activation energy
relatively small diffusion coefficient. Table 5.2 lists D0 and Qd values for several diffusion systems.
Taking natural logarithms of Equation 5.8 yields ln D = ln D0 - Qd
R a1
Tb (5.9a)
or, in terms of logarithms to the base 10,
log D = log D0 - Qd
2.3Ra1
Tb (5.9b)
Because D0, Qd, and R are all constants, Equation 5.9b takes on the form of an equation of a straight line:
y = b + mx Table 5.2
A Tabulation of Diffusion Data
Diffusing
Species Host Metal D0(m2/s) Qd(J/mol) Interstitial Diffusion
Cb Fe (a or BCC)a 1.1106 87,400
Cc Fe (g or FCC)a 2.3105 148,000
Nb Fe (a or BCC)a 5.0 107 77,000
Nc Fe (g or FCC)a 9.1105 168,000
Self-Diffusion
Fec Fe (a or BCC)a 2.8 104 251,000
Fec Fe (g or FCC)a 5.0105 284,000
Cud Cu (FCC) 2.5105 200,000
Alc Al (FCC) 2.3104 144,000
Mgc Mg (HCP) 1.5 104 136,000
Znc Zn (HCP) 1.5105 94,000
Mod Mo (BCC) 1.8104 461,000
Nid Ni (FCC) 1.9104 285,000
Interdiffusion (Vacancy)
Znc Cu (FCC) 2.4105 189,000
Cuc Zn (HCP) 2.1104 124,000
Cuc Al (FCC) 6.5 105 136,000
Mgc Al (FCC) 1.2104 130,000
Cuc Ni (FCC) 2.7105 256,000
Nid Cu (FCC) 1.9104 230,000
aThere are two sets of diffusion coefficients for iron because iron experiences a phase trans- formation at 912C; at temperatures less than 912C, BCC a-iron exists; at temperatures higher than 912C, FCC g-iron is the stable phase.
bY. Adda and J. Philibert, Diffusion Dans Les Solides, Universitaires de France, Paris, 1966.
cE. A. Brandes and G. B. Brook (Editors), Smithells Metals Reference Book, 7th edition, Butterworth-Heinemann, Oxford, 1992.
dJ. Askill, Tracer Diffusion Data for Metals, Alloys, and Simple Oxides, IFI/Plenum, New York, 1970.
Tutorial Video:
Diffusion Tables How to Use Diffusion Data Found in Table 5.2
where y and x are analogous, respectively, to the variables log D and 1/T. Thus, if log D is plotted versus the reciprocal of the absolute temperature, a straight line should result, having slope and intercept of Qd/2.3R and log D0, respectively. This is, in fact, the manner in which the values of Qd and D0 are determined experimentally. From such a plot for several alloy systems (Figure 5.7), it may be noted that linear relationships exist for all cases shown.
Figure 5.7 Plot of the logarithm of the diffusion coefficient versus the reciprocal of absolute temperature for several metals.
[Data taken from E. A. Brandes and G. B. Brook (Editors), Smithells Metals Reference Book, 7th edition, Butterworth-Heinemann, Oxford, 1992.]
1500 1200 1000 800 600 500 400 300
Diffusion coefficient (m2/s) 10–8
10–10
10–12
10–14
10–16
10–18
10–20
Temperature (°C)
Reciprocal temperature (1000/K)
0.5 1.0 1.5 2.0
Al in Al Zn in Cu
Cu in Cu C in -Fe␥
Fe in -Fe␥
C in -Fe␣
Fe in -Fe␣
Concept Check 5.1 Rank the magnitudes of the diffusion coefficients from greatest to least for the following systems:
N in Fe at 700C Cr in Fe at 700C
N in Fe at 900C Cr in Fe at 900C
Now justify this ranking. (Note: Both Fe and Cr have the BCC crystal structure, and the atomic radii for Fe, Cr, and N are 0.124, 0.125, and 0.065 nm, respectively. You may also want to refer to Section 4.3.)
[The answer may be found at www.wiley.com/college/callister (Student Companion Site).]
Concept Check 5.2 Consider the self-diffusion of two hypothetical metals A and B. On a schematic graph of ln D versus 1/T, plot (and label) lines for both metals, given that D0(A) D0(B) and Qd(A) Qd(B).
[The answer may be found at www.wiley.com/college/callister (Student Companion Site).]
EXAMPLE PROBLEM 5.4
Diffusion Coefficient Determination
Using the data in Table 5.2, compute the diffusion coefficient for magnesium in aluminum at 550C.
Solution
This diffusion coefficient may be determined by applying Equation 5.8; the values of D0 and Qd from Table 5.2 are 1.2 104 m2/s and 130 kJ/mol, respectively. Thus,
D = (1.2 * 10-4 m2/s) expc- (130,000 J/mol) (8.31 J/mol#K)(550 + 273 K)d
= 6.7 * 10-13 m2/s
EXAMPLE PROBLEM 5.5
Diffusion Coefficient Activation Energy and Preexponential Calculations Figure 5.8 shows a plot of the logarithm (to the base 10) of the diffusion coefficient versus re- ciprocal of absolute temperature for the diffusion of copper in gold. Determine values for the activation energy and the preexponential.
Solution
From Equation 5.9b the slope of the line segment in Figure 5.8 is equal to Qd/2.3R, and the intercept at 1/T 0 gives the value of log D0. Thus, the activation energy may be determined as
Qd = -2.3R(slope) = -2.3R £(log D) a1
Tb §
= -2.3R £log D1 - log D2
1 T1 - 1
T2
§ where D1 and D2 are the diffusion coeffi-
cient values at 1/T1 and 1/T2, respectively.
