SME
Thermal & Statistical Physics · Problem type
The Fermi energy of a metal
Use this for “Gold has 5.90 × 10²⁸ free electrons per cubic meter. Find its Fermi energy, Fermi temperature and Fermi speed.”
Table
Electron density (n)
(m⁻³)
Fermi energy (EF)
(eV)
This is an example. Type your own values to start a new problem.
ħ²
2m
(3π²
n (m⁻³)
)
2
÷
3
=
EF (eV)
TF
Fermi temperature
calculated · K
vF
Fermi speed
calculated · m/s
Formulas
EF
=
3.810 ×
10
-20
×
(3 ×
π
2
×
n
)
2
3
7.0328
=
3.810 ×
10
-20
×
(3 ×
π
2
× (8.47 ×
10
28
))
2
3
TF
=
EF
8.617 ×
10
-5
81,615
=
7.0328
8.617 ×
10
-5
vF
=
√(2
×
EF
×
1.602 ×
10
-19
9.109 ×
10
-31
)
1.5728 ×
10
6
=
√(2
×
7.0328
×
1.602 ×
10
-19
9.109 ×
10
-31
)
Assumptions
•
Free electrons in a box at T = 0: each state up to EF holds two electrons, spin up and down.
•
n counts the conduction electrons: one per atom for sodium and copper, three for aluminum.
•
ħ² ÷ 2m = 3.810 × 10⁻²⁰ eV·m² for an electron; TF = EF ÷ kB, kB = 8.617 × 10⁻⁵ eV/K.
•
The fastest electrons have ½mvF² = EF, m = 9.109 × 10⁻³¹ kg; 1 eV = 1.602 × 10⁻¹⁹ J.
Step-by-step
WE KNOW
n = 8.47 × 10²⁸ m⁻³
FIND
Fermi energy (EF), Fermi temperature (TF), Fermi speed (vF)
Step 1 · Find Fermi energy (EF)
Use EF = 3.810 × 10⁻²⁰ × (3 × π² × n)2/3
The
states
up
to
the
Fermi
sphere’s
radius
kF
=
(3π²
n
)
1
3
hold
all
n
electrons;
EF
=
ħ
2
kF
2
2m
,
in
eV.
EF
=
3.810 ×
10
-20
×
(3 ×
π
2
× (8.47 ×
10
28
))
2
3
EF
=
(3.810
×
10
-20
)
×
(3 × 9.87 × (8.47 ×
10
28
))
2
3
EF
=
(3.810
×
10
-20
)
×
(29.61 × (8.47 ×
10
28
))
2
3
EF
=
(3.810
×
10
-20
)
×
(2.508 ×
10
30
)
2
3
EF
=
(3.810
×
10
-20
)
×
(1.846
×
10
20
)
EF = 7.0328 eV
Step 2 · Find Fermi temperature (TF)
Use TF = EF ÷ (8.617 × 10⁻⁵)
The temperature at which kBT would equal EF; for a metal it is tens of thousands of kelvins, so its electrons stay degenerate.
TF
=
7.0328
8.617 ×
10
-5
TF = 81,615 K
Step 3 · Find Fermi speed (vF)
Use vF = √(2 × EF × 1.602 × 10⁻¹⁹ ÷ (9.109 × 10⁻³¹))
The
electrons
at
the
top
of
the
sea
have
kinetic
energy
EF:
1
2
mv²
=
EF,
with
EF
turned
into
joules.
vF
=
√(2
×
7.0328
×
1.602 ×
10
-19
9.109 ×
10
-31
)
vF
=
√(14.0656
×
1.602 ×
10
-19
9.109 ×
10
-31
)
vF
=
√(
2.2533 ×
10
-18
9.109 ×
10
-31
)
vF
=
√
2.474 ×
10
12
vF
=
1.5728 ×
10
6
m/s
Check
Put the numbers back in. Both sides should match.
7.0328
=
3.810 ×
10
-20
×
(3 ×
π
2
× (8.47 ×
10
28
))
2
3
✓
81,615
=
7.0328
8.617 ×
10
-5
✓
1.5728 ×
10
6
=
√(2
×
7.0328
×
1.602 ×
10
-19
9.109 ×
10
-31
)
✓
Topic check
See what you know: 5 questions on this topic
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