Average Energy Electron Relation to Fermi Energy

This question asks for the specific relationship between the average energy of an electron ($\bar{E}$) and the Fermi energy ($E_F$) when the temperature is absolute zero (0 Kelvin). This is a key concept in understanding electron behavior in metals within the free electron model.

Fermi Energy and Electron States at 0 K

At absolute zero, electrons occupy the lowest available energy states. The Fermi energy ($E_F$) is defined as the energy of the highest occupied state at this temperature. All states below $E_F$ are filled, and all states above $E_F$ are empty.

Derivation of Average Energy

The average energy ($\bar{E}$) is determined by integrating the energy ($E$) over all occupied states, weighted by the density of states, and dividing by the total number of occupied states. For a 3D free electron gas at absolute zero, the density of states $g(E)$ is proportional to $\sqrt{E}$.

The calculation involves the following integral relation: $ \bar{E} = \frac{\int_0^{E_F} E \cdot g(E) \, dE}{\int_0^{E_F} g(E) \, dE} $ Substituting $g(E) \propto E^{1/2}$: $ \bar{E} = \frac{\int_0^{E_F} E \cdot E^{1/2} \, dE}{\int_0^{E_F} E^{1/2} \, dE} = \frac{\int_0^{E_F} E^{3/2} \, dE}{\int_0^{E_F} E^{1/2} \, dE} $ Evaluating the integrals gives: $ \bar{E} = \frac{\left[ \frac{E^{5/2}}{5/2} \right]_0^{E_F}}{\left[ \frac{E^{3/2}}{3/2} \right]_0^{E_F}} = \frac{\frac{2}{5} E_F^{5/2}}{\frac{2}{3} E_F^{3/2}} $ Simplifying the expression yields: $ \bar{E} = \frac{3}{5} E_F $

Conclusion on Energy Relationship

The average energy of an electron ($\bar{E}$) is directly proportional to the Fermi energy ($E_F$) at absolute zero, with the constant of proportionality being 3/5.

The established relation is: $ \bar{E} = \frac{3}{5} E_{F} $