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: Understanding the Concept:
The specific heat of a metal has contributions from both lattice vibrations (phonons) and the conduction electrons. The electronic contribution (\(C_{el}\)) is significant only at very low temperatures but can be calculated for any temperature using the free electron model. It arises because only electrons within an energy range of about \(k_B T\) of the Fermi level can be thermally excited.
Step 2: Key Formula or Approach:
The molar electronic specific heat is given by the formula:
\[ C_{el} = \frac{\pi^2}{2} \left( \frac{k_B T}{E_F} \right) R \]
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: Detailed Explanation:
First, it is convenient to calculate the thermal energy \(k_B T\) in eV.
\[ k_B T = (8.617 \times 10^{-5} \text{ eV/K}) \times (300 \text{ K}) = 0.02585 \text{ eV} \]
Now, substitute the values into the formula for molar specific heat:
\[ C_{el} = \frac{\pi^2}{2} \left( \frac{0.02585 \text{ eV}}{7.05 \text{ eV}} \right) (8.314 \text{ J mol}^{-1} \text{ K}^{-1}) \]
\[ C_{el} = \frac{9.8696}{2} \times (0.003667) \times (8.314 \text{ J mol}^{-1} \text{ K}^{-1}) \]
\[ C_{el} = (4.9348) \times (0.003667) \times (8.314 \text{ J mol}^{-1} \text{ K}^{-1}) \]
\[ C_{el} \approx 0.1504 \text{ J mol}^{-1} \text{ K}^{-1} \]
The question asks for the value in units of J kmol\(^{-1}\) K\(^{-1}\). To convert from mol to kmol, we multiply by 1000.
\[ C_{el} = 0.1504 \times 1000 = 150.4 \text{ J kmol}^{-1} \text{ K}^{-1} \]
Step 4: Final Answer:
The calculated value is approximately 150 J kmol\(^{-1}\) K\(^{-1}\).