chapter 5: phonons ii – thermal properties. what is a phonon? we’ve seen that the physics of...

Chapter 5: Phonons II – Thermal Properties

What is a Phonon?• We’ve seen that the physics of lattice vibrations in a

crystalline solidReduces to a CLASSICAL normal mode problem.

The goal of the ENTIRE DISCUSSION so far has been to

find the normal mode vibrational frequencies ofthe crystalline solid.

• In the harmonic approximation, this is achieved by first writing the solid’s vibrational energy as a system of coupled simple harmonic oscillators & then finding the classical normal mode frequencies & ion displacements for that system.

• Given the results of the classical normal mode calculation for the lattice vibrations, in order to treat some properties of the solid,

It is necessary to QUANTIZEthese normal modes.

• These quantized normal modes of vibration are called

PHONONS• PHONONS are massless quantum mechanical

“particles” which have NO CLASSICAL ANALOGUE.– They behave like particles in momentum space or k space.




• Phonons are one example of many like this in many areas of physics. Such quantum mechanical particles are often called

“Quasiparticles”





“Quasiparticles”Examples of other Quasiparticles:

Photons: Quantized Normal Modes of electromagnetic waves.






Photons: Quantized Normal Modes of electromagnetic waves.Rotons: Quantized Normal Modes of molecular rotational excitations.

Magnons: Quantized Normal Modes of magnetic excitations in magnetic solids






Photons: Quantized Normal Modes of electromagnetic waves.Rotons: Quantized Normal Modes of molecular rotational excitations.

Magnons: Quantized Normal Modes of magnetic excitations in magnetic solids

Excitons: Quantized Normal Modes of electron-hole pairsPolaritons: Quantized Normal Modes of electric polarization excitations in solids

+ Many Others!!!

sphonon

hE

Phonon Wavelength:λphonon ≈ a ≈ 10-10 m

phonon

hp

photon

hcE

photon

hp

Comparison of Phonons & Photons

(visible) Photon Wavelength:λphoton ≈ 10-6 m >> a

PHONONS • Quantized normal modes

of lattice vibrations. The energies & momenta of phonons are quantized

PHOTONS • Quantized normal modes

of electromagnetic waves. The energies & momenta of photons are quantized

2

1nn

n = 0,1,2,3,..

E

Quantum Mechanical Simple Harmonic Oscillator• Quantum mechanical results for a simple

harmonic oscillator with classical frequency ω:

The energy is quantized.

En

The energylevels are

equally spaced!

Often, we consider En as being constructed by adding n excitation quanta of energy to the ground state.

2

10

If the system makes a transition from a lower energy level to a higher energy level, it is always true that the change in energy is an integer multiple of

Phonon Absorptionor Emission

Oscillator Ground State(or “zero point”) Energy.

ΔE = (n – n΄)

n & n ΄ = integers

In complicated processes, such as phonons interacting with electrons or photons, it is known that

The number of phonons is NOT conserved.That is, phonons can be created & destroyed during such

interactions.

E0 =

Thermal Energy & Lattice VibrationsAs was already discussed in detail, the atoms in a crystal vibrate about their equilibrium positions.

This motion produces vibrational waves.The amplitude of this vibrational motion increases as the temperature increases.

In a solid, the energy associated with these vibrations is called the

Thermal Energy

• A knowledge of the thermal energy is fundamental to obtaining an understanding many of the basic properties (thermodynamic properties & others!) of solids.

ExamplesHeat Capacity, Entropy, Helmholtz Free Energy, Equation of State, etc....

• A relevant question is how is this thermal energy calculated? We would like to know how much thermal energy is available to scatter a conduction electron in a metal or a semiconductor. This is important because this scattering contributes to electrical resistance & other transport properties.

• How is this thermal energy calculated? We would like to know how much thermal energy is available to scatter a conduction electron in a metal or a semiconductor. This is important because this scattering contributes to electrical resistance & other transport properties. Most important, the thermal energy plays a fundamental role in determining the

Thermal (Thermodynamic)Properties of a Solid

• Knowledge of how the thermal energy changes wih temperature gives an understanding of heat energy necessary to raise the temperature of the material. An important, measureable property of a solid is its

Specific Heat or Heat Capacity

The Thermal Energy is the dominant contribution to the heat capacity in most solids. In non-magnetic insulators, it is the only contribution. Other contributions: Conduction Electrons in metals & semiconductors.Magnetic ordering in magnetic materials. The calculation of the vibrational contribution to the thermal energy & heat capacity of a solid has 2 parts:

1. Evaluation of the contribution of a single vibrational mode.

2. Summation over the frequency distribution of the modes.

Lattice Vibrational Contribution tothe Heat Capacity

Vibrational Specific Heat of Solidscp Data at T = 298 K

17

• In 1819, using room temperature data, Dulong and Petit empirically found thatthe molar heat capacity for solids is approximately

3VC RHere, R is the gas constant.

