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Solution Key to Problem Set 6 –––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––
1. Consider an ordered alloy containing a total number of N lattice sites divided into NX^ ^ sites and NX^ ^ sites. In the fully ordered state the sites are occupied by only A atoms and the sites by only B atoms so that the relative fractions of A and B atoms in the stoichiometric composition are X (^) A and X (^) B. A long range structural order parameter, , may defined as:
= ( r^ X A) ( 1 X A) or = 1 ( X B^ ^ X B)
where r^ is the probability that an site is occupied by the “right” atom, A, and X (^) B^ ^ is the fraction of sites occupied by the “wrong” atom, B.
1.1. Show that the expression on the left, proposed by Bragg & Williams, is equivalent to the expression on the right, and that the latter expression has the right limits for an order pa- rameter, i.e. = 1 for the fully ordered state and = 0 for the random state, for the L1 2 compound AuCu 3.
r^ is the number of sites occupied by the right atom divided by the number of sites.
r^ =
NX^ X B^ NX
NX^
= 1 X B^
Neglecting vacancies the fraction of and sites is that of the A and B atoms for the stoichiometric compound, and from mass balance: X (^) A + X (^) B = 1
Substitution for r^ and 1-XA = XB in to the LHS of the original equation yields:
1 X B^ ^ X A
1 X A
X B X B^
X B
X B
X B
We note now that for the stoichiometric compound X (^) B^ ^ varies from 0 for full order to XB for full disorder (random occupancy) since the probability of any site being occupied by a B atom in the disordered state is essentially that corresponding to the composition. Hence, must vary from 1 to zero as X B^ varies from 0 to XB.
1.2. Sketch the expected variation in the order parameter as a function of temperature for the order/disorder transition in AuCu 3 and comment on the physical significance of the shape of the curve below T (^) c. The phase diagram indicates a first order transition for AuCu 3 between the fcc and L1 2 structures, as reflected by the presence of two-phase fields between the ordered and disor- dered states for off-stoichiometric compositions. The nature of the transition is also exten- sively discussed in the literature, so it is not too difficult to find in a quick internet search. Hence, the variation of order parameter with temperature is shown below. At 0K the struc- ture would be completely ordered. As temperature increases, departs from 1 due the in- creasing influence of entropy, which reduces the free energy of the system if some disorder is allowed. At the stoichiometric composition the drop is relatively modest as the enthalpy effect is still strong in this case. At the transition temperature the long-range order parame- ter drops abruptly from some value near 1 to zero, as indicated in the plot.
Solution Key to Problem Set 6 –––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––
1.3. (*) Comment on the implications of deviations from stoichiometry (X (^) Cu :X (^) Au 3) on the or- der parameter as written above. If the composition deviates from stoichiometry one cannot have a fully ordered L1 2 struc- ture, but some interesting effects arise because of the AB 3 composition of the intermetallic compound. For XAu < 0.25 at 0K the Cu sites can be “fully ordered”, i.e. occupied by the right type of atom, but the Au sites will be partially disordered because of the presence of the excess Cu atoms. Conversely, for XAu > 0.25 one may have the Au sites fully ordered but not the Cu sites, which accommodate the excess Au. If is defined with for the Cu sites and B as Au then =1 at 0K for excess Cu even though the Au site has Cu atoms in it, but <1 for the same conditions if are the Au sites and B is Cu. A similar argument can be made for the Au excess side. Hence, the above definition of the order parameter can be misleading but, as you can see, it depends on what we call the “ordered” state. This can have subtle implications in the analysis of diffraction patterns in XRD and TEM.
2. Recall the ordering transformations that lead to the formation of liquid crystals (see Allen & Thomas, Ch. 4), which are characterized by two different order parameters that reflect orientational and translational order.
2.1. For a system that exhibits both smectic and nematic forms, sketch the orientational and translational parameters on the same graph using the transition to the isotropic state as the reference temperature. Sketch a similar graph for the change in order parameter across the melting transition in a metal and discuss the similarities and differences in be- havior with the liquid crystals. The diagram below, on the left, can be drawn from the discussion of ordering in liquid crystals in Allen & Thomas, section 4.2, which is posted on the web site for the course. The transition between nematic and isotropic phases can be compared to the melting transi- tion of a crystalline solid. In the crystalline phase, all atoms are completely ordered. At a critical temperature, the bonds that hold the crystal together are broken, and the crystal be- comes a liquid with no order at all (right figure).
