Physical Chemistry Third Edition

28.4 Electrical Resistance in Solids 1181

This equation conforms to Ohm’s law with the conductivity given by

σ

Ne^2 τ

m

(28.4-9)

EXAMPLE28.9

From the density of gold and assuming that one conduction electron comes from each atom,

the density of conduction electrons in gold is equal to 5.90× 1028 m−^3. The resistivity

(reciprocal of the conductivity) is equal to 2.24 microohm cm at 20◦C. Find the value ofτ

that corresponds to these values.

Solution

σ

1

2 .24 microohm cm

(

106 microohm

1 ohm

)(

100 cm

1m

)

 4. 46 × 107 ohm−^1 m−^1

τ

mσ

Ne^2



(9. 1 × 10 −^31 kg)(4. 46 × 107 ohm−^1 m−^1 )

(5. 90 × 1028 m−^3 )(1. 60 × 10 −^19 C)^2

 2. 7 × 10 −^14 s

EXAMPLE28.10

a.In classical gas kinetic theory, the root-mean-square speed of particles of massmis

given by

vrms

√

3 kBT

m

Find the mean speed of electrons at 293 K using this formula.

b.If a current per unit area of 1.00× 106 Am−^2 is flowing in a sample of gold at 293 K,

find the mean drift speed. Find the ratio of the mean drift speed to the root-mean-square

speed of electrons at this temperature.

Solution

a.vrms

(

3(1. 3807 × 10 −^23 JK−^1 )(293 K)

9. 1094 × 10 −^31 kg

) 1 / 2

 1. 154 × 105 ms−^1

b.From Example 28.9 we take the density of mobile electrons in gold to be 5.90× 1028

m−^3. The current in electrons m−^2 s−^1 is

(1. 00 × 106 Cs−^1 m−^2 )

(

1 electron

1. 6022 × 10 −^19 C

)

 6. 24 × 1024 electrons m−^2 s−^1

The drift speed times the density of mobile electrons equals the number of electrons

passing per second.

vdrift

6. 24 × 1024 electrons m−^2 s−^1

5. 90 × 1028 m−^3

 1. 06 × 10 −^4 ms−^1