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How much energy? How fast do the molecules/atoms move?
Translation
Thermodynamics, Maxwell-Boltzmann, equipartition theorem, translational kinetic
energy of gas = (3/2) k T per molecule. (CM2004)
At room temperature, T = 300 K, Etrans = 6.21 x10-21 J molecule-1 = m v2 / 2
For He atom, m = 4 amu,
For C6H6
m = 78 amu
v = 1.36 x103 ms-1 (~3000 mph!!!)
v = 3.05 x102 ms-1
This is the velocity on average, not of any particular molecule.
Rotation
Similarly the equipartition theorem states that the average rotation energy of a diatomic
molecule is (1/2) k T per molecule per rotational direction.
At T = 300 K,
Erot,x = μ Re2 ωx2 /2 = (1/2) k T = 2.07 x10-21 J molecule-1
for H – 35Cl:
Re = 1.2746x10-10 m (spectroscopic data)
μ = mH mCl / (mH + mCl) ~ (1 x 35)/(1+35) amu ~ 1.614x10-27 kg
ωx ~ 6.288x1012 radians s-1 = 1.000x1012 rev s-1
a revolution every 1.0 ps on average.
Vibration
If we assume a vibrational energy of 10-19 J molecule-1 for NO which has a force constant
k = 1.54x103 Nm-1 and Re = 1.1508x10-10 m, can estimate (classically) the extent to which
the bond stretches.
At maximum (or minimum) R all the energy is Potential Energy (the atoms come to rest
at the limits of the vibration, no Kinetic Energy) therefore
(k/2)(R – Re)2 = 10-19 J, from which (R – Re) = 1.14x10-11 m
about 10% of Re
When R = Re all the energy is kinetic energy, therefore
μ vR2/2 = 10-19 J , from which vR = 4016 ms-1
and from the classical equations of motion, estimate that a vibration period is 2.26x10-13 s
or the vibrational frequency is 4.40x1014 s-1.
Quantum Mechanical Molecular Energy of the Nuclei.
The motion of nuclei in molecules does not obey classical mechanics. The energies of
translational, rotational and vibrations motion of nuclei obey Quantum Mechanics.
In all three cases quantum mechanics finds the energy to be quantized. Only certain
values of energy are allowed. The equations for the energy involve quantum numbers (1
quantum number for every independent coordinate).
Translation. The ‘Particle in the box’ problem. The energy of motion of a particle of
mass m in 1 D (say along x) and restricted to a path length of L (e.g. a molecule contained
in a box; an electron traveling along a copper wire) is found to be:
En = n2 h 2 / (8mL2) Joules
n = 1, 2, 3, … all positive integer values >0
h = 6.626x10-34 Js
Planck’s constant
m
mass, kg
L
length of ‘box’, m
Units (Js)2/(kg m2)  J
In 3 D, a box of lengths A, B, C in the x, y, z directions, the energy is specified by three
quantum numbers.
En,p,q = n2 h 2 /(8m A2) + p2 h 2 /(8m B2) + q2 h 2 /(8m C2) Joules
n = 1, 2, 3, …; p = 1, 2, 3, …; q = 1, 2, 3, …
The sum of the three 1 D energies.
The 3D case demonstrates ‘degeneracy’ if the ‘box’ is a cube, A = B = C:
En,p,q = (n2+ p2 + q2 )h 2 /(8m A2) J
n = 1, 2, 3, …; p = 1, 2, 3, …; q = 1, 2, 3, …
n = 1, p = 2, q = 3 has energy E1,2,3 = (1+ 4 + 9 )h 2 /(8m A2) = 14 h 2 /(8m A2) J
as do: E1,3,2 E2,1,3 E2,3,1 E3,1,2 E3,2,1
6 different energy levels with the same energy – 6 degenerate levels.
Consider a molecule in a (1D) room: O2 (g) , m = 32 amu = 5.314x 10-26 kg, L = 10 m
En = 1.03283x10-44 n2 Joules
How this relates to the Classical translational energy is a matter for CM3109/CM4113
(but it does suggest that, on average, n ~ 1011)
Points to note about the ‘Particle in the Box’ problem:
1) n ≠ 0. The particle cannot have zero translational energy. This has to do with the
Heisenberg uncertainty principle.
2) Energy levels ~ n2, get further apart with increasing n.
3) 1D box is non-degenerate (no 2 energy levels have the same energy), 3D box has
degenerate levels under special circumstances.
Quantized Rotational Energy for a diatomic molecule
Rotation. The ‘Rigid Rotator’ problem. The energy of two masses, mA and mB at a fixed
distance Re rotating about their centre of mass, is found to be:
Erot = (ħ 2 / 2 I) J (J + 1) Joules
J = 0, 1, 2, 3 …
I = μ Re2
momentum of inertia, kg m2
ħ = h / 2π
Js
μ = mA mB / (mA + mB)
reduced mass, kg
Each energy level is degenerate. There are (2 J + 1) levels with the same energy. There is
a second quantum number mJ which has integer values between – J and + J , that is
(2 J + 1) values all with the same energy determined by J. 2 quantum numbers as there
are 2 separate rotational axes/motions.
For H – 35Cl (data given previously)
Erot = (ħ 2 / 2 I) J (J + 1) = 2.12x10-22 J (J + 1) Joules
(again the connection with the Classical estimate is complicated and is covered in
CM3109/CM4113 but shows that J ~ 3 on average)
In spectroscopy of diatomic molecules, it is usual to write
Erot = B J (J + 1)
B = (ħ 2 / 2 I) = (ħ 2 / 2 μ Re2) Joules is called the rotational constant and in spectroscopy
it is usual to give the value in wavenumbers, cm-1
E (J) = h c E(cm-1)
c
velocity of light, cm s-1 !! For B = 2.12x10-22 ≡ 10.7 cm-1
Er
qu. number energy
degeneracy
J = 4, E4 = 20B, (2J+1) = 9
levels
J = 3, E3 = 12B, (2J+1) = 7
levels
J = 2, E2 = 6B, (2J+1) = 5 levels
J = 1, E1 = 2B, (2J+1) = 3 levels
J = 0, E0 = 0B, (2J+1) = 1 level
Points to note:
1) Energy levels ~(J2 + J), get
further apart with increasing J
2) J = 0 is allowed.
3) Degeneracy (2 J + 1)
4) Excellent agreement with
spectroscopy in the microwave
region.