Physical Chemistry Third Edition

28.2 Crystal Vibrations 1167

EXAMPLE28.4

CalculateνDfor copper, which has a density of 8.96 g cm−^1. The speed of sound in copper at

room temperature is 4760 m s−^1 for the longitudinal wave and 2325 m s−^1 for the transverse

waves.

Solution

The molar volume is

Vm

63 .456 g mol−^1

8 .96 g cm−^3

 7 .082 cm^3 mol−^1  7. 082 × 10 −^6 m^3 mol−^1

N

V



6. 02214 × 1023 mol−^1

7. 082 × 10 −^6 m^3 mol−^1

 8. 503 × 1028 m−^3

We take as an average value ofc

c

1

3

[

c(longitudinal)+ 2 c(transverse)

]



1

3

[

(4760 m s−^1 )+2(2325 m s−^1 )

]

3137 m s−^1

ν^3 D

3 Nc^3

4 πV



(3)(8. 503 × 1028 m−^3 )(3137 m s−^1 )^3

4 π

 6. 2645 × 1038 s−^3

νD(6. 2645 ×1038 s−^3 )^1 /^3  8. 557 × 1012 s−^1

If this value is multiplied by 2π, the result is 5.38× 1013 radians s−^1 , in rough agreement

with the value in Figure 28.8.

Exercise 28.5

Find the value of the speed of sound in copper that corresponds to the cutoff frequency in

Figure 28.8, 4. 6 × 1013 radians s−^1.

The canonical partition function is given by Eq. (28.2-3) except that each normal

mode has its own frequency,νi.

Ze−U^0 /kBT

∏^3 N

i 1

⎛

⎝

∑∞

vi 0

e−hνivi/kBT

⎞

⎠e−U^0 /kBT

∏^3 N

i 1

zi (28.2-19)

whereziis the vibrational partition function for normal mode numberi:

zi

∑∞

v 0

e−hνiv/kBT

1

1 −e−hνi/kBT

(28.2-20)

The logarithm of the partition function of the crystal is given by

ln(Z)−

U 0

kBT

+

∑^3 N

i 1

ln(zi)−

U 0

kBT

−

∑^3 N

i 1

ln

(

1 −e−hνi/kBT

)

(28.2-21)