Download Solutions to Exercise sheet 2

Einführung in die Physikalische Chemie, Herbstsemester 2013
(EPC 2013)
Solutions to Exercise sheet 2
Prof. Dr. Anatole von Lilienfeld
20 November 2013
1. Counting particle
Number of ways to distribute 2 fermions in 3 energy levels:
Let us represent the states of the collective system (of two fermions) using the occupation
numbers of the energy levels. In this representation, (n1 ,n2 ,n3 ) means that we have
in energy level 1,
n2
particles in energy level 2, and
n3
n1
particles
particles in energy level 3. The following
matrix shows all the possible states for the system of two fermions.
(2,0,0)
(0,2,0)
(0,0,2)
(1,1,0)
(1,0,1)
(0,1,1)
Notice that by using the occupation number representation, we have already accounted for
indistinguishability of the fermions, hence eliminated duplicate states.
For ex.
the state of
having the fermion-1 in the energy level 1 and fermion-2 in energy level 2 is the same as having
fermion-1 in energy level 2 and fermion-2 in energy level 1.
Both the states are same and
considered only once as (1,1,0).
The states (2,0,0), (0,2,0), and (0,0,2) violate Pauli's exclusion principle, i.e., we cannot have
more than one fermion in an energy level. Hence we are left with the three allowed states (1,1,0),
(1,0,1), and (0,1,1).
Number of ways to distribute 3 fermions in 3 energy levels:
We have 10 unique states as shown below, where the rst 9 states violate Pauli's exclusion
principle and we are left with only one allowed state, which is (1,1,1).
(3,0,0)
(0,3,0)
(0,0,3)
(2,1,0)
(2,0,1)
(1,2,0)
(0,2,1)
(1,0,2)
(0,1,2)
(1,1,1)
Number of ways to distribute 2 bosons in 3 energy levels:
We do not have the Pauli restriction for bosons, i.e., we can have any number of bosons in an
energy level, hence all the six states shown for the case of 2 fermions are allowed for 2 bosons.
Number of ways to distribute 3 bosons in 3 energy levels:
All the 10 states shown for the case of 3 fermions are allowed.
General case, femions:
The expression for the number of ways to distribute N fermions in M energy levels is
M!
N !(M − N )!
General case, bosons:
The expression for the number of ways to distribute N bosons in M energy levels is
(N + M − 1)!
N !(M − 1)!
1
2. Separating partition functions
Mono-atomic ideal gas:
The energy expectation value is given by (see eq. (3.6), p.13 in the lecture notes)
hEi = −∂β ln[Q]
Let us substitute
qA
and
qB in Q
and isolate the terms that are dependent only on
β.


3NA
3NB
2
2πm 2
1
2πm
1
V NA ·
V NB 
hEi = −∂β ln 
2
2
NA !
h β
NB !
h β
1
3NA
1
3NB
2πm
2πm
= −∂β ln
+
+ NA ln V + ln
+
+ NB ln V
ln
ln
NA !
2
h2 β
NB !
2
h2 β
2πm
3
(NA + NB ) ln
= −∂β
2
h2 β
3
3
3
2
= −∂β
+ ∂β
(NA + NB ) ln (2πm) + ∂β
(NA + NB ) ln h
(NA + NB ) ln (β)
2
2
2
3
= ∂β
(NA + NB ) ln (β)
2
1
3
3
(NA + NB ) = (NA + NB ) kB T.
=
2
β
2
The pressure expectation value is given by (see eq. (3.50), p.21 in the lecture notes)
P = hP i =
Using the expansions of
Q
1
∂V ln[Q]
β
givenabove, we can extract the terms that depend only on
P
⇒ PV
1
∂V ln[V NA · V NB ]
β
1
1
=
(NA + NB )
β
V
= kB T (NA + NN ) .
=
Harmonic oscillator:
q vib (T ) =
∞
X
e−(i+1/2)βhv
i=0
= e−βhν/2
∞
X
e−iβhv
i=0
= e−βhν/2
=
∞ X
i=0
−βhν/2
e
1 − e−βhν
2
e−βhv
i
V
Population of electronic levels of Na (gas):
The expression for the population of a quantum state is
gk e−βεk
Pk = P
−βεj
j gj e
−1 , let us use
Since the problem gives energies in the units of cm
kB cm−1 K−1 =
kB
−1 K−1 .
in cm
kB J K−1
= 0.6950356300864200 cm−1 K−1
h [J s] c [cm s−1 ]
Using the formula given above (where we use
β = kB T
in the units of cm
−1 ) and the temperatures
and energies given in the problem, the populations of the electronic states can be computed by
direct substitutions.
Population
P1
P2
P3
P4
P1 +P2 +P3 +P4
T =1000K
T =2500K
0.9999999999250311(≈
1.00)
2.54 × 10−11 )
−11 )
0.0000000000495654(≈ 4.96 × 10
−17 )
0.0000000000000001(≈ 1.00 × 10
0.9998273879050946(≈
0.0000000000254035(≈
0.0000577939423357(≈
1.0000000000000000
1.0000000000000000
3
1.00)
5.78 × 10−5 )
−4 )
0.0001144496142361(≈ 1.14 × 10
−7 )
0.0000003685383336(≈ 3.69 × 10