Let us arbitrarily take 1/T1 0.8 103 (K)1 and 1/T2 1.1 103 (K)1. We may now read the corresponding log D1
and log D2 values from the line segment in Figure 5.8.
[Before this is done, however, a note of caution is offered: The vertical axis in Figure 5.8 is scaled logarithmi- cally (to the base 10); however, the actual diffusion coefficient values are noted on this axis. For example, for D 1014 m2/s, the logarithm of D is 14.0, not 1014. Furthermore, this logarithmic scaling affects the readings between decade values; for example, at a location midway between 1014 and 1015, the value is not 5 1015 but, rather, 1014.5 3.2 1015.]
: VMSE D0 and Qd from Experimental Data
Figure 5.8 Plot of the logarithm of the diffusion coefficient versus the reciprocal of absolute temperature for the diffusion of copper in gold.
0.7 0.8 0.9 1.0 1.1 1.2
Diffusion coefficient (m2/s) 10–12
10–13
10–15 10–14
10–16
10–17
Reciprocal temperature (1000/K)
Thus, from Figure 5.8, at 1/T1 0.8 103 (K)1, log D112.40, whereas for 1/T2 1.1 103 (K)1, log D215.45, and the activation energy, as determined from the slope of the line segment in Figure 5.8, is
Qd = -2.3R£
log D1 - log D2
1 T1 - 1
T2
§
= -2.3(8.31 J/mol#K)c -12.40 - (-15.45)
0.8 * 10-3(K)-1 - 1.1 * 10-3(K)-1d
= 194,000 J/mol = 194 kJ/mol
Now, rather than try to make a graphical extrapolation to determine D0, we can obtain a more accurate value analytically using Equation 5.9b, and we obtain a specific value of D (or log D) and its corresponding T (or 1/T) from Figure 5.8. Because we know that log D 15.45 at 1/T 1.1 103 (K)1, then
log D0 = log D + Qd
2.3Ra1 Tb
= -15.45 + (194,000 J/mol)(1.1 * 10-3 [K]-1) (2.3)(8.31 J/mol#K)
= -4.28 Thus, D0 104.28 m2/s 5.2 105 m2/s.
DESIGN EXAMPLE 5.1
Diffusion Temperature–Time Heat Treatment Specification
The wear resistance of a steel gear is to be improved by hardening its surface. This is to be accomplished by increasing the carbon content within an outer surface layer as a result of car- bon diffusion into the steel; the carbon is to be supplied from an external carbon-rich gaseous atmosphere at an elevated and constant temperature. The initial carbon content of the steel is 0.20 wt%, whereas the surface concentration is to be maintained at 1.00 wt%. For this treatment to be effective, a carbon content of 0.60 wt% must be established at a position 0.75 mm below the surface. Specify an appropriate heat treatment in terms of temperature and time for tem- peratures between 900 and 1050C. Use data in Table 5.2 for the diffusion of carbon in g-iron.
Solution
Because this is a nonsteady-state diffusion situation, let us first employ Equation 5.5, using the following values for the concentration parameters:
C0 = 0.20 wt% C Cs = 1.00 wt% C Cx = 0.60 wt% C Therefore,
Cx - C0
Cs - C0 = 0.60 - 0.20
1.00 - 0.20 = 1 - erfa x 21Dtb and thus,
0.5 = erfa x 21Dtb
Using an interpolation technique as demonstrated in Example Problem 5.2 and the data presented in Table 5.1, we find
x
21Dt = 0.4747 (5.10)
The problem stipulates that x 0.75 mm 7.5 104 m. Therefore, 7.5 * 10-4 m
21Dt = 0.4747 This leads to
Dt = 6.24 * 10-7 m2
Furthermore, the diffusion coefficient depends on temperature according to Equation 5.8, and, from Table 5.2 for the diffusion of carbon in g-iron, D0 2.3 105 m2/s and Qd 148,000 J/mol.
Hence,
Dt = D0expa-Qd
RTb(t) = 6.24 * 10-7 m2 (2.3 * 10-5 m2/s) exp c- 148,000 J/mol
(8.31 J/mol#K)(T)d(t) = 6.24 * 10-7 m2 and, solving for the time t, we obtain
t (in s) = 0.0271 expa-17,810
T b
Thus, the required diffusion time may be computed for some specified temperature (in K). The following table gives t values for four different temperatures that lie within the range stipulated in the problem.
Temperature Time
(ⴗC) s h
900 106,400 29.6
950 57,200 15.9
1000 32,300 9.0
1050 19,000 5.3
One technology that applies solid-state diffusion is the fabrication of semiconductor integrated circuits (ICs) (Section 18.15). Each integrated circuit chip is a thin square wafer having dimensions on the order of 6 mm 6 mm 0.4 mm; furthermore, millions of interconnected electronic devices and circuits are embedded in one of the chip faces.
Single-crystal silicon is the base material for most ICs. In order for these IC devices to function satisfactorily, very precise concentrations of an impurity (or impurities) must be incorporated into minute spatial regions in a very intricate and detailed pattern on the silicon chip; one way this is accomplished is by atomic diffusion.
Typically, two heat treatments are used in this process. In the first, or predeposition step, impurity atoms are diffused into the silicon, often from a gas phase, the partial pressure of which is maintained constant. Thus, the surface composition of the impurity