This relationship is now known as the

Dulong-Petit “Law”

Historical Background

18

•Assume that the heat supplied to a solid istransformed into the vibrational kinetic

& potential energies of the lattice.•To explain the Dulong-Petit “Law” theoretically, usingClassical, Maxwell-Boltzmann Statistical Mechanics, a knowledge of how the heat is divided up among the degrees of freedom of the solid is needed.

The Molar Heat Capacity

19

•The Dulong-Petit “Law” can be explained using Classical Maxwell-Boltzmann statistical physics.Specifically,

The Equipartition Theorem can be used.•This theorem states that, for a system in thermal equilibrium with a heat reservoir at temperature T,

The thermal average energy per degree of freedom is (½)kTIf each atom has 6 degrees of freedom: 3 translational &

3 vibrational, then

13

26AE N kT RT

R = NA k

The Molar Heat CapacityThe Molar Energy of a Solid

20

VV

EC

T

By definition, the heat capacity of a substanceat constant volume is

Classical physics therefore predicts:

3VC R

A value independent of temperature

The Molar Heat CapacityHeat Capacity at Constant Volume

21

The Molar Heat CapacityExperimentally, the Dulong-Petit Law, however, is found to be valid only at high temperatures.

22

Einstein Model of a Vibrating SolidIn 1907, Einstein extended Planck’s ideas to matter:he proposed that the energy values of atoms are quantized and proposed the following simple model of a vibrating solid:

, 0,1,2,...nE n n

hf

Each atom is independentEach vibrates in 3-dimensionsEach vibrational normal modehas energy:

23

In effect, Einstein modeled one mole of a solidas an assembly of 3NA distinguishable oscillators. He used the Canonical Ensembleto calculate the average energy of an oscillator in this model.

0

/

0

( )

n

n B nn

E kTn

n

E E f E

A E e

24

To compute the average, note that it can be written as 1 dZ

EZ d

0 0

( ) nB

n n

Z f n e

Here, [1/(kT)]Z is called

The Partition Function

Z has the form:

25

0

1

1n

n

Z ee

This follows from thegeometric series result

0

1, | | 1

1n

n

S x xx

With [1/(kT)] and En = n, , the partition function Z for the Einstein Model is

26

Differentiating Z withrespect to gives:

1

0

(1 )n

n

Z e e

2(1 )

dZ e

d e

1 dZE

Z d

Multiplying by –1/Z gives:

1

1

dZE

Z d e

This is the Einstein Model Result for the averagethermal energy of an oscillator. The total vibrationalenergy of the solid is just 3NA times this result.

27

2 /

/ 23

( 1)

E

E

T TE

V T T

TE eC R

T T e

The Heat Capacity in the Einstein Modelis given by:

VV

EC

T

Do the derivative & define TE /k. TE is called The Einstein Temperature Finally, in the Einstein Model, CV has

the form:

28

The Einstein Model of a Vibrating SolidEinstein, Annalen der Physik 22 (4), 180 (1907)

CV for Diamond

exp( / )n Bk T

Thermal Energy & Heat Capacity:The Einstein Model: Another Derivation

•The following assumes that you know enoughstatistical physics to have seen the CannonicalEnsemble & the Boltzmann Distribution!•The Quantized Energy of a Single Oscillator hasthe form:

•If the oscillator interacts with a heat reservoir atabsolute temperature T, the probability Pn of itbeing in quantum level n is proportional to theBoltzmann Factor: Pn

nn

nP _

•In the Cannonical Ensemble, a formal expression for the

average energy of the harmonic oscillator & therefore ofa lattice normal mode of angular frequency ω attemperature T is given by:

•The probability Pn of the oscillator being in quantumlevel n has the form:

Pn [exp (-β)/Z]where the partition function Z is given by:

0 0

( ) nB

n n

Z f n e

Quantized Energy of a Single Oscillator: Quantized Energy of a Single Oscillator:

0

/ 2 3 / 2 5 / 2

/ 2 / 2 /

/ 2 / 1

1exp[ ( ) ]

2

.....

