatures well below the Debye temperature. At the lowest temperatures the linear electronic term dominates, and a plot of \(C/T\) against \(T^{2}\) is a straight line.

\(\beta T^{3}\) alone describes an insulator, with no free electrons. \(\exp(-\alpha/T)\) is the behaviour of a system with an energy gap, such as the Einstein model or a superconductor. There is no \(T^{5}\) term in the specific heat; a \(T^{5}\) law belongs to the electron-phonon resistivity.

Answer: B

The plot is a straight line with a positive intercept, graph (B).

At low temperature the heat capacity of a non-magnetic metal is the sum of an electronic term, linear in \(T\), and the Debye phonon term, proportional to \(T^{3}\):

\[ \begin{aligned} C &= \gamma T + \beta T^{3} \\ \frac{C}{T} &= \gamma + \beta T^{2} \end{aligned} \]

Against \(T^{2}\) this is a straight line of slope \(\beta\) (from the phonons) cutting the axis at \(\gamma\) (from the conduction electrons). This is how the two contributions are separated experimentally.

A line through the origin (D) would describe an insulator with no electronic term. The two curved graphs do not follow from either term.

Answer: C

The entropy is \(\alpha T + \frac{\beta T^{3}}{3}\).

At constant volume \(dS = C_{V}\,dT/T\). Integrating from absolute zero, where S = 0 by the third law:

\[ \begin{aligned} S &= \int_{0}^{T}\frac{\alpha T' + \beta T'^{3}}{T'}\,dT' \\ &= \int_{0}^{T}(\alpha + \beta T'^{2})\,dT' \\ &= \alpha T + \frac{\beta T^{3}}{3} \end{aligned} \]

The \(\alpha T\) term is the electronic contribution and the \(T^{3}\) term the Debye lattice contribution. The options \(\alpha + 3\beta T^{2}\) and \(\alpha T + 3\beta T^{2}\) differentiate instead of integrating, and \(\alpha + \beta T^{3}/3\) does not vanish at T = 0.

Answer: B

The molar heat capacity is 0.175 Jkmol\(^{-1}\)K\(^{-1}\).

At 10 K, far below the Debye temperature, the Debye \(T^{3}\) law applies:

\[ \begin{aligned} C &= \frac{12\pi^{4}}{5}R\left(\frac{T}{\theta_{D}}\right)^{3} \\ &= 1944 \times \left(\frac{10}{2230}\right)^{3} \\ &= 1944 \times 9.02 \times 10^{-8} = 1.75 \times 10^{-4} \end{aligned} \]

in J mol\(^{-1}\) K\(^{-1}\), which is 0.175 J per kilomole per kelvin, the unit used in the options. The tiny value shows how completely the vibrations of stiff diamond are frozen out at low temperature; the other options do not follow from the \(T^{3}\) law with these numbers.

Answer: B

The Debye temperature is about 240 K.

First the number of atoms per unit volume, then the Debye wave number \(k_{D} = (6\pi^{2}n)^{1/3}\), and \(\theta_{D} = \hbar v k_{D}/k_{B}\):

\[ \begin{aligned} n &= \frac{1.9 \times 10^{4} \times 6.02 \times 10^{23}}{0.197} \\ &= 5.8 \times 10^{28}\ \text{m}^{-3} \\ k_{D} &= (6\pi^{2}n)^{1/3} = 1.51 \times 10^{10}\ \text{m}^{-1} \\ \frac{\hbar v}{k_{B}} &= \frac{1.055 \times 10^{-34}}{1.38 \times 10^{-23}} \times 2100 \\ &= 1.60 \times 10^{-8}\ \text{K m} \\ \theta_{D} &= 1.60 \times 10^{-8} \times 1.51 \times 10^{10} \\ &\approx 242\ \text{K} \end{aligned} \]

The atomic weight 197 must be converted to 0.197 kg/mol. The other values do not follow from this sound speed; the tabulated Debye temperature of gold, about 165 K, is lower because a single sound speed overestimates it.

Answer: B

The plot of \(C_{v}/T\) against T is quadratic in T.

At low temperatures the electronic heat capacity of a metal is linear in T (Sommerfeld) and the lattice heat capacity follows the Debye \(T^{3}\) law:

\[ \begin{aligned} C_{v} &= \gamma T + AT^{3} \\ \frac{C_{v}}{T} &= \gamma + AT^{2} \end{aligned} \]

So \(C_{v}/T\) is a quadratic function of T, a straight line when plotted against \(T^{2}\), with intercept \(\gamma\) and slope A. This is the standard way of separating the two contributions experimentally.

'Linear in T' would hold only if one plotted against \(T^{2}\); 'cubic' is the lattice term in \(C_{v}\) itself, before dividing by T, and nothing in it is inversely proportional to T.

Answer: A

The ratio is about 3.75 × 10^-8 (option A, the only one of the right order).

The equilibrium vacancy fraction is \(n/N = \exp(-H_{v}/RT)\). Quenching freezes in the 800 K concentration, so the ratio is:

\[ \begin{aligned} \frac{n_{300}}{n_{800}} &= \exp\left[-\frac{H_{v}}{R}\left(\frac{1}{300} - \frac{1}{800}\right)\right] \\ \frac{H_{v}}{R} &= \frac{68000}{8.314} = 8179\ \text{K} \\ \text{ratio} &= e^{-8179 \times 0.002083} \\ &= e^{-17.04} \approx 4 \times 10^{-8} \end{aligned} \]

This works out at about 4.0 × 10^-8; the keyed 3.75 × 10^-8 is the only option of that order, the small difference coming from rounding in the constants. The options of order 10^-10 would need an exponent near -22, which these data do not give.