Thermal Physics Lecture 26

7/30/2019 Thermal Physics Lecture 26

1/8

Physics 301 18-Nov-2002 26-1

Reading

Finish K&K Chapter 11. Chapter 12. Start on Chapter 13.

Also, Im passing out a recent article Development of One-Dimensional Band Struc-ture in Articial Gold Chains, N. Nilius, T. M. Wallis, and W. Ho, 2002, Science , 297 ,1853. This article describes observations of the transition of energy levels in single atomsinto energy bands in a linear crystal of atoms.

Examples of Positive Mixing Energy

We saw last time that a solubility gap occurs in a two component mixture when thereis a positive mixing energy and when the temperature is low enough that the mixing energy

contribution to the free energy wins over the mixing entropy contribution. The mixingentropy term is relatively straightforward. The mixing energy depends on the details of the atomic interactions. Well consider some examples.

The rst example has to do with a difference in binding energy between atoms of typeA and type B . We hinted at this possibility last time. Suppose that when two A atomsare nearest neighbors, the interaction energy is uAA (probably less than 0, indicating abinding energy) and similarly two B atoms have an interaction energy uBB and an A-B pair has interaction energy uAB . Assume that atoms are randomly distributed in themixture. Recall that x is the fraction of B atoms. A nearest neighbor of an A atom isanother A atom with probability 1

x and a B atom with probability x. The interaction

energy per nearest neighbor for an A atom is

uA = (1 x)uAA + xu AB ,

and similarly, the interaction energy per nearest neighbor for a B atom is

uB = (1 x)uAB + xu BB .

Suppose that each type of atom has p nearest neighbors. Then the average energy peratom (the factor of 1 / 2 is a double counting correction) is

u = 12 p[(1 x)uA + xu B ] ,

=12 p[(1 x)

2 uAA + 2 x(1 x)uAB + x2 uBB ] ,

=12 p[(1 x)uAA + xu BB ] +

12 px(1 x)[2uAB (uAA + uBB )]

u M .

Copyright c 2002, Princeton University Physics Department, Edward J. Groth


2/8

Physics 301 18-Nov-2002 26-2

In the last line, weve separated out the mixing energy from the energy of the 1 x typeA atoms plus x type B atoms, which we would have for the separated phase. The mixingenergy has the form of a parabola so we can get a bump in the middle of the plot of freeenergy versus composition. We need a positive bump to get a solubility gap. This requires

that the parabola opens downward, which means that

uAB >12

(uAA + uBB ) .

In other words, the strength of the A-B interaction must be less than the average strengthof the A-A and B -B interactions in order to have a solubility gap.

In this case, we can relate the temperature at which a solubility gap appears to theenergies per atom. The idea is that the as the temperature is lowered, the entropy of mixingcontribution to the free energy decreases until a positive bump appears in the free energy

which indicates a solubility gap. The border between homogeneous and heterogeneousmixtures occurs when the free energy is at (linear) as a function of composition. In thiscase, the second derivative of the free energy, with a negative contribution due to theenergy of mixing, and a positive contribution due to the entropy of mixing, must be zero.So,

0 =d2 F M dx2

=d2

dx2(uM M ) ,

=d2

dx212 px(1 x)[2uAB (uAA + uBB )] + [(1 x)log(1 x) + x log x] ,

= p[2uAB (uAA + uBB )] + 1

1 x +1x .

The peak of the bump occurs at x = 1 / 2, so the maximum temperature for which aheterogeneous mixture is the stable equilibrium is

M =p4

[2uAB (uAA + uBB )] .

Another example relates to the gold-silicon mixture we used as an example last time.In this case, the two pure substances have different crystal structures. So it is easy to

imagine that starting from pure gold and adding a small amount of silicon, silicon atoms just substitute for gold atoms in the gold lattice which is a face centered cubic lattice.Similarly, starting from pure silicon and adding gold, the gold just substitutes for siliconin the silicon lattice which is a diamond lattice.

Aside: what are face centered cubic and diamond lattices? Answer: to make a facecentered cubic lattice, start with a cube and put an atom at each corner (8 atoms) andan atom in the center of each face (6 atoms). Now replicate this cube throughout space.



