Question

The electronic contribution of specific heat of copper at 300K is:
(Given that the fermi energy of copper is 7.05eV, and it is assumed to be temperature independent)

• A. 210 J kmol\(^{-1}\) K\(^{-1}\)
• B. 165 J kmol\(^{-1}\) K\(^{-1}\)
• C. 190 J kmol\(^{-1}\) K\(^{-1}\)
• D. 150 J kmol\(^{-1}\) K\(^{-1}\)

Hint:

For these problems, the ratio \(k_B T / E_F\) is always very small for metals at room temperature. The electronic specific heat is linear in T and much smaller than the classical value of \(\frac{3}{2}R\). Memorizing the formula \(C_{el} = (\pi^2/2)(k_BT/E_F)R\) is key.

Answer 1

The Correct Option is D

Step 1: Concept Overview:
The specific heat of a metal is influenced by lattice vibrations (phonons) and conduction electrons. The electronic contribution (\(C_{el}\)) is most noticeable at very low temperatures but can be determined at any temperature using the free electron model. This contribution is due to the thermal excitation of electrons near the Fermi level, within an energy range of approximately \(k_B T\).
Step 2: Formula:
The molar electronic specific heat is calculated as:
\[ C_{el} = \frac{\pi^2}{2} \left( \frac{k_B T}{E_F} \right) R \]
Where:

\(R\) is the universal gas constant (\(8.314 \text{ J mol}^{-1} \text{ K}^{-1}\))
\(k_B\) is the Boltzmann constant (\(1.38 \times 10^{-23} \text{ J/K}\) or \(8.617 \times 10^{-5} \text{ eV/K}\))
\(T\) is the absolute temperature (300 K)
\(E_F\) is the Fermi energy (7.05 eV)

Step 3: Calculation:
First, calculate the thermal energy \(k_B T\) in eV:
\[ k_B T = (8.617 \times 10^{-5} \text{ eV/K}) \times (300 \text{ K}) = 0.02585 \text{ eV} \]
Substitute the values into the molar specific heat formula:
\[ C_{el} = \frac{\pi^2}{2} \left( \frac{0.02585 \text{ eV}}{7.05 \text{ eV}} \right) (8.314 \text{ J mol}^{-1} \text{ K}^{-1}) \]
\[ C_{el} = \frac{9.8696}{2} \times (0.003667) \times (8.314 \text{ J mol}^{-1} \text{ K}^{-1}) \]
\[ C_{el} = (4.9348) \times (0.003667) \times (8.314 \text{ J mol}^{-1} \text{ K}^{-1}) \]
\[ C_{el} \approx 0.1504 \text{ J mol}^{-1} \text{ K}^{-1} \]
Convert the result to J kmol\(^{-1}\) K\(^{-1}\) by multiplying by 1000:
\[ C_{el} = 0.1504 \times 1000 = 150.4 \text{ J kmol}^{-1} \text{ K}^{-1} \]
Step 4: Answer:
The calculated electronic specific heat is approximately 150 J kmol\(^{-1}\) K\(^{-1}\).