---
title: "Al-Cu thermocouple potential difference"
subject: "Electrical Engineering Material"
chapter: 2
key: "semester_3/electrical_engineering_material:al-cu-thermocouple"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/al-cu-thermocouple
site: BCT Engineering
---

# Al-Cu thermocouple potential difference

Electrical Engineering Material · Chapter 2: Free Electron Theory of Conduction in Metal

- Asked in: 2073 Chaitra, 4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/al-cu-thermocouple
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-2

## Question

Consider a Al-Cu thermocouple pair, Estimate the potential difference available from this thermo-couple if one junctions is held at 0°C and other at 100°C.

| Metal | Fermi Energy, EF (eV) | Constant (x) |
|---|---|---|
| Al | 11.6 | 2.78 |
| Cu | 7.0 | -1.79 |

## Short answer

Use V_{AB} = −(π²k²/6e)[x_A/E_{FA} − x_B/E_{FB}](T² − T₀²) with T₀=273 K, T=373 K and given data. |V_{AB}| ≈ 0.39 mV (Al negative relative to Cu).

## Answer

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).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The free-electron thermocouple emf has the form
   A. V = -(π^2k^2/6e)[x_A E_FA - x_B E_FB](T - T0)
   B. V = -(π^2k^2/6e)[x_A/E_FA - x_B/E_FB](T^2 - T0^2)
   C. V = (kT/e) ln(E_FA/E_FB)
   D. V = -(π^2k^2/6e)[x_A/E_FA + x_B/E_FB](T^2 - T0^2)
   Answer: B. Mott-Jones result, difference of x/E_F for the two metals and difference of T squared. (Key point: Formula)

2. Cu has x = -1.79, so the bracket is
   A. 2.78/11.6 + 1.79/7.0 = 0.495
   B. 2.78/11.6 - 1.79/7.0 = -0.016
   C. 2.78/11.6 = 0.24
   D. 0.495 with E_F in eV changed to J
   Answer: A. Subtracting a negative term adds. (Key point: Sign of Cu)

3. T^2 - T0^2 for 0°C and 100°C is
   A. 10,000 K^2
   B. 100 K^2
   C. 373^2 + 273^2
   D. about 64,630 K^2
   Answer: D. 373.15^2 - 273.15^2 = 64,630. (Key point: Temperature term)

4. The emf magnitude is about
   A. 39 mV
   B. 3.9 µV
   C. 0.39 mV
   D. 0.39 V
   Answer: C. Calculation gives 3.91e-4 V. (Key point: Result)

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