3/8

Physics 301 18-Nov-2002 26-3

Note that each atom in a corner position is shared among 8 cubes, and each atom in a facecenter is shared among two cubes, so there are 4 atoms per cube. Each atom in the facecentered cubic structure has 12 nearest neighbors at distance a/ 2, where a is the sideof the cube. Consider an atom in the center of a face. Its nearest neighbors are the four

corner atoms in the same face and the other face atoms in the two cubes its in, exceptfor the face atoms on the opposite sides of the cubes. Note that if you start with a simplecubic lattice (atoms at the corner positions only) and add three more, each offset by a/ 2along one side of the cube and a/ 2 along another side (there are three possibilities fordoing this), you wind up with the face centered cubic lattice. So corner and face positionsare equivalent.

How about the diamond lattice? Start with a face centered cubic lattice. Add an-other face centered cubic lattice offset by a/ 4 along each side of the cube. Each atomin this lattice has four nearest neighbors at distance 3a/ 4. For example, the atom at(a/ 4, a/ 4, a/ 4) has neighbors at (0 , 0, 0), (0, a/ 2, a/ 2), (a/ 2, 0, a/ 2), and ( a/ 2, a/ 2, 0). Anatom can be thought of as being at the center of regular tetrahedron and its four nearestneighbors can be thought of as being at the vertices of the tetrahedron. The tetrahedralstructure matches the four covalent bonds that carbon (in a diamond!), silicon, germanium,and tin can form.

To complete the story on the crystal structures of gold and silicon, golds face centeredcubic lattice has a = 4 .07 A, a nearest neighbor distance of 2 .88 A and an atomic volumeof 10.2 cm3 mole 1 . Silicons diamond lattice has a = 5 .43 A, a nearest neighbor distanceof 2.35 A and an atomic volume of 12 .0 cm3 mole 1 . All these numbers are for roomtemperature. The bond lengths arent all that different, which means it isnt that hard tosubstitute a few of one kind of atom in the other atoms lattice.

On the other hand, the bond lengths are different, and as the amount of substitutionincreases, the energy will go up. If the energy (or free energy) for silicon substituted in thegold lattice and gold substituted in a silicon lattice are plotted versus composition, theywill cross at some intermediate composition. So not only is there a bump at intermediatecompositions, theres a kink! This will lead to a solubility gap when the temperature islow enough that the entropy of mixing doesnt win. Note that this argument depends onthe fact that the crystal lattices are sufficiently different that they cant be continuouslydeformed into one another.

A third example is a mixture of liquid3

He and4

He. Below 0.87 K, such a mixtureseparates into two components: liquid 3 He containing some dissolved 4 He oats on topof liquid 4 He containing some dissolved 3 He. As the temperature is lowered, almost allthe 4 He condenses into the superuid (BE) ground state with about 6% dissolved 3 He,while the remaining 3 He becomes almost pure. The phase diagram for this mixture isshown in K&K, gure 11.7. The fact that 3 He and 4 He become almost immiscible at lowtemperatures is the basis for operation of the helium dilution refrigerator. In addition, theoverall reason for this effect is amusing.



4/8

Physics 301 18-Nov-2002 26-4

At these low temperatures, the 4 He is essentially all in the superuid ground state (asalready noted). In this state, its energy is all potential energy of interaction among atoms.(That is, no kinetic energy except zero point energy.) The 3 He is a degenerate Fermiliquid with all kinetic energy states up to the Fermi energy populated. Since 3 He and 4 He

have the same electronic structure, one expects similar interaction energies between themolecules. The big difference is the kinetic energy. The kinetic energy per atom is 3 F / 5and the Fermi energy depends on concentration as n 2 / 3 . (See lecture 16.) If we consider ahomogeneous mixture of 3 He and 4 He with x denoting the fraction of the 4 He and 1 xthe fraction of 3 He, then the energy of mixing is proportional to the concentration of 3 Heto the 2 / 3 power or,

uM

(1 x)2 / 3 ,

which starts at 1 at x = 0 and ends up at 0 at x = 1, and this may not seem like a bump,but remember, were concerned with a rise over the straight line connecting the energyper atom when its all 3 He and when its all 4 He. Relative to this line (the straight line

from (0, 1) to (1 , 0)), there is a bump. Another way to see this is to note that the secondderivative with respect to concentration is negative,

d2 uM dx2

29

(1 x) 4 / 3 .

