T=800 C- a Si T Xmax a + 1 1/2 X = Xp P GS I Si Xmax T = 800 C 1/ X = Xp -G(x) -Ga(x) G = XsiGsi + XpGp + RT(xsi ln(xsi) + xp ln(xp) + xp ln(y)) P Go -G (/2) The phase diagram of silicon-phosphorus (Si-P) is given above, together with the Gibbs free energy curves of the a and 3 phases at 800 C. Our goal is to compute the solubility Xmax of P in Si. For the a phase, the Gibbs free energy can be written as where y is a constant equal to y = 2. For the phase, the Gibbs free energy is very sharp around xp = /2 and its minimum value is GB(/2) = /2Gsi +/2Gp + AG where AG = -20,600 J/mol. (1) Give the expression of the chemical potential of Si in the a phase usi in terms of Gsi and xsi? Show your derivation. (2) Give the expression of the chemical potential of P in the a phase up in terms of Gp, xp, and y? Show your derivation. (3) By examining the Gibbs free energy curves, it is seen that the tangent to Gat x = xmax must intersect with the tip of G. From this observation, show that Xxmax must fulfill 2AG = RT In(max(1-Xmax)). (4) Since the solubility is expected to be small, it is safe to make the approximation Xmax(1-Xmax)~Xmax Using this approximation, calculate the solubility Xmax (y=/2, AG = -20,600 J/mol, T = 800 C)