(1 .....

(1 )

B B B

B B B

B B

n B

k T k T k T

k T k T k T

k T k T

z nk T

z e e e

z e e e

z e e

_0

0

1 1exp /

2 2

1exp /

2

Bn

Bn

n n k T

n k T

Putting in the explicit form gives:

According to the Binomial expansion, for x << 1 where

/ Bx k T

nn

nP _

Now, some straightforward math manipulation!

Average Energy:

The equation for ε can be rewritten:

_

/

1

2 1Bk Te

_2 2

/ 2_2

/

_/ 2 /2

_/2

/2 2_

22 2 /

1(ln )

ln1

ln ln 1

ln 12

2

4 1

B

B

B B

B

B

B

B B

k T

B k T

k T k TB

k TB

B

k TB

B BB k T

B

zk T k T z

z T T

ek T

T e

k T e eT

k T eT k T T

ke

k k Tk T

k T e

/

/

1

2 1

B

B

k T

k T

e

e

Finally, theresult is:

_

/

1

2 1Bk Te

•This is the Mean Phonon Energy. The first term in(1) is called the Zero-Point Energy. As mentionedbefore, even at 0ºK the atoms vibrate in the crystal &have a zero-point energy. This is the minimum energyof the system.•The thermal average number of phonons n(ω) atTemperature T is given by The Bose-EinsteinDistribution, & the denominator of the secondterm in (1) is often written:

1

1)(

TBken

(1)

_

/

1

2 1Bk Te

•By using (2) in (1), (1) can be rewritten:

<> = ћω[n() + ½]•In this form, the mean energy <> looks analogous to aquantum mechanical energy level for a simple harmonicoscillator. That is, it looks similar to:

1

1)(

TBken

•So the second term in the mean energy (1) is interpreted as

The number of phonons at temperatureT & frequency ω.

(1) (2)

At high T, <> is independent of ω. This high T limit is equivalent to the classical limit, (the energy steps are small compared to the total energy). So, in this case, <> is the thermal energy of the classical 1D harmonic oscillator (given by the equipartition theorem).

..........!2

12

x

xex

Tke

B

TBk

1 112

1_

TkB

_ 1

2 Bk T

_

Bk T

High Temperature Limit:ħω << kBT

T

2

1 TkB

Temperature dependence of the mean energy <> of a

quantum harmonic oscillator.

Taylor’s series expansion of ex for x << 1

_

/

1

2 1Bk Te

Low Temperature Limit:ħω > > kBT

T

2

1 TkB

Temperature dependence of the mean energy <> of a

quantum harmonic oscillator.

_

/

1

2 1Bk Te

“Zero Point Energy”

2

1_

At low T, the exponential in the denominator of the 2nd term gets larger as T gets smaller. At small enough T, neglect 1 in the denominator. Then, the 2nd term is e-x, x = (ħω/(kBT). At very small T, e-x 0. So, in this case, <> is

independent of T: <> (½)ħω

Einstein Heat Capacity CV

• The heat capacity CV is found by differentiating the average phonon energy:

2

2

1

k TB

k TB

B

Bv

ke

k TdC

dTe

2

2 2

1

k TB

k TB

v B

B

eC k

k T e

k

2

2

1

T

T

v B

eC k

T e

Let

_

/

1

2 1Bk Te

Area =2

T

Bk

Bk

The specific heat CV in this approximation vanishes exponentially at low T & tends to the classical value at high T. These features are common to all quantum systems; the energy tends to the zero-point-energy at low T & to the classical value at high T.

vC

where

2

2

1

T

T

v B

eC k

T e

k

Einstein Heat Capacity CV

,T K

3R The figure shows that

Cv = 3Rat high temperatures for all substances. This is the classical

Dulong-Petit law.This states that specific heat of a given number of atoms of any solid is independent of temperature & is the same for all materials!

vC

•The specific heat at constant volume Cv dependsqualitatively on temperature T as shown in thefigure below. For high temperatures, Cv (per mole) isclose to 3R (R = universal gas constant. R 2 cal/K- mole).

