### Exercise 8: Superlattices of Spherical Au NanoparticlesSpherical gold (Au) nanoparticles, each with a diameter of 4 nm, are known to form face-centered cubic (fcc) superlattices. This exercise involves calculations related to these superlattices:#### a) Maximum Diameter for Octahedral Holes- Calculate the largest possible diameter of particles that can fit into the octahedral holes within the fcc superlattice.#### b) Maximum Diameter for Tetrahedral Holes- Calculate the largest possible diameter of particles that can fit into the tetrahedral holes within the fcc superlattice.#### c) Density of the Nanoparticle Superlattice- Determine the density associated with the gold nanoparticle superlattice structure.These calculations involve understanding the geometry and arrