Question

A one-dimensional nanowire has a linear electron density of $10^8 \, electrons \, cm^{-1}$. The Fermi energy of the system is (in eV) ............ (rounded off to two decimal places). Given: \[ \frac{\hbar^2}{2m} = 0.24 \, (eV)^2 \, s^2 \, kg^{-1} \] where $m$ is the electron mass.

Hint:

In 1D: $n \propto k_F$, while in 3D: $n \propto k_F^3$. This makes low-dimensional systems highly sensitive to carrier concentration when computing Fermi energy.

Answer 1

Correct Answer: 9.2

Step 1: Recall formula for Fermi wavevector in 1D.
For a one-dimensional electron gas, the density of states fills from $-k_F$ to $+k_F$. Taking spin degeneracy into account: \[ n = \frac{2k_F}{2\pi/L} . \frac{1}{L} = \frac{2k_F}{\pi} \] where $n$ is the electron density per unit length.
Step 2: Rearrangement.
\[ k_F = \frac{\pi n}{2} \]
Step 3: Substitute the given density.
Given $n = 10^8 \, electrons/cm = 10^{10} \, electrons/m$: \[ k_F = \frac{\pi \times 10^{10}}{2} \approx 1.57 \times 10^{10} \, m^{-1} \]
Step 4: Fermi energy formula.
\[ E_F = \frac{\hbar^2 k_F^2}{2m} \] Substituting the given constant: \[ E_F = 0.24 \times (k_F)^2 \] Note: $k_F$ must be expressed in $nm^{-1}$ to match unit convention. Convert $k_F$: \[ k_F = 1.57 \times 10^{10} \, m^{-1} = 15.7 \, nm^{-1} \]

Step 5: Final calculation.
\[ E_F = 0.24 \times (15.7)^2 = 0.24 \times 246.49 = 59.16 \, eV \] Final Answer: \[ \boxed{59.16 \, eV} \]