Distribution Over Quantum States:
Using the kinetic-molecular derivation of PV = nRT we saw,
Avg. translational energies per molecule = kT /2
and
Avg. translational energies per mole = RT /2
In order to understand, how the molecules are distributed in each allowed
energy states, we must first look at their probability of distribution.
Let us consider an example of throwing marble randomly in to a box with
equal sized compartment.
Let us consider the four-marble, two compartment distribution as shown
below:
Refer Figure 2-13
The number of ways W, of distribution of these marbles is given as
W
=
4 ! / 1! 2! 3! 4!
The same analogy can be extended to the distribution of molecules through
out the set of quantum states.
W
=
N ! / N1! N2! N3!……..
This gives an expression with which we can calculate the number of ways a
particular distribution can be achieved.
This number is proportional to the probability of distribution.
Qualitative Consideration:
Let us find out the values of W for various distributions subject to following
constraints:
1. The total number of particles is fixed.
2. The system of the particles has some fixed total energy.
e.g. The most probable distribution of 14 particles with an average energy as
2 units
Since N and Ni values are small, we can calculate the value of W.
The largest value for W is obtained when the particles are spread out as
much as possible.
Refer figure 2-14
e.g. The most probable distribution of 14 particles with varying average
energy of the system.
All distributions have same general shape.
Lower the average energy, the more particles are crowded in lower energy
state.
Higher the average energy, more particles are spread out into higher energy
level.
Refer Figure 2-15
Boltzmann Distribution:
This a mathematical expression, for a large number of molecules, which
gives the most probable distribution.
The Boltzmann distribution gives the number if particles per state at one
energy εi compared to that of another energy εj as
Ni / Nj = e –(
εi -εj) / kT
More conveniently, we compare the population of the states at one energy
level with the population of the sates of lowest energy level. Then we get,
Ni / N0 = e –(
εi -ε0) / kT
-------- (1)
Thus, according to Boltzmann distribution,
ε -ε0) to kT.
The population depends upon the ratio of ( i
The population of ith state is always less than that of the lowest-energy state.
At low temperature:
The exponential factor makes population decrease rapidly with increasing
energy and particles are crowded into the states of lower energy.
At high temperature:
The exponential factor fall-off is less rapid and the particles can spread out
into the higher-energy states.
Problem 1.
Use the Boltzmann distribution expression to calculate the ratio of the
number of molecules per quantum state at an energy equal to (1/2)kT, the
average energy for one-dimensional motion, to the number per quantum
state at the lowest allowed energy.
Ni / N0 = e –(
εi -ε0) / kT
ε = (1/2) kT
ε=0
i
0
Ni / N0 = e –((1/2) kT -0) / kT
Ni / N0 = e –(1/2)
Ni / N0 = 0.606
In most cases, we are given the total number of particle, i.e. N
The Boltzmann distribution relates the number of particles Ni in the quantum
states at any energy to the number N0 in the states at the lowest energy.
Ni / N0 = e –(
Ni = N0 e –(
εi -ε0) / kT
εi -ε0) / kT
The total number of particles is given as
N = ∑ Ni
N = ∑ [N0 e –(
N = N0 ∑ [e –(
εi -ε0) / kT
εi -ε0) / kT
]
]
N = N0 q
--------- (2)
Where q is the partition function and its given as
q = ∑ [e –(
εi -ε0) / kT
= ∑ [e –(
……
N0 = N / q
]
ε1 -ε0) / kT
] + ∑ [e –(
--------- (3)
ε2 -ε0) / kT
] + ∑ [e –(
ε3 -ε0) / kT
]
--------- (4)
This relation gives us how many molecules are in the lowest-energy state if
we know the total number of molecules in a given sample.
One-dimensional Translational Motion:
The Boltzmann distrubution can be used to find the distribution of the
molecules of a gas over the allowed energies and speeds.
Let us consider the molecule of a gas with mass m, moves along in x
direction and is confined to a line-segment of length a.
Then its translational energy is given as
εn
x
= nx2 h2 / (8 m a2)
where nx = 1,2,3…
The energy difference is given as
εn - εn
x
0
= nx2 h2 / (8 m a2) - n02 h2 / (8 m a2)
= nx2 h2 / (8 m a2) - (1)2 h2 / (8 m a2)
= ( nx2 – 1) h2 / (8 m a2)
Since most of the nx values are very much larger than 1 and thus
( nx2 – 1) ≅ nx2
εn - εn ≅
x
0
nx2 h2 / (8 m a2)
------- (5)
The partition function q from equation (3) can be written as
qx = ∑ [e
-(εnx -εn0) / kT
]
Summation can be replaced by integration, since summation is over so many
closely spaced energy levels.
