Question:medium

In 3-dimensional system, the mean energy of an electron in electron gas at absolute zero is __________________________ of fermi energy, \(E_f(0)\) at absolute zero.

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This is a standard result from the free electron model and is worth memorizing. The factor 3/5 comes directly from the integration of the 3D density of states (\(\propto E^{1/2}\)).
Updated On: Feb 18, 2026
  • \( \frac{4}{3} \)
  • \( \frac{5}{3} \)
  • \( \frac{3}{5} \)
  • \( \frac{3}{4} \)
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The Correct Option is C

Solution and Explanation

Step 1: Problem Definition:
Determine the average energy (\(\langle E \rangle\)) of an electron within a 3D free electron gas at absolute zero (T=0 K), expressed as a fraction of the Fermi energy (\(E_F\)). At T=0 K, all energy levels up to \(E_F\) are filled, while those above are empty.
Step 2: Methodology:
Calculate the average energy \(\langle E \rangle\) by integrating the product of energy \(E\) and the density of states \(D(E)\) from 0 to \(E_F\), then dividing by the total number of electrons \(N\).
\[ \langle E \rangle = \frac{1}{N} \int_0^{E_F} E \cdot D(E) \, dE \]
For a 3D free electron gas, \( D(E) = C E^{1/2} \), where \(C\) is a constant.
The total number of electrons is given by \( N = \int_0^{E_F} D(E) \, dE \).
Step 3: Solution:
First, compute the total number of electrons:
\[ N = \int_0^{E_F} C E^{1/2} \, dE = C \left[ \frac{E^{3/2}}{3/2} \right]_0^{E_F} = \frac{2}{3} C (E_F)^{3/2} \]
Next, compute the total energy:
\[ E_{\text{total}} = \int_0^{E_F} E \cdot (C E^{1/2}) \, dE = C \int_0^{E_F} E^{3/2} \, dE = C \left[ \frac{E^{5/2}}{5/2} \right]_0^{E_F} = \frac{2}{5} C (E_F)^{5/2} \]
Then, determine the average energy per electron:
\[ \langle E \rangle = \frac{E_{\text{total}}}{N} = \frac{\frac{2}{5} C (E_F)^{5/2}}{\frac{2}{3} C (E_F)^{3/2}} \]
\[ \langle E \rangle = \frac{2/5}{2/3} \cdot \frac{(E_F)^{5/2}}{(E_F)^{3/2}} = \frac{3}{5} E_F \]
Step 4: Conclusion:
The average energy of an electron in a 3D free electron gas at T=0 K is \( \frac{3}{5} \) of the Fermi energy.
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