Solution Key to Problem Set 6 –––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––
force for recrystallization, Grxtl, i.e. the difference in strain energy density (strain energy per unit volume) between the small and large grains, as a function of cw, neglecting any capillarity effects. (Do not insert numbers yet.) The driving force for the process is the difference in dislocation energy between the cold worked matrix and the recrystallized grain. The energy associated with the dislocation density, , is, from HW1:
E =
(^2 )μb^2
ln
R
rc
( 2 )μb^2
12 ln ln 2b( )
and the driving force for recrystallization is: G (^) rxtl E (^) annann E (^) cwcw
G (^) rxtl =
(^2 )μb^2
{ ann 12 lnann ^ ln 2b(^ ) cw ^12 lncw ^ ln 2b(^ )}
G (^) rxtl =
( 2 )μb^2
1 { 2 [^ cw lncw ann lnann] +^ [^ cw ann]^ ln 2b(^ )}
3.2. Write now an expression for the net change in free energy associated with the formation of the small grain in terms of cw and its radius. Hint: first write G in terms of Grxtl( cw) and insert the expression for the latter at the end. Clearly identify the nature of each term in your expression (what drives the process, what hinders it, etc.). Derive expressions for the critical radius of nucleation and the energy barrier for the formation of the nucleus from the expression above. Again, write first the expressions in terms of Grxtl and gb , and then insert the expression for Grxtl( cw). In forming the spherical nucleus, we add a new grain boundary between the recrystallized region and the surrounding cold worked matrix. At the same time, we eliminate the seg- ments of the grain boundaries that lie within the recrystallized volume. The total area re- moved for the 6 grain boundaries (between grains 1-4, 2-4, 3-4, 1-2, 2-3, and 1-3), taking them to be circular segments with an angle equal to the characteristic angle between the in- tersecting edges, i.e. 109.5°, i.e. 6 r (^2) ( 109.5 ° 360 °) = 1.825 r 2 Note that this is less than the area associated with the new boundary between the cold worked regions and the spherical grain ( 4 r^2 ), so there is still a barrier to nucleation. The free energy change associated with the formation of a nucleus is then:
G =
r 3
G (^) rxtl + 4 r 2 GB 1.825r 2 GB =
r 3
G (^) rxtl + 2.175r 2 GB
Finding the maximum:
dG
dr
= 4 r 2 G rxtl + 4.35rGB = 0
whereupon r *^ =
1.0875GB
G (^) rxtl
and G *^ =
0.857GB^3
G (^) rxtl^2
Solution Key to Problem Set 6 –––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––
3.3. Plot G as a function of r for values of cw of 10^16 , 10^14 and 10 12 m-2^. Comment on the changes in r* and G* as cw increases. Given are: = 48GPa, = 0.34, ann ~10^6 m -2^ , a = 361pm, gb = 625mJ·m -2^ , b=a2/2, r (^) o =2b (cw) [m-2^ ] 1.00E+16 1.00E+14 1.00E+ G (^) rxtl (J/m^3 ) -9.31E+06 -1.65E+05 -2.37E+ r* [m] 7.30E-08 4.11E-06 2.86E- G* [J] 7.58E-15 2.41E-11 1.17E-
Net Change in Energy due to Formation of Small Grain
-1.0E-
-8.0E-
-6.0E-
-4.0E-
-2.0E-
0.0E+
2.0E-
4.0E-
6.0E-
8.0E-
1.0E-
0.0E+00 2.0E-10 4.0E-10 6.0E-10 8.0E-10 1.0E-09 1.2E-09 1.4E- r (m)
G (J)
1e16/m
Net Change in Energy due to Formation of Small Grain
-2.00E-
-1.50E-
-1.00E-
-5.00E-
0.00E+
5.00E-
1.00E-
1.50E-
2.00E-
2.50E-
3.00E-
0.0E+00 1.0E-06 2.0E-06 3.0E-06 4.0E-06 5.0E-06 6.0E-06 7.0E- r (m)
G (J)
1e14/m
Net Change in Energy due to Formation of Small Grain
-3.00E-
0.00E+
3.00E-
6.00E-
9.00E-
1.20E-
1.50E-
0.0E+00 1.0E-04 2.0E-04 3.0E-04 4.0E-04 5.0E-04 6.0E- r (m)
G (J)
1e12/m
As the dislocation density decreases the driving force for the recrystallization decreases as well. Since both r* and G* are inversely proportional to Grxtl, as the recrystallization driving force decreases the critical radius for nucleation must increase. Which also means that the energy barrier for the formation of the nucleus increases with decreasing disloca- tion density. Note that the critical radius of nucleation is unrealistically large for the lower dislocation densities, suggesting that the degree of cold work must be high to promote nu- cleation of the recrystallized grain in a reasonable time.
Solution Key to Problem Set 6 –––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––
3.5. (*) What would happen if the material were a Cu-alloy instead of pure Cu? Explain.
In principle there would be effects of the composition on the energy of the dislocations be- cause of the tendency for dissociation, as well as an effect on the boundary energies. The mobility of the interface is likely to be affected through the activation barrier across the in- terface, especially if there is a tendency for segregation at the grain boundaries.
