Thermal & Kinetic Lecture 7

Equipartition, specific heats, and

blackbody radiation

Recap….

Blackbody radiation: the ultraviolet catastrophe

Heat transfer: radiation and Stefan’s law

LECTURE 7 OVERVIEW

Equipartition and specific heats contd.

Last time….

Degrees of freedom and equipartition theorem

Specific heats

CP, CV and problems with classical theory

Not only can diatomic molecules rotate – they can vibrate

Consider kinetic and potential energy of SHO, as before. This gives another contribution of kT according to equipartition.

So, classically, we expect Cv = 7R/2.

?? If Cv = 7R/2, what is the value of CP ?

ANS: Cp = 9R/2

?? …and so is ? ANS: = 9/7

This is very puzzling. We seem to get the correct value for H2 and O2 if we ignore vibrational energy?! However, it gets worse……..

“I’ve done the math enough to know the dangers of our second guessing……”, MJ Keenan, ‘Schism’, © Toolshed music (2002)

CP, CV and problems with classical theory

From classical equipartition theory, the value of should be temperature independent…..

Variation of ratio of specific heats with temperature

1

1.1

1.2

1.3

1.4

1.5

1.6

1.7

-250 250 750 1250 1750

Temperature (C)

H2

O2

Classicalprediction

…but it’s not!

500 1000 15000 2000

Towards quantum theory: the first ‘cracks’ in the classical physics framework

“I have now put before you what I consider to be the greatest difficulty yetencountered by the molecular theory…….” James Clerk Maxwell, 1869

If we can eliminate vibrations or rotations by cooling then the system effectively loses some degrees of freedom and they won’t contribute to the specific heat.

However, eliminating vibrations or rotations at certain finite temperatures is not possible classically because classically the energies are distributed continuously.

To explain the discrepancies between classical equipartition of energy theory and the experimental data, Jeans suggested that at as the temperature falls some types of motion must ‘freeze out’.

Quantum levels and specific heats

Just as vibrational motion of a diatomic moleculeIs quantised, so too is its rotational motion.

In both cases, the occupation of a given energy level is proportional to a Boltzmann factor: exp(-E/kT).

Rotational levels have energy spacings of a few hundredths of an eV.Vibrational levels have energy spacings of a few tenths of an eV.

kT at room temperature (293 K) = 0.025 eV

When the temperature is much greater than a few hundredths of an eV (e.g. RT) then a very large number of rotational energy levels are occupied and the classical approximation is reasonably valid.

However, we need to go much higher in temperature before vibrational energy levels start to contribute.

Quantum levels and specific heats

Electronic energy levels

Vibrational energy levels

Rotational energy levels

Vibrational energy levels given by: 0)21( nEn

Rotational energy levels given by: IllEl 2/)1( 2 where I is the moment of inertia of the molecule.

CP

T (K)100 1000 10000

5R/2

7R/2

9R/2Schematic illustration of the variation of Cp with temperature for a hypothetical diatomic molecule.

Of all the diatomic molecules, only H2 is gaseous at low enough temperatures to ‘freeze out’ rotational motion.

Hence, only H2 and monatomic gases can have a value of CV corresponding to 3 degrees of freedom.

Quantum levels and specific heats

NB We can use the equipartition theorem to estimate the mean square angular frequency associated with a molecule at a given temperature.

The energy associated with the rotation of the molecule is given by: I<2>/2

Equipartition tells us that the average energy is ½ kT.

Equate these two expressions to get an estimate for the rate at which a molecule spins.

See Worked Example 12.5 on p. 448 of G & P.

Black-body radiation and the ultraviolet catastrophe

A solid surface at high temperature emits radiation spread across a wide range of frequencies with a peak wavelength dependent on the temperature of the object.

The temp. dependence of the spectrum explains why the colour of a tungsten filament changes from red-hot to white as its temperature is increased.

A surface is described as ideally black if it absorbs all the radiation that falls on it. In equilibrium it emits the same amount of energy as it absorbs.

Can only have equilibrium if we have a closed box (otherwise light energy is radiated away) with ideally black walls. The equilibrium radiation within the box is called black-body radiation.

The spectrum of black-body radiation proved to be a particularly sharp thorn in the side of classical physics…………………….

Diagrams taken from Hyperphysics websitehttp://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

I strongly recommend you visit this website!

Black-body radiation and the ultraviolet catastrophe

kTdvc

d3

28)(

(i) Note that at any temperature, the intensity in a unit frequency range depends on the square of the frequency. So, there’s a lot of X-rays in the box at any temperature!

(ii) Total energy in the box = area under the infinite curve.

“This is the prediction of classical physics….it is fundamentally, powerfully and absolutely wrong”RP Feynmann, Lectures in Physics

Black-body radiation and the ultraviolet catastrophe

The radiation in the box arises fundamentally from oscillations of the atoms. (The electrons are ‘shaken’ and emit radiation of a given frequency).

Classically the oscillators each have kT of energy on average.

The classical curve ‘blows up’ for precisely the same reason that the classical description of specific heats fails – we haven’t considered the energies of the oscillators correctly.

“Mending” the ultraviolet catastrophe

We can work out the correct quantum mechanical formula for <E> by considering the appropriate Boltzmann factors for the populations of the energy levels.

E0=0 (we’re ignoring zero point energy)

E1=

E2= 2

E3= 3

Number of oscillators, nn , in energy level En =

where n0 is the number of oscillators in the ground state.

)exp(0 kT

wnn

Let x = )exp(kT

w Therefore, n1 = n0x, n2 = n0x2, ………, nn = n0xn

“Mending” the ultraviolet catastrophe

Now we need to work out the total energy of all the oscillators.

If oscillator is in the first excited state (E1) its energy is . The total number of oscillators in energy level 1 is n1. Therefore the oscillators in level 1 contribute an energy of n1 .

However, n1 = n0x

If an oscillator is in the second level (E2) its energy is . The total number of oscillators in energy level 2 is n2. Therefore the oscillators in level 2 contribute an energy of n2 .

However, n2 = n0x2

2

2

2 2

E0

E1

Add these all together to get: ETOTAL = n0 (x + 2x2 + 3x3 + ….)

Similarly: NTOTAL = n0 (1 + x + x2 + x3 +….)

“Mending” the ultraviolet catastrophe

The average energy, <E> is: ......)1(

....)32(32

0

320

xxxn

xxxn

N

EE

TOTAL

TOTAL

?? Show that the equation above reduces to the classical value of average energy in the limit of either very high temperature or very low oscillator frequency.

By evaluating the series in the numerator and denominator and substituting in the expression for x given before, we get the following very important equation:

1)exp(

kT

E

This is the quantum mechanical expression for the average energy of a collection of oscillators.

Heat transfer: Radiation and Stefan’s law

Graph taken from the “Hyperphysics” website

The total energy density is obtained by integrating under the curve over all frequencies.

The result is:

What about the total radiated power ?

433

45

15

8T

ch

ktotal

Radiated power per unit area, J, is given by Stefan’s law: 4TJ

33

45

15

2

ch

k where

Later in the module we’ll return to a derivation ofStefan’s law based on a treatment of radiation as

a photon gas.