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I can't understand why in this phase diagram

http://upload.wikimedia.org/wikipedia/commons/thumb/5/53/Eutectic_system_phase_diagram.svg/350px-Eutectic_system_phase_diagram.svg.png,

in the area where the system is approaching the 100% $\alpha$ or 100% $\beta$ composition, the state is just a pure solid. I mean, where does the other component go? If I am interpreting it right, there's no liquid-phase mixture left, and there's neither any solid rest from the other substance.

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This used to confuse me too! The key is that the $\alpha$ and $\beta$ phases are not necessarily pure component A or B. For example $\alpha$ is a single phase which consists of mostly component A with small amounts of component B dissolved in it - it is a single "A-rich" phase. The B atoms are distributed throughout the A crystal but do not form into separate regions. If you looked into a microscope you would see only one kind of crystal throughout. In contrast, in the $\alpha + \beta$ region you will see two different kinds of crystal one where there is mostly A with B dissolved in it and one where there is B with a little A dissolved in it.

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Thanks! That completely answers my question. However, what will happen if we were to lower even more the temperature of the almost-pure sample? According to the diagram it would turn into $\alpha$ + $\beta$, does it mean that the atoms of the low-concentration component would "get closer" until we could see some of them with the naked eye? How could this even happen at an atomic level if the high-concentration element particles have already organized into a solid body? –  carllacan Jan 14 '13 at 21:07
    
As you cool down into the two phase region the A-rich crystal can no longer hold the B in solution. The B will start to diffuse out of the A rich crystals and form new B rich crystals. This will normally happen at the grain boundaries and so you end up with lots of tiny B-rich crystals around the larger A-rich crystals. This does indeed all happen in the solid phase. Even in a solid, atoms can diffuse around the lattice. But if you cool faster than the atoms can diffuse you can force a material into a single phase. But then the standard phase diagram in the original question would not apply. –  James Hoyland Oct 3 '13 at 9:54
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At 100 \% there is only one component present, right? So there is no missing component.

I am guessing that your problem lies in the "approaching 100 %" part of the question. Now how can that happen. Either one is fishing one component out of the system, leaving the other alone, or one is adding huge amounts of the other component making the percentage of the first smaller and smaller. In either case only one component is left and the other component has either been removed from the solution or is present at zero relative concentration.

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Suppose we have a number of mixtures of $\alpha$ and $\beta$. If we pick one with a 99% concentration of, say $\alpha$ at a temperature above $\alpha$'s fusion point. When the temperature drops to below the L+$\alpha$ - Solid $\alpha$ equilibrium line, what happens? According to the diagram we have pure solid $\alpha$. Where does the other component go? –  carllacan Jan 14 '13 at 4:07
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It doesn't mean that you've only got $\beta$ material, what it means is that you've got a material which has the structure of a pure $\beta$ material with some impurities of $\alpha$ material in a given proportion. If you looked at a good chunk of it with the naked eye (except in some particular cases), you would never tell the difference between a pure $\beta$ material and this non-pure alloy. This case is akin to the case of the aliovalent impurities in a semiconductor. It's not exactly, rigorously the same idea, but it's a similar one and helps you getting it right. –  user17581 Jan 14 '13 at 10:26
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