Al-Cu thermocouple potential difference

The Seebeck (thermoelectric) emf of a thermocouple made of metals A and B, with junctions at $T_0$ and $T$, is given by the Mott-Jones free-electron result:

$$V_{AB} = -\frac{\pi^2 k^2}{6e}\left[\frac{x_A}{E_{FA}} - \frac{x_B}{E_{FB}}\right](T^2 - T_0^2)$$

where $x$ is the constant for each metal and $E_F$ its Fermi energy. (Each metal has Seebeck coefficient $S = -\frac{\pi^2k^2T}{3eE_F}x$.)

Given: A = Al: $E_F = 11.6$ eV, $x = 2.78$; B = Cu: $E_F = 7.0$ eV, $x = -1.79$; $T_0 = 273.15$ K, $T = 373.15$ K.

Constant term

$$\frac{\pi^2 k^2}{6e} = \frac{\pi^2 \times (1.381 \times 10^{-23})^2}{6 \times 1.602 \times 10^{-19}} = 1.958 \times 10^{-27}\ \text{J}^2\text{K}^{-2}\text{C}^{-1}$$

Bracket term (E_F in joules)

$$\begin{aligned} \frac{x_A}{E_{FA}} - \frac{x_B}{E_{FB}} &= \frac{1}{1.602 \times 10^{-19}}\left[\frac{2.78}{11.6} - \frac{-1.79}{7.0}\right] \\ &= \frac{0.2397 + 0.2557}{1.602 \times 10^{-19}} = \frac{0.4954}{1.602 \times 10^{-19}}\ \text{J}^{-1} \end{aligned}$$

Temperature term

$$T^2 - T_0^2 = 373.15^2 - 273.15^2 = 64\,630\ \text{K}^2$$

Emf

$$\begin{aligned} V_{AB} &= -1.958 \times 10^{-27} \times \frac{0.4954}{1.602 \times 10^{-19}} \times 64\,630 \\ &= -3.91 \times 10^{-4}\ \text{V} \end{aligned}$$

The negative sign means Al is negative with respect to Cu in this sign convention; the magnitude is what a voltmeter reads.

Answer: $|V_{AB}| \approx 0.39\ \text{mV}$ (about 391 µV, $V_{Al\text{-}Cu} = -0.391$ mV).