So, at high temperatures Cv 6 cal/K-mole

Einstein Model for Lattice Vibrations in a SolidCv vs T for Diamond

Einstein, Annalen der Physik 22 (4), 180 (1907)

Points:ExperimentCurve: Einstein Model

Prediction

Einstein Model of Heat Capacity of Solids• The Einstein Model was the first quantum theory of lattice

vibrations in solids. He made the assumption that all 3N vibrational modes of a 3D solid of N atoms had the same frequency, so that the whole solid had a heat capacity 3N times

3 3BNk R

• In this model, the atoms are treated as independent oscillators, but the energies of the oscillators are the quantum mechanical energies. This assumes that the atoms are each isolated oscillators, which is not at all realistic. In reality, they are a huge number of coupled oscillators. Even this crude model gives the correct limit at high temperatures, where it reproduces the Dulong-Petit law of 3R per mole.

2

2

1

T

T

v B

eC k

T e

At high temperatures, all crystalline solids have a vibrational specific heat of 6 cal/K per mole; they require 6 calories per mole to raise their temperature 1 K. This arrangement between observation and classical theory breaks down if the temperature is not high. Observations show that at room temperatures and below the specific heat of crystalline solids is not a universal constant.

6cal

Kmol

Bk

vC

T3vC R

In each of these materials (Pb,Al, Si,and Diamond) specific heat approaches a constant value asymptotically at high T. But at low T, the specific heat decreases towards zero which is in a complete contradiction with the above classical result.

• The Einstein model also gives correctly a specific heat tending to zero at absolute zero, but the temperature dependence near T= 0 does not agree with experiment.

• Taking into account the actual distribution of vibration frequencies in a solid this discrepancy can be accounted using one dimensional model of monoatomic lattice

Density of StatesAccording to Quantum Mechanics if a particle is constrained;

• the energy of particle can only have special discrete energy values.• it cannot increase infinitely from one value to another.• it has to go up in steps.

Thermal Energy & Heat Capacity Debye Model

• These steps can be so small depending on the system that the energy can be considered as continuous.

• This is the case of classical mechanics.• But on atomic scale the energy can only jump by a

discrete amount from one value to another.

Definite energy levels

Steps get small Energy is continuous

• In some cases, each particular energy level can be associated with more than one different state (or wavefunction )

• This energy level is said to be degenerate.

• The density of states is the number of discrete states per unit energy interval, and so that the number of states between and will be .

( )

( )d d

There are two sets of waves for solution;

• Running waves

• Standing waves

0

2

L

2

L

4

L

4

L

6

L

k

These allowed k wavenumbers correspond to the running waves; all positive and negative values of k are allowed. By means of periodic boundary condition

2 2 2NaL Na p k p k p

p k Na L

an integer

Length of the 1D chain

Running waves:

These allowed wavenumbers are uniformly distibuted in k at a density of between k and k+dk. R k

running waves 2R

Lk dk dk

In some cases it is more suitable to use standing waves,i.e. chain with fixed ends. Therefore we will have an integral number of half wavelengths in the chain;

Standing waves:

0L

2

L

6

L

4

L

5

L

2 2;

2 2

n n nL k k k

L L

3

L

7

L

k0

L

2

L

3

L

These are the allowed wavenumbers for standing waves; only positive values are allowed.

2k p

L

for

running wavesk p

L

for

standing waves

These allowed k’s are uniformly distributed between k and k+dk at a density of

( )S

Lk dk dk

2R

Lk dk dk

DOS of standing wave

DOS of running wave

( )S k

The density of standing wave states is twice that of the running waves.

However in the case of standing waves only positive values are allowed

Then the total number of states for both running and standing waves will be the same in a range dk of the magnitude k

The standing waves have the same dispersion relation as running waves, and for a chain containing N atoms there are exactly N distinct states with k values in the range 0 to ./ a

modes with frequency from to +d corresponds to

modes with wavenumber from k to k+dk

The density of states per unit frequency range g():

The number of modes with frequencies

& + d will be g()d.

g() can be written in terms of S(k) and R(k).

