SME
University Physics III: Waves, Optics & Modern
Condensed matter physics
Use this for “Find the electron density and Fermi energy of copper (8.96 g/cm³, 63.55 g/mol, one free electron per atom).”
Chart
EF(8.96) ≈ 7.044
This is an example. Type your own values to start a new problem.
n (m⁻³)
=
ρ (g/cm³)
×
N
A
×
M (g/mol)
EF
Fermi energy
calculated · eV
vF
Fermi speed
worked out (not typed) · m/s
TF
Fermi temperature
worked out (not typed) · K
Formulas
n
=
ρ
×
10
6
×
6.022 ×
10
23
×
Z
M
8.4905 ×
10
28
=
8.96 ×
10
6
×
6.022 ×
10
23
×
1
63.55
EF
=
3.810 ×
10
-20
×
(3 ×
π
2
×
n
)
2
3
7.0441
=
3.810 ×
10
-20
×
(3 ×
π
2
× (8.4905 ×
10
28
))
2
3
vF
=
√(2
×
EF
×
1.602 ×
10
-19
9.109 ×
10
-31
)
1.5741 ×
10
6
=
√(2
×
7.0441
×
1.602 ×
10
-19
9.109 ×
10
-31
)
TF
=
EF
8.617 ×
10
-5
81,747
=
7.0441
8.617 ×
10
-5
Assumptions
•
The valence electrons move freely through the metal as in a box; NA = 6.022 × 10²³ /mol and ħ² ÷ 2mₑ = 3.810 × 10⁻²⁰ eV·m².
•
At 0 K they fill every level up to EF, two to each state (spin up and down).
•
TF = EF ÷ kB (kB = 8.617 × 10⁻⁵ eV/K) is far above room temperature, so warming the metal barely changes the sea.
•
The fastest electrons have ½mₑvF² = EF, with mₑ = 9.109 × 10⁻³¹ kg and 1 eV = 1.602 × 10⁻¹⁹ J.
Step-by-step
WE KNOW
ρ = 8.96 g/cm³, M = 63.55 g/mol, Z = 1
FIND
electron density (n), Fermi energy (EF), Fermi speed (vF), Fermi temperature (TF)
Step 1 · Find electron density (n)
Use n = ρ × 10⁶ × 6.022 × 10²³ × Z ÷ M
ρ
M
counts
moles
in
each
cm³;
times
10
6
cm³
in
each
m
3
,
then
atoms
per
mole,
then
free
electrons
per
atom.
n
=
8.96 ×
10
6
×
6.022 ×
10
23
×
1
63.55
n
=
5.3957 ×
10
30
63.55
n
=
8.4905 ×
10
28
m
-3
Step 2 · Find Fermi energy (EF)
Use EF = 3.810 × 10⁻²⁰ × (3 × π² × n)2/3
Filled
two
to
each
state,
the
n
electrons
per
m
3
reach
kF
=
(3π²
n
)
1
3
;
EF
=
ħ
2
kF
2
2m
ₑ,
in
eV.
EF
=
3.810 ×
10
-20
×
(3 ×
π
2
× (8.4905 ×
10
28
))
2
3
EF
=
(3.810
×
10
-20
)
×
(3 × 9.87 × (8.4905 ×
10
28
))
2
3
EF
=
(3.810
×
10
-20
)
×
(29.61 × (8.4905 ×
10
28
))
2
3
EF
=
(3.810
×
10
-20
)
×
(2.514 ×
10
30
)
2
3
EF
=
(3.810
×
10
-20
)
×
(1.849
×
10
20
)
EF = 7.0441 eV
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
mₑv²
=
EF,
with
EF
in
joules.
vF
=
√(2
×
7.0441
×
1.602 ×
10
-19
9.109 ×
10
-31
)
vF
=
√(14.0882
×
1.602 ×
10
-19
9.109 ×
10
-31
)
vF
=
√(
2.2569 ×
10
-18
9.109 ×
10
-31
)
vF
=
√
2.478 ×
10
12
vF
=
1.5741 ×
10
6
m/s
Step 4 · Find Fermi temperature (TF)
Use TF = EF ÷ (8.617 × 10⁻⁵)
The temperature at which kBT would reach EF: for a metal tens of thousands of kelvin, so at room temperature only the top of the sea stirs.
TF
=
7.0441
8.617 ×
10
-5
TF = 81,747 K
Check
Put the numbers back in. Both sides should match.
8.4905 ×
10
28
=
8.96 ×
10
6
×
6.022 ×
10
23
×
1
63.55
✓
7.0441
=
3.810 ×
10
-20
×
(3 ×
π
2
× (8.4905 ×
10
28
))
2
3
✓
1.5741 ×
10
6
=
√(2
×
7.0441
×
1.602 ×
10
-19
9.109 ×
10
-31
)
✓
81,747
=
7.0441
8.617 ×
10
-5
✓
Topic check
See what you know: 5 questions on this topic
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