Fundamentals of Materials Science and Engineering: An Integrated Approach, 3e

GTBL042-06 GTBL042-Callister-v3 September 28, 2007 21:46

2nd Revise Page

6.5 Factors That Influence Diffusion • 175

0.7 0.8 0.9 1.0 1.1 1.2

Diffusion coefficient (m

2 /s)

10 –12

10 –13

10 –15

10 –14

10 –16

10 –17

Reciprocal temperature (1000/K)

Figure 6.8 Plot of the

logarithm of the diffusion

coefficient versus the

reciprocal of absolute

temperature for the

diffusion of copper in gold.

D= 10 −^14 m^2 /s, the logarithm ofDis –14.0,not 10 −^14. Furthermore, this log-

arithmic scaling affects the readings between decade values; for example, at a

location midway between 10−^14 and 10−^15 , the value is not 5× 10 −^15 but, rather,

10 −^14.^5 =3.2× 10 −^15 .]

Thus, from Figure 6.8, at 1/T 1 =0.8× 10 −^3 (K)−^1 , logD 1 =–12.40, while

for 1/T 2 =1.1× 10 −^3 (K)−^1 , logD 2 =–15.45, and the activation energy, as

determined from the slope of the line segment in Figure 6.8, is

Qd=− 2. 3 R

⎡

⎢

⎢

⎣

logD 1 −logD 2

1

T 1

−

1

T 2

⎤

⎥

⎥

⎦

=− 2 .3(8.31 J/mol-K)

[

− 12. 40 −(− 15 .45)

0. 8 × 10 −^3 (K)−^1 − 1. 1 × 10 −^3 (K)−^1

]

= 194 ,000 J/mol=194 kJ/mol

Now, rather than trying to make a graphical extrapolation to determine

D 0 , a more accurate value is obtained analytically using Equation 6.9b and a

specific value ofD(or logD) and its correspondingT(or 1/T) from Figure 6.8.

Since we know that logD=–15.45 at 1/T=1.1× 10 −^3 (K)−^1 , then

logD 0 =logD+

Qd

2. 3 R

(

1

T

)

=− 15. 45 +

(194,000 J/mol)(1. 1 × 10 −^3 [K]−^1 )

(2.3)(8.31 J/mol-K)

=− 4. 28

Thus,D 0 = 10 −^4.^28 m^2 /s=5.2× 10 −^5 m^2 /s.