dn

dR

Choose standing waves to obtain ( )g

Let’s remember dispertion relation for 1D monoatomic lattice

2 24sin

2

K ka

m 2 sin

2

K ka

m

dk

ddk

d

d

dk

2cos

2 2

a K ka

m 1

cos2

K kaa

m

;( ) ( )Rdn k dk g d ( ) ( )Sdn k dk g d

( ) ( )Sg k

( ) ( )Sg k

( ) ( )Sg k 1 1

cos / 2

m

a K ka

2cos 1 sin2 2

ka ka

2 2 2sin cos 1 cos 1 sinx x x x

( ) ( )Sg k 2

1 1 4

41 sin

2

m

a K ka

Multibly and divide

( ) ( )Sg k 2

1 2

4 4sin

2

a K K kam m

( )S

Lk dk dk

Let’s remember:

( )gL

2 2max

2 1

a

L Na

2max

4K

m

2 24sin

2

K ka

m

True density of states

constant density of states

N m

K

max 2K

m

K

m

True density of states by means of above equation

1/ 22 2max

2( )

Ng

True DOS(density of states) tends to infinity at ,

since the group velocity goes to zero at this value of .

Constant density of states can be obtained by ignoring the dispersion of sound at wavelengths comparable to atomic spacing.

max 2K

m

/d dk

( )g

The energy of lattice vibrations will then be found by integrating the energy of single oscillator over the distribution of vibration frequencies. Thus

/0

1

2 1kTg d

e

1/ 22 2max

2N

Mean energy of a harmonic oscillator

One can obtain same expression of by means of using running waves.

for 1D

It should be better to find 3D DOS in order to compare the results with experiment.

( )g

3D DOS• Let’s do it first for 2D, then for 3D.

• Consider a crystal in the shape of 2D box with crystal sides L.

+

+

+ -

-

-

L0

L

y

x

L

Standing wave pattern for a 2D box

Configuration in k-space

xk

yk

L

Let’s calculate the number of modes within a range of wavevector k.

Standing waves are choosen but running waves will lead same expressions.

Standing waves will be of the form

Assuming the boundary conditions of

Vibration amplitude should vanish at edges of

Choosing

0 sin sinx yU U k x k y

0; 0; ;x y x L y L

;x y

p qk k

L L

positive integer

+

+

+ -

-

-

L0

L

y

x

The allowed k values lie on a square lattice of side in the positive quadrant of k-space.

These values will so be distributed uniformly with a density of per unit area.

This result can be extended to 3D.

Standing wave pattern for a 2D box

L

L

Configuration in k-space

/ L

2/L

yk

xk

L

L

L

Octant of the crystal:

kx,ky,kz(all have positive values)

The number of standing waves;

3

3 3 33s

L Vk d k d k d k

/L

k

dk

zk

yk

xk

214

8k dk

3 23

14

8s

Vk d k k dk

2

322s

Vkk d k dk

2

22S

Vkk

• is a new density of states defined as the number

of states per unit magnitude of in 3D.This eqn can be obtained

by using running waves as well.

• (frequency) space can be related to k-space:

2

22

Vkk

g d k dk dkg k

d

Let’s find C at low and high temperature by means of using the

expression of . g

B

Tk

High and Low Temperature Limits

This result is true only if at low T’s only lattice modes having low frequencies can be excited from their ground states;

3 BNk T Each of the 3N lattice modes of a crystal containing N atoms

dC

dT

3 BC Nk

k

a0

Low frequency long

sound waves

sv k svk

=

depends on the direction and there are two transverse, one longitudinal acoustic branch:

2

22

Vk dkg

d

1 1

ss s

k dkv

k v d v

and

2

2

2

1

2s

s

Vv

gv

at low T

sv

2 2

2 3 2 3 3

1 1 2

2 2s L T

V Vg g

v v v

Velocities of sound in longitudinal and transverse direction

/0

1

2 1kTg d

e

33

3

/0 01 1

B

BkT x

k Tx

k Td dx

e e

Zero point energy = z

2

/ 2 3 30

1 1 2

2 1 2kTL T

Vd

e v v

3

2 3 3 /0

1 2

2 1z kT

L T

Vd

v v e

B

xk T

Bk Tx

Bk T

d dx

4

43 3

/ 30 0

15

1 1B

kT x

k T xd dx

e e

4 4

2 3 3 3

1 2

2 15B

zL T

k TV

v v

32

3 3

2 1 2

15B

v BL T

k TdC V k

dT v v

at low temperatures

Debye Model of Heat Capacity of Solids

How good is the Debye approximation at low T?3

23 3

2 1 2

15B

v BL T

k TdC V k

dT v v

The lattice heat capacity of solids thus varies as T3 at low temperatures; this is referred to as the Debye T3 law. The figure illustrates the excellent agreement of this prediction with experiment for a non-magnetic insulator. The heat capacity vanishes more slowly than the exponential behaviour of a single harmonic oscillator because the vibration spectrum extends down to zero frequency.

