Question

(a) The sublimation energy of bulk gold is 334 Kjmol-1 and the surface energy is 1.5 Jm2. What percentage of
the bulk bonding energy is lost by atoms at the (111) surface of gold.? The unit cell length of gold is 4.08 Ao

b) Further (to a) calculate the melting temperature for a 4 nm gold particle given a bulk melting temperature
of 1273 K and a density of 19500 kgm-3 and a latent heat of fusion of 12.2 Kjmol-1

Answer 1

(a) Gold has the face-centered cubic structure, and the unit cell edge length is 4.08 Å

                Au-Au distance is = 4.08/1.414 = 2.88 Å

In a hexagonal array of gold atoms with this interatomic spacing, the surface area per atom is

                                  (2.88 Å)^² x 0.866 = 7.2 Å^{^2}

the area per mole of Au surface atoms on the (111) crystal face is 4.3*10^4 m^2

multiply this area by the surface energy

               (4.3*10^4)*(1.5J/mol)(1kJ/1000J) = 65 kJ per mole of Au surface atoms

                                                              = 65 kJ/334 kJ x 100% = 19 %