SME
Thermal & Statistical Physics
Quantum statistics
Use this for “A state lies 0.1 eV above the chemical potential at 500 K. Find its mean occupancy for fermions, bosons and classical particles.”
Chart
At x = 1.934: Fermi–Dirac 0.1263, Bose–Einstein 0.169, Boltzmann 0.1445
Dashed:
Bose–Einstein
1
eˣ − 1
;
dotted:
Boltzmann
e⁻ˣ.
All three agree when x ≫ 1
This is an example. Type your own values to start a new problem.
x
=
E − μ (eV)
÷
k
B
T (K)
=
f
FD
=
1
e
+
1
=
fFD
fBE
Occupancy for bosons
calculated
fMB
Boltzmann occupancy
calculated
Formulas
x
=
E − μ
8.617 ×
10
-5
×
T
1.9342
=
0.05
8.617 ×
10
-5
× 300
fFD
=
1
e
x
+ 1
0.1263
=
1
e
1.9342
+ 1
fBE
=
1
e
x
− 1
0.169
=
1
e
1.9342
− 1
fMB
=
e
−
x
0.1445
=
e
−1.9342
Assumptions
•
f is the mean number of particles in one state of energy E, at temperature T and chemical potential μ.
•
Fermions never exceed 1 (Pauli); bosons need E above μ (x > 0) and pile up as x → 0.
•
All three agree when x = (E − μ) ÷ kBT ≫ 1, the classical limit; kB = 8.617 × 10⁻⁵ eV/K.
Step-by-step
WE KNOW
E − μ = +0.05 eV, T = 300 K
FIND
energy in kB T (x), Fermi–Dirac occupancy (fFD), occupancy for bosons (fBE), Boltzmann occupancy (fMB)
Step 1 · Find energy in kB T (x)
Use x = (E − μ) ÷ (8.617 × 10⁻⁵ × T)
Measure the energy above μ in units of kBT: every occupancy depends on this ratio alone.
x
=
0.05
8.617 ×
10
-5
× 300
x
=
0.05
0.02585
x
=
1.9342
Step 2 · Find Fermi–Dirac occupancy (fFD)
Use fFD = 1 ÷ (eˣ + 1)
Fermi–Dirac:
the
+1
keeps
a
state
at
most
singly
filled,
and
f
=
1
2
at
E
=
μ.
fFD
=
1
e
1.9342
+ 1
fFD
=
1
6.919 + 1
fFD
=
1
7.919
fFD = 0.1263
Step 3 · Find occupancy for bosons (fBE)
Use fBE = 1 ÷ (eˣ − 1)
Bose–Einstein: the −1 lets the occupancy grow past 1 as E nears μ.
fBE
=
1
e
1.9342
− 1
fBE
=
1
6.9185 − 1
fBE
=
1
5.9185
fBE = 0.169
Step 4 · Find Boltzmann occupancy (fMB)
Use fMB = e⁻ˣ
Boltzmann: the classical factor, which lies between the other two.
fMB
=
e
−1.9342
fMB = 0.1445
Check
Put the numbers back in. Both sides should match.
1.9342
=
0.05
8.617 ×
10
-5
× 300
✓
0.1263
=
1
e
1.9342
+ 1
✓
0.169
=
1
e
1.9342
− 1
✓
0.1445
=
e
−1.9342
✓
← Previous lesson
Maxwell’s speeds: most probable, mean and rms
Next lesson →
The Fermi energy of a metal
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
More problem types
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.”
A blackbody: Wien’s and the Stefan–Boltzmann laws
Use this for “A star’s surface is at 3200 K and its area is 1.5 × 10¹⁷ m². Where does its spectrum peak, and what total power does it radiate?”
Bose–Einstein condensation temperature
Use this for “A cloud of sodium-23 atoms holds 5 × 10¹⁹ atoms per cubic meter. Below what temperature does it form a Bose–Einstein condensate?”