IGNOU MPH 12 2026 - In Sommerfeld free electron theory, show that at a temperature T(i)&

Question

In Sommerfeld free electron theory, show that at a temperature T
(i) $\mu = \varepsilon_F \left[ 1 - \frac{1}{3} \left( \frac{\pi k_B T}{2 \varepsilon_F} \right)^2 \right]$ and (ii) $c_v = \frac{\pi^2}{2} \left( \frac{k_B T}{\varepsilon_F} \right) n k_B$

19 Jan 2026

Answer :

Word Count : 414

In the Sommerfeld free electron theory, the electrons in a metal are treated as a degenerate Fermi gas. At finite temperature (T), the chemical potential (\mu) and electronic specific heat (c_v) can be obtained using the Fermi–Dirac distribution. The total number of electrons is: [ N = \int_0^\infty g(\varepsilon) f(\varepsilon) d\varepsilon ] where (g(\varepsilon) = \frac{3N}{2\varepsilon_F} \sqrt{\frac{\varepsilon}{\varepsilon_F}}) is the density of states and (f(\varepsilon) = \frac{1}{e^{(\varepsilon-\mu)/k_B T}+1}) is the Fermi–Dirac distribution. At low temperatures ((k_B T \ll \varepsilon_F)), we can expand integrals using the Sommerfeld expansion: [ \int_0^\infty \frac{\phi(\varepsilon)}{e^{(\varepsilon-\mu)/k_B T}+1} d\varepsilon = \int_0^\mu \phi(\varepsilon) d\varepsilon + \frac{\pi^2}{6} (k_B T)^2 \phi'(\mu) + \dots ] --- (i) Chemical potential (\mu(T)): Using (N = \int_0^\infty g(\varepsilon) f(\varepsilon) d\varepsilon) and expanding: [ N = \int_0^\mu g(\varepsilon) d\varepsilon + \frac{\pi^2}{6} (k_B T)^2 g'(\mu) ] At (T=0), (\mu = \varepsilon_F) and (N = \int_0^{\varepsilon_F} g(\varepsilon) d\varepsilon). For

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