Since we have a bump, we have a solubility gap which, as 0, extends from x = 0 tox = 0 .94.

Liquid and Solid Mixtures Without a Solubility Gap

Weve been considering a two component system thats completely solid or completelyliquid. If we consider the liquid and solid phases of two component systems, things geteven more interesting.

To start with, suppose that neither the liquid nor the solid phase has a solubility gap.Then we plot the free energy for the liquid and the free energy for the solid as a functionof composition and neither curve will exhibit a local maximum. That is, neither curve willhave a bump in the middle which would lead to a solubility gap. At high temperatures, theentropy wins and the liquid free energy curves lies below the solid free energy curve for the

entire range of compositions and any composition will be a liquid at high temperatures. Atlow temperatures, the energy wins, and the solid free energy curve will lie below the liquidcurve for all compositions and any composition is a homogeneous alloy at low temperatures.(These are basically just a restatement that there are no solubility gaps!) There must bea temperature range over which the solid and liquid free energy curves intersect. In thisrange, the appropriate free energy curve is the minimum of the two curves. Since there isan intersection, the free energy curve will be the liquid curve for one range of compositionsand the solid curve for another range. (K&K gure 11.9 would be good a good thing to



5/8

Physics 301 18-Nov-2002 26-5

study at this point.) If the overall minimum curve has a bump in it, then we have thesame kind of situation that we had with the solubility gap. Draw the tangent curve belowthe bump; this is tangent to the minimum free energy curve near the two local minima.One local minimum is in the liquid part of the minimum free energy curve and the other

is in the solid part. Anywhere between the tangent points, the system can lower its freeenergy by splitting into a liquid phase with the composition of the liquid tangent pointand a solid phase with the composition of the solid tangent point. The overall amounts of liquid and solid are determined by the overall composition. (This is essentially the sameargument we had with the solubility gap.)

We can produce a phase diagram for thissituation: a schematic is shown in the gure.The upper curve in the gure is known as theliquidus curve. Above this curve, the systemis a homogeneous liquid. The lower curve iscalled the solidus curve and below this curvethe system is a homogeneous solid. In be-tween the two curves, the equilibrium cong-uration of the system is a mixture of homoge-neous liquid and homogeneous solid, but withdifferent compositions. The composition of the liquid is determined by the liquidus curveat the given temperature and the compositionof the solid by the solidus curve. The relative amounts of liquid and solid are determinedby the overall composition.

Now imagine the following experiment. Start with a homogeneous liquid of somecomposition. Lower the temperature. Eventually the liquidus curve is reached. At thispoint, a further lowering of the temperature causes some of the material to solidify with thecomposition of the solidus curvein general, this will have a greater fraction of the highmelting temperature material than contained in the initial composition. The remainingliquid has more of the lower melting temperature material, so to continue to solidify whatsleft, the temperature has to be lowered still more. In other words, this system does not havea well dened freezing point. It will solidify over a range of temperatures. Furthermore,the composition of the solid changes as solidication proceeds.



6/8

Physics 301 18-Nov-2002 26-6

Liquid and Solid Mixtures With a Solubility Gap

As you can imagine, the number of things that can happen with mixtures is quitelarge. As a nal example, we will consider a two component mixture with liquid and solid

phases. Well suppose there is no solubility gap in the liquid phase (whenever we havea liquid, its homogeneous) but well suppose theres a solubility gap in the solid phase.At low enough temperatures, where theres no liquid phase, we have the same situationas weve discussed before. For intermediate compositions (within the solubility gap) theequilibrium conguration is a heterogeneous mixture of solid A containing some dissolvedB and solid B containing some dissolved A.

