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Statistical Physics
Statistical Distribution
Understanding the distribution of certain energy among particle members and their
most probable behavior.
n(E) = g(E) f(E)
E3
E3
E2
E1
n(E) : number of particles with energy E
g(E) : number of state with energy E
F(E) : distribution function
average number of particles in each
states with energy E
probability of occupancy of each state
with energy E
Statistical Distribution Law
1. Maxwell –Boltzmann Statistics
Distribution of identical particles which are widely separated. Example: gas molecules
𝑛 𝐸 = 𝐴 𝑔 𝐸 𝑒 −𝐸 /𝑘𝐵 𝑇
2. Bose Einstein Distribution
For identical particles with spin 0 or integer/boson. Example: photon, phonon.
𝑛 𝐸 =𝐴
𝑔 𝐸
𝑒 𝛼 𝑒 −𝐸 /𝑘𝐵 𝑇 − 1
3. Fermi Diract Distribution
For particles with odd half-integer spin/fermion (1/2, 3/2, ...). Example: electron
𝑛 𝐸 =𝐴
𝑔 𝐸
𝑒 𝛼 𝑒 −𝐸 /𝑘𝐵 𝑇 + 1
Example:
A cointainer has 1025 hydrogen atoms at 00 and atmospheric pressure. If possible
quntum states for each qunatum number n is 𝑔 𝐸𝑛 = 2𝑛2 . Find the number of
atoms at first excited states (n=2)
Molecular Energy in Ideal Gas
Following Maxwell-Botlzmann Distribution
𝑛 𝐸 𝑑𝐸 = 𝐴 𝑔 𝐸 𝑒 −𝐸 /𝑘𝐵 𝑇 𝑑𝐸
The easiest way is to find the relation between momentum and energy because
𝑝2 = 2𝑚𝐸
Imagine the particles are distributed within a sphere with radius p. Each particle in
certain state with energy E has momentul p
Number of momentum states
𝑔 𝑝 𝑑𝑝 = 𝐵𝑝2 𝑑𝑝
Number of energy states
𝑔 𝐸 𝑑𝐸 = 𝑔 𝑝 𝑑𝑝 = 𝐵𝑝2 𝑑𝑝
= 𝐵 2𝑚𝐸 𝑚 𝑑𝐸/ 2𝑚𝐸
Number of molecules in energy state between E and dE
𝑛 𝐸 𝑑 𝐸 = 𝐶 𝐸𝑒 −𝐸/𝑘𝐵 𝑇 𝑑𝐸
To calculate C, use normalization for total molecules N
∞
𝐶 𝐸𝑒 −𝐸/𝑘𝐵𝑇 𝑑𝐸
𝑁=
0
Remember
∞
0
𝐶
𝑁=
𝜋 𝑘𝐵 𝑇
2
𝑥𝑒 −𝑎𝑥 𝑑𝑥 =
1
2𝑎
𝜋
𝑎
3/2
Molecular Energy Distribution
𝐸
2𝜋𝑁
−𝑘 𝑇
𝑛 𝐸 𝑑𝐸 =
𝐸𝑒 𝐵 𝑑𝐸
3/2
𝜋𝑘𝐵 𝑇
2𝜋𝑁
𝐶=
𝜋𝑘𝐵 𝑇 3/2
Total energy of N molecule
∞
𝐸=
0
2𝜋𝑁
𝐸 𝑛 𝐸 𝑑𝐸 =
𝜋𝑘𝐵 𝑇 3/2
Remember
∞ 3/2 𝑎𝑥
𝑥 𝑒 𝑑𝑥
0
3
= 4𝑎2
Average Molecular Energy
𝐸 3
𝐸 = = 𝑘𝐵 𝑇
𝑁 2
∞
𝐸 3/2 𝑒 −𝐸/𝑘𝐵 𝑇 𝑑𝐸
0
𝜋
𝑎
3
= 𝑁𝑘𝐵 𝑇
2
Distribution of Molecular Speed
𝐸=
1
𝑚𝑣 2
2
𝑑𝐸 = 𝑚𝑣𝑑𝑣
𝑚
𝑛 𝑣 𝑑𝑣 = 𝑛 𝐸 𝑑𝐸 = 4𝜋𝑁
2𝜋𝑘𝑇
3/2
𝑣 2 𝑒 −𝑚𝑣
2 /2𝑘 𝑇
𝐵
𝑑𝑣
Average molecular speed
1
𝑣=
𝑁
∞
𝑣 𝑛 𝑣 𝑑𝑣 =
0
8𝑘𝐵 𝑇
𝜋𝑚
Remember
∞ 3 −𝑎𝑥
𝑥 𝑒
𝑑𝑥
0
The RMS speed
𝑣𝑟𝑚𝑠 = 𝑣^2 =
3𝑘𝐵 𝑇
𝑚
Remember
1
2
𝑚𝑣
2
3
= 2 𝑘𝐵 𝑇
1
= 2𝑎2
Planck Radiation
∈=
ℎ𝑣
𝑒 ℎ𝑣 /𝑘𝐵 𝑇 − 1
Wien’s displacement
𝜆𝑚𝑎𝑥 𝑇 = 2.898 × 10−3 m.K
Stefan-Boltzaman Law
Energi radiated by an object per second
𝑅 = 𝑒𝜎𝑇 4 , with 𝜎 = 5.67 × 10−8 W/m^2K^4
Specific Heat of Solids
Classical internal Energy of solid
𝐸 = 3𝑁𝑘𝐵 𝑇 = 3𝑅𝑇
𝑅 = 𝑁𝐾𝐵 = 8,31 × 103 J/mol.K
Specific Heat at constant volume
𝐶𝑣 =
𝜕𝐸
𝜕𝑇 𝑉
= 3𝑅 = 24.93 J/mol.K
Dulong Petit law
Einstein Formula
Probability for an oscilator to have frequency 𝑣
𝑓 𝑣 =
1
ℎ𝑣
𝑒 𝑘𝐵 𝑇
−1
Average energy per oscilator
𝐸=
Internal Energy of Solid
ℎ𝑣
ℎ𝑣
𝑒 𝑘𝐵 𝑇
𝐸=
−1
Einstein specific heat formula
𝜕𝐸
𝑐𝑉 =
𝜕𝑇
𝑉
ℎ𝑣
= 3𝑅
𝑘𝐵 𝑇
2
𝑒 ℎ𝑣/𝑘𝐵 𝑇
2
ℎ𝑣
𝑒 𝑘𝐵 𝑇 −1
3𝑁ℎ𝑣
ℎ𝑣
𝑘
𝑒 𝐵𝑇
−1
Einstein Model at high temperature
ℎ𝑣 ≪ 𝑘𝐵 𝑇
𝑥
Remember : 𝑒 ≈ 1 + 𝑥 +
ℎ𝑣
𝑘
𝑒 𝐵𝑇
𝐸=
3𝑁ℎ𝑣
ℎ𝑣
𝑒 𝑘𝐵 𝑇
𝐶𝑣 =
≈1+
𝑥2
2!
+ ⋯.
ℎ𝑣
𝑘𝐵 𝑇
= 3𝑁𝑘𝐵 𝑇
−1
𝜕𝐸
𝜕𝑇 𝑉
= 3𝑅 = 24.93 J/mol.K
Free electron contribution to specific heat