The Debye interpolation scheme The calculation of is a very heavy calculation for 3D, so it must be

calculated numerically.

Debye obtained a good approximation to the resulting heat capacity by neglecting the dispersion of the acoustic waves, i.e. assuming

for arbitrary wavenumber. In a one dimensional crystal this is equivalent to taking as given by the broken line of density of states figure rather than full curve. Debye’s approximation gives the correct answer in either the high and low temperature limits, and the language associated with it is still widely used today.

( )g

sk

( )g

1. Approximate the dispersion relation of any branch by a linear extrapolation of the small k behaviour:

Einstein approximation to the dispersion

Debye approximation to the dispersion

vk

The Debye approximation has two main steps:

23

9( )

D

Ng

2 3 3 3 3

1 2 3 9( ) 3

2 L T D D

V N N

v v

Debye cut-off frequency

2. E nsure the correct number of modes by imposing a cut-off frequency , above which there are no modes. The cut-off freqency is chosen to make the total number of lattice modes correct. Since there are 3N lattice vibration modes in a crystal having N atoms, we choose so that

0

( ) 3D

g d N

2

2 3 3

1 2( ) ( )

2 L T

Vg

v v

22 3 3

0

1 2( ) 3

2

D

L T

Vd N

v v

32 3 3

1 2( ) 3

6 DL T

VN

v v

D

D

D

2( ) /g

The lattice vibration energy of

becomes

and,

First term is the estimate of the zero point energy, and all T dependence is in the second term. The heat capacity is obtained by differentiating above eqn wrt temperature.

/0

1( ) ( )2 1Bk T

E g de

3 32

/ /3 30 0 0

9 1 9( )2 1 2 1

D D D

B Bk T k TD D

N NE d d d

e e

3

/30

9 9

8 1

D

BD k TD

N dE N

e

dEC

dT

/2 4

23 2 /0

9

1

D B

B

k T

Dk T

D B

dE N eC d

dT k T e

3

/30

9 9

8 1

D

BD k TD

N dE N

e

Let’s convert this complicated integral into an expression for the specific heat changing variables to

and define the Debye temperature

B

xk T

DD

Bk

The heat capacity is

d kT

dx

kTx

4 /2 4

23 20

9

1

D T xB B

Dx

D B

k T k TdE N x eC dx

dT k T e

3 / 4

20

91

D T x

D Bx

D

T x eC Nk dx

e

The Debye prediction for lattice specific heat

DD

Bk

where

3 /2

0

9 3D T

D D B BD

TT C Nk x dx Nk

How does limit at high and low temperatures?

High temperature

DC

DT 2 3

12! 3!

x x xe x

4 4 42

2 2 2

(1 ) (1 )

1 11

x

x

x e x x x xx

xxe

x is always small

DT

3 / 4

20

91

D T x

D D Bx

D

T x eT C Nk dx

e

3412

5B

DD

Nk TC

How does limit at high and low temperatures?

Low temperature

44 /15

DC

For low temperature the upper limit of the integral is infinite; the integral is then a known integral of .

We obtain the Debye law in the form3T

T < <

Lattice heat capacity due to Debye interpolation scheme

Figure shows the heat capacity between the two limits of high and low T as predicted by the Debye interpolation formula. 3 B

C

Nk T

Because it is exact in both high and low T limits the Debye formula gives quite a good representation of the heat capacity of most solids, even though the actual phonon-density of states curve may differ appreciably from the Debye assumption.

Debye frequency and Debye temperature scale with the velocity of sound in the solid. So solids with low densities and large elastic moduli have high . Values of for various solids is given in table. Debye energy can be used to estimate the maximum phonon energy in a solid.

Lattice heat capacity of a solid as predicted by the Debye interpolation scheme

3 / 4

20

91

D T x

D B xD

T x eC Nk dx

e

/ DT

1

1

Solid Ar Na Cs Fe Cu Pb C KCl

93 158 38 457 343 105 2230 235( )D K

D

D D

chapter 5: phonons ii – thermal properties. what is a phonon? we’ve seen that the physics of...

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