Now suppose we have a high enough temperature that the liquid phase is the equi-

librium conguration for some compositions. The free energy curves for this temperature

might look something like those shown (very schematically) in the gure. Starting from theleft we have the free energy curve for a homogeneous solid of mostly A atoms with some Batoms substituted at lattice sites. Next is the free energy curve for a homogeneous liquidof A and B atoms. Finally, the curve on the right is for a homogeneous solid of mostlyB atoms with some A atoms substituted at lattice sites. If the mixture were to remainhomogeneous, its phase (liquid or solid) would be determined by the lowest of the threefree energy curves. But once again, we can draw tangents to the curves and whenever thecomposition lies in a range between tangent points, the equilibrium conguration is a two



7/8

Physics 301 18-Nov-2002 26-7

phase (in this case, liquid and solid) mixture. We can draw two tangents: one connectsthe mostly A solid and the liquid curves and the other connects the liquid and mostly Bsolid curves. The tangent points are indicated in the gure.

At this temperature there are ve possibilities. If the composition is less than xa , wehave a homogeneous solid of mostly A atoms but with some B atoms. Between a and b(xa < x < x b), the equilibrium consists of a homogeneous solid of composition xa and ahomogeneous liquid of composition xb. The relative amounts of liquid and solid determinethe overall composition within the range ( xa , xb). Between b and c (xb < x < x c ), wehave a homogeneous liquid. Between c and d (xc < x < x d ), the equilibrium consists of a homogeneous liquid of composition xc and a homogeneous solid of composition xd . Therelative amounts of liquid and solid determine the overall composition within the range(xc , xd ). Finally, to the right of d, (xd < x ), we have a homogeneous solid of mostly Batoms with some A atoms.

The phase diagram might look something like that shown schematically in the gure.

At high temperatures we have a homogeneous liquid. At low temperatures we have, de-pending on the composition, either homogeneous nearly pure solids, or a heterogeneoussolid mixture. This is just what we discussed last time. At intermediate temperatures, wecan have the ve cases we discussed in the previous paragraph. (If the melting tempera-



8/8

Physics 301 18-Nov-2002 26-8

tures of the two solids arent the same, we can also have a range of temperatures wherethere are three cases.) A mixture with this kind of phase diagram, that is, two liquidusbranches, is called a eutectic system. The minimum temperature at which the mixture canexist as a liquid is the eutectic temperature ( e ) and the composition that remains liquid

at e is called the eutectic composition, xe .

Now suppose we start out in the liquid state with a composition x > x e . We lowerthe temperature. Eventually we reach the right liquidus curve. As we lower the temper-ature still further, the system splits into two phases. The solid phase solidies with thecomposition of the right solidus curve, while the remaining liquid follows the compositionof the liquidus curve. What happens when the composition and temperature reach xe and e ? Answer, the remaining liquid solidies at that temperature. Of course, it solidiesinto the heterogeneous solid weve been discussing. Similarly, if we start with a liquid withcomposition x < x e and lower the temperature, the solid that forms will follow the leftsolidus curve, while the remaining liquid follows the left liquidus curve until it reaches xeand e where it solidies. If we started with the eutectic composition, then the entire sys-tem would reach the eutectic temperature as a liquid, where it would solidify all at once.In this respect (having a single freezing and melting temperature), a eutectic behaves verymuch like a pure substance.

You may be wondering about the fact that a eutectic system can have a melting tem-perature lower than the melting temperature of either pure component. This is somewhatnon-intuitive. The key idea is that the system must come to some compromise betweenminimizing its energy and maximizing its entropy. If we have a pure substance as acrystalline solid, minimizing the energy has won over maximizing the entropy. With twocomponents, with different crystal structures, the energy might still win over the entropyeven allowing for the additional entropy of mixing. However, if we allow the system tobe a liquid, then the energy of mixing may mostly disappear since we are not trying toforce atoms into the wrong crystal structure. Then the entropy of mixing can have asubstantial effect and make the free energy of the liquid lower than that of the solid attemperatures substantially below the melting temperature of either pure substance.


Documents

Thermal Physics Lecture 26