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

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

2nd Revise Page

174 • Chapter 6 / Diffusion

EXAMPLE PROBLEM 6.4

Diffusion Coefficient Determination

Using the data in Table 6.2, compute the diffusion coefficient for magnesium in

aluminum at 550◦C.

Solution

This diffusion coefficient may be determined by applying Equation 6.8; the val-

ues ofD 0 andQdfrom Table 6.2 are 1.2× 10 −^4 m^2 /s and 131 kJ/mol, respectively.

Thus,

D=(1. 2 × 10 −^4 m^2 /s)exp

[

−

(131,000 J/mol)

(8.31 J/mol-K)(550+273 K)

]

= 5. 8 × 10 −^13 m^2 /s

EXAMPLE PROBLEM 6.5

Diffusion Coefficient Activation Energy and

Preexponential Calculations

In Figure 6.8 is shown a plot of the logarithm (to the base 10) of the diffusion

coefficient versus reciprocal of absolute temperature, for the diffusion of cop-

per in gold. Determine values for the activation energy and the preexponen-

tial.

Solution

From Equation 6.9b, the slope of the line segment in Figure 6.8 is equal to

VMSE

D 0 and Qd from

Experimental Data

• Qd/2.3R, and the intercept at 1/T=0 gives the value of logD 0. Thus, the
  activation energy may be determined as

Qd=− 2. 3 R(slope)=− 2. 3 R

⎡

⎢

⎢

⎣

(logD)



(

1

T

)

⎤

⎥

⎥

⎦

=− 2. 3 R

⎡

⎢

⎢

⎣

logD 1 −logD 2

1

T 1

−

1

T 2

⎤

⎥

⎥

⎦

whereD 1 andD 2 are the diffusion coefficient values at 1/T 1 and 1/T 2 , re-

spectively. Let us arbitrarily take 1/T 1 =0.8× 10 −^3 (K)−^1 and 1/T 2 =1.1×

10 −^3 (K)−^1. We may now read the corresponding logD 1 and logD 2 values from

the line segment in Figure 6.8.

[Before this is done, however, a parenthetic note of caution is offered. The

vertical axis in Figure 6.8 is scaled logarithmically (to the base 10); however,

the actual diffusion coefficient values are noted on this axis. For example, for