Experiment 4Bench 4MZU-PE / EEA-4Ready

Fermi-Dirac Statistics and the Free Electron Model

Electron Energy Analyser

Cool and heat an electron gas in the cryostat and watch the occupation edge sharpen and smear.

Duration: 3 hours
Kittel: Chapter 6
Topics: Free electron modelDensity of statesFermi energy

Energy analyser — occupation spectrum

Copper · E_F = 7.033 eV · T = 300 K

Running

f(E), g(E) AND THE OCCUPIED DENSITY OF STATES

f(E) — occupationg(E) — density of statesf(E)·g(E) — occupied statesAnalyser slit

Fermi edge — expanded

Analyser zoomed onto the occupation edge

SPAN ±0.155 eV

Edge width (90→10%)

113.61meV

k_B from W / 4.394T

8.617e-5eV/K

Analyser readout

Channeltron detector · 0.5 s integration

Sample temperature

Fermi energy E_F

Analyser energy

f(E) at slit

Detector counts

c/sGAP

k_BT

meV

Derived

Fermi temperature

81612K

k_BT / E_F

0.368 %

Electronic C (Sommerfeld)

0.151J/mol·K

The standard reference metal, E_F ≈ 7 eV. Compare k_BT at 300 K with that: only a sliver of the distribution is disturbed.

Cryostat & analyser control

MZU-PE / EEA-4

Ready

Liquid helium cryostat. Do not open the shroud while the bath is below 120 K.

Sample

Temperature

Sample temperature300 K

CryoNormalHot

Analyser

Analyser energy7.000 eV

At any temperature f(E_F) = 0.5 exactly. Use it to check the analyser calibration.

Key Equations

f(E)=\frac{1}{e^{(E-E_F)/k_BT}+1}

At $T=0$ this is a step. At finite $T$ it smears over a width of order $k_BT$ ; measured between the 90% and 10% points the width is exactly

W = 2k_BT\ln 9 = 4.394\,k_BT

E_F=\frac{\hbar^{2}}{2m}\left(3\pi^{2}n\right)^{2/3},\quad g(E)\propto\sqrt{E}

Procedure

0/4

AThe step at absolute zeroLoad copper and cool the sample to 1 K. Describe the shape of f(E) below and above E_F, and note what the detector counts do as you drive the slit across the edge.
BHeat the electron gasTake the sample to 300, 1000, 3000 and 5000 K. At each temperature record f(E_F) and the 90→10% edge width. Check that f(E_F) never moves off 0.5.
CMeasure Boltzmann's constantPlot your edge widths against temperature. The gradient is 4.394 k_B — compare the value you get with 8.617 × 10⁻⁵ eV/K.
DDifferent metals, and a semiconductorStep through K, Na, Ag, Cu and Al at 300 K and record E_F for each. Then load intrinsic silicon: note where E_F sits and what the detector reads when the slit is inside the gap.

Table 4 — Fermi edge measurements

0 entries

Park the slit on E_F to check f(E_F) = 0.5 before recording each new temperature.

#	Sample	n/ m⁻³	E_F/ eV	T/ K	k_BT/ meV	f(E_F)	Edge width/ meV	k_B from W/ eV/K	k_BT/E_F/ %
No readings yet — set the controls, then press “Record reading”.

Analysis & Reflection

Analysis questions

The electronic heat capacity of a metal is $C_{el}=\gamma T$ with $\gamma \propto 1/E_F$ . Using the fact that only electrons within about $k_BT$ of $E_F$ can be excited, argue why $C_{el}\propto T/E_F$ , and why that is so much smaller than the classical $\tfrac{3}{2}Nk_B$ .
For copper at room temperature the bench reads $k_BT/E_F$ of well under 1%. Explain why that means the conductivity of a metal changes only slightly with temperature, and say which part of the distribution actually carries the current.
Compare potassium, sodium, silver, copper and aluminium. Aluminium contributes three electrons per atom rather than one. How does that show up in E_F?
With silicon loaded, the counter reads background when the slit is inside the gap even though f(E) there is not zero. Why?
Explain why raising the temperature increases the conductivity of a semiconductor but decreases it in a metal.

Physics problems

Copper has $E_F = 7.04$ eV. Calculate the Fermi temperature $T_F=E_F/k_B$ and explain why it shows that quantum statistics dominate in metals even at room temperature.
Why can the Maxwell-Boltzmann distribution not be used for electrons in a metal? Which physical principle forces Fermi-Dirac statistics?
Show that the mean kinetic energy of a free electron gas at $T=0$ is $\bar E = \tfrac{3}{5}E_F$ , not zero. Why do the electrons have kinetic energy at absolute zero at all?
From $E_F=\frac{\hbar^2}{2m}(3\pi^2 n)^{2/3}$ , work out the electron density that would give $E_F = 5$ eV, and compare it with the samples in the rack.