Download 12.3 Assembly of distinguishable Particles

12.3 Assembly of distinguishable
particles
• An isolated system consists of N
distinguishable particles.
• The macrostate of the system is defined by (N,
V, U).
• Particles interact sufficiently, despite very
weakly, so that the system is in thermal
equilibrium.
• Two restrictive conditions apply here
n
(conservation of particles)
N N

j 1
n
j
(conservation of energy)
N
E

U
 j j
j 1
where Nj is the number of particles on the energy
level j with the energy Ej.
Example: Three distinguishable particles labeled A, B,
and C, are distributed among four energy levels, 0, E,
2E, and 3E. The total energy is 3E. Calculate the
possible microstates and macrostates.
Solution: The number of particles and their total
energy must satisfy
3
N
j 0
j
3
(here the index j starts from 0)
3
N E
j 0
j
j
 3E
# particles on
Level 0
# Particles on
Level 1 E
# particles on
Level 2E
# particles on
Level 3E
Case 1
(Macrostate 1)
2
0
0
1
Case 2
(Macrostate 2)
1
1
1
0
Case 3
(Macrostate 3)
0
3
0
0
So far, there are only THREE macrostates
satisfying the conditions provided.
Detailed configurations (i.e. microstate) for case 1
Level 0
Level 1E
Level 2E
Level 3E
A, B
C
A, C
B
B, C
A
Thus, thermodynamic probability for case 1 is 3
Microstates for case 2
Level 0
Level 1E
Level 2E
A
B
C
A
C
B
B
A
C
B
C
A
C
A
B
C
B
A
Level 3E
Microstates for case 3
Level 0
Level 1E
Level 2E
Level 3E
A, B and C
Therefore, W1 = 3, W2 = 6, W3 = 1, and Ω = 10
• The most “disordered” macrostate is the state with the
highest probability.
• The macrostate with the highest thermodynamic
probability will be the observed equilibrium state of
the system.
• The statistical model suggests that systems tend to
change spontaneously from states with low
thermodynamic probability to states with high
thermodynamic probability.
• The second law of thermodynamics is a consequence
of the theory of probability: the world changes the
way it does because it seeks a state of higher
probability.
12.4 Thermodynamic Probability
and Entropy
Boltzman made the connection between the classical
concept of entropy and the thermodynamic probability
S = f (w)
f (w) is a single-valued, monotonically increasing function
(because S increases monotonically)
For a system which consists of two subsystems A and B
Or…
Stotal = SA + SB
(S is extensive)
f (Wtotal) = f (WA) + f (WB)
Configurations of the total system are
calculated as Wtotal = WA x WB
thus: f (WA x WB) = f (WA) + f (WB)
The only function for which the above
relationship is true is the logarithm.
Therefore:
S = k · lnW
where k is the Boltzman constant with the
units of entropy.
12.5 Quantum States and Energy
Levels
To each energy level, there is one or more quantum
states described by a wave function Ф.
When there are several quantum states that have the
same energy, the states are said to be degenerate.
The quantum state associated with the lowest energy
level is called the ground state of the system.
Those that correspond to higher energies are
called excited states.
The energy levels can be thought of as a set of
shelves of different heights, while the quantum
states correspond to a set of boxes on each
shelf
•
•
•
•
E3
g3 = 5
E2
g2 = 3
E1
g1 = 1
For each energy level the number of quantum
states is given by the degeneracy gj
a particle in a one-dimensional box
with infinitely high walls
• A particle of mass m
• The time-independent part of the wave function Ф (x)
is a measure of the probability of finding the particle
at a position x in the 1-D box.
• Ф (x) = A*sin(kx)
0≤x≤1
with k = n (π/L), n = 1, 2, 3
n=1
n=2
The momentum of the particle is
(de Broglie relationship)
where k is the wave number and h is the Planck
constant.
P=
·k
The kinetic energy is:
2
2
2
1 2 1P
1 hk
E  mv 

2
2
2 m 2m 4
• Plug-in k  n 
L
h 2 n 2 2
E
8m 2 L2
h2n2

8mL2
Therefore, the energy E is proportional to the square
of the quantum number, n.
In a three-dimensional box
E = h2 ( +
+
)
8m
where any particular quantum state is designated by
three quantum numbers x, y, and z.
If Lx=Ly=Lz=L
then, we say nj2 = nx2 + ny2 + nz2
Where is the total quantum number for states
whose energy level is EJ.
An important result is that energy levels depends
only on the values of nj2 and not on the individual
values of integers (nx, ny, nz).
(nj is the total quantum number for energy level Ej)
Since
h2 1 2
2
2
Ej 
(
n

n

n
x
y
z)
2
8m L
L3 = V
L2 = V ⅔
therefore, E  h 2 1
j
8m V
2
2
2
(
n

n

n
x
y
z)
2/3
For the ground state, nx = 1, ny = 1, nz = 1
There is only one set of quantum number leads to this
energy, therefore the ground state is non-degenerate.
For the excited state, such as nx2 + ny2 + nz2 = 6
it could be
nx = 1
ny = 1
nz = 2
or
nx = 1
ny = 2
nz = 1
or
nx = 2
ny = 1
nz = 1
They all lead to
h2 1
Ej 
6
2/3
8m V
Note that increasing the volume would reduce the
energy difference between adjacent energy levels!
For a one liter volume of helium gas
nJ ≈ 2 x 109
12.6. Density of Quantum States
When the quantum numbers are large and the
energy levels are very close together, we can
treat n’s and E’s as forming a continuous
function.
2
h
1
For
E 
(n 2  n 2  n 2 )
j
8m V
2/3
x
y
z
nx2 + ny2 + nz2 =
· V⅔ ≡ R2
The possible values of the nx, ny, nz correspond to
points in a cubic lattice in (nx, ny, nz) space
Let g(E) · dE = n (E + dE) – n(E) ≈
dn( E )
dE
dE
Within the octant of the sphere of Radius R;
n(E) = ⅛ · πR³ = V( ) · E
Therefore,
dn( E )
4 2V 3 / 2 1/ 2
g ( E )dE 
dE
dE 
h
3
m E dE
Note that the above discussion takes into account
translational motion only.
4 2V 3 / 2 1/ 2
g ( E )dE   s
m E dE
3
h
Where γs is the spin factor, equals 1 for zero
spin bosons and equals 2 for spin one-half
fermions
• Example (12.8) Two distinguishable particles are to
be distributed among nondegenerate energy levels, 0,
ε, 2ε , such that the total energy U = 2 ε.
(a)What is the entropy of the assembly?
(b)If a distinguishable particle with ZERO energy is
added to the system, show that the entropy of the
assembly is increased by a factor of 1.63.