Integration can be evaluated to give
π
qx
kT
1/2
=
2
h2 / (8ma2)
----------- (6)
Thus the partition function q for a particle of a gas with mass m, moving in
x-direction confined to a line-segment of length a at any temperature can be
calculated.
From equation (4) we can write,
N0 = N / q x
--------- (7)
Thus, now we can calculate the number of molecules present in the lowest
energy state.
Problem 2.
Calculate the value for qx and the value for N0 for 1-mol sample of nitrogen
gas in 1-L container at 250C.
Mass of one N2 molecule = m = (0.028 kg) / (6.02 × 1023) = 4.65 × 10-26 kg
Assuming the container to be cubic, the length of container is
a = 10 cm = 0.1 m
k = 1.38066 × 10-23 J k-1
T = 25 + 273.15 = 298.15 K
h = 6.626 × 10-34 J s
π
qx
kT
=
2
1/2
= 5.23 × 109
h2 / (8ma2)
and
N0 = N / q x
With a total 6.02 × 1023 molecuels in this 1-mol sample.
N0 = 6.02 × 1023 / 5.23 × 109 = 1.15 × 1014
Distribution of Quantum States:
The number of molecules per unit nx interval can be expressed as dN
/ dnx
Then,
The fraction of the molecules per unit nx interval as (dN
/ dnx) / N
Then
The Boltzmann distribution gives the number of molecules in a unit
quantum-number interval as
Nn = (dN / dnx) = N0 e –(
εnx -ε0) / kT
The fraction of the molecules per unit quantum state can be expressed as:
(dN / dnx) / N = (N0 / N )e –(
εnx -ε0) / kT
--------- (8)
For one-dimensional translational motion, (from equation 7)
N / N0 = q x
And from equation (5)
εn - εn
x
0
= nx2 h2 / (8 m a2)
Then,
Fraction per unit nx interval = (dN / dnx) / N
= (N0 / N )e –(
εnx -ε0) / kT
= (1/ qx) e –(
2
=
εnx -ε0) / kT
h2 / (8ma2)
π
1/2
kT
--------(9)
Problem 3:
Use the equation for fraction per unit nx interval to calculate he number of
molecules per unit quantum-number interval at nx= 0 for he system of 1-mol
sample of nitrogen in a 1-L container at 250C.
Mass of one N2 molecule = m = (0.028 kg) / (6.02 × 1023) = 4.65 × 10-26 kg
Assuming the container to be cubic, the length of container is
a = 10 cm = 0.1 m
k = 1.38066 × 10-23 J k-1
T = 25 + 273.15 = 298.15 K
h = 6.626 × 10-34 J s
Thus, the fraction per unit interval at nx=0 is
1
dN
2
=
N
dnx
nx = 0
π
h2 / (8ma2)
kT
= 1.91 × 10-10
1/2
This is the fraction of the molecules that are in the first quantum-number
unit interval.
The number of molecules in a 1-mol sample that are in this interval is
obtained by multiplying this fraction by Avogadro’s number.
Then we get
dN
N
dnx
NA ×
1
= NA × 1.91 × 10-10
nx = 0
= (6.02 × 1023) (1.91 × 10-10)
= 1.15 × 1014
Distribution Over Energies:
To obtain the distribution molecules over the allowed energies, we convert
the derivative dN/dnx (which gives the number of molecules per unit
interval nx) to dN/dεx (which gives the number of molecules per unit
energy interval εx)
We can write
dN/dεx
= (dN/dnx) (dnx/dεx)
We have
ε
x
= nx2 h2 / (8 m a2)
nx2 = εx (8 m a2) / h2
nx =
εx
(2
2
m
a) / h
nx = {(2 a) / h } 2 mεx
dnx/dεx = ( a / h ) 2 m / εx
--------(10)
Now, the fraction per unit energy interval εx cn be written as
Fraction per unit energy interval
= (1/N) dN/dεx
= (1/N) (dN/dnx) (dnx/dεx)
From equation (9) and (10)
1
1
e-
=
πkT
εx
εx / (kT)
Distribution over Speeds:
We know
εx = (1/2) m υx2
Differentiate with respect υx
We get
(1/2) m dυx2/ dυx
dεx /dυx =
= mυx
=
2m εx
Using equation (11) and (12) we can write
Fraction per unit speed interval
= (1/N) dN/dεx
m
=2
e
2πkT
-(1/2) mυx2 / (kT)
---------(11)