T= 800 C Gx) GS -GP G 61 G(x) T= 800 C -GATV) a a+B Si P 12 Xmax Si Xmax 12 X = Xp X = XP The phase diagram of silicon-phosphorus (Si-P) is given above, together with the Gibbs free energy curves of the a and B 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 G = Xs: Gsi + XpGp + RT( xsi In(xsi) + xp ln(xp) + xp In(y)) where y is a constant equal to y = 1/2 For the B phase, the Gibbs free energy is very sharp around xp = 1/2 and its minimum value is G(1/2) = 1/2Gsi +/2Gp + AG where AG = -20,600 J/mol. (1) Give the expression of the chemical potential of Si in the a phase u si in terms of Gsi and xsi? Show your derivation. (2) Give the expression of the chemical potential of P in the a phase 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 G at x = Xmax must intersect with the tip of G. From this observation, show that Xmax must fulfill 2AG = RT ln(yXmax(1-Xmax.)). (4) Since the solubility is expected to be small, it is safe to make the approximation Xmax(1-xmax) Xmas Using this approximation, calculate the solubility Xmax. (y = 1/2, AG =-20,600 J/mol, T = 800 C) T= 800 C Gx) GS -GP G 61 G(x) T= 800 C -GATV) a a+B Si P 12 Xmax Si Xmax 12 X = Xp X = XP The phase diagram of silicon-phosphorus (Si-P) is given above, together with the Gibbs free energy curves of the a and B 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 G = Xs: Gsi + XpGp + RT( xsi In(xsi) + xp ln(xp) + xp In(y)) where y is a constant equal to y = 1/2 For the B phase, the Gibbs free energy is very sharp around xp = 1/2 and its minimum value is G(1/2) = 1/2Gsi +/2Gp + AG where AG = -20,600 J/mol. (1) Give the expression of the chemical potential of Si in the a phase u si in terms of Gsi and xsi? Show your derivation. (2) Give the expression of the chemical potential of P in the a phase 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 G at x = Xmax must intersect with the tip of G. From this observation, show that Xmax must fulfill 2AG = RT ln(yXmax(1-Xmax.)). (4) Since the solubility is expected to be small, it is safe to make the approximation Xmax(1-xmax) Xmas Using this approximation, calculate the solubility Xmax. (y = 1/2, AG =-20,600 J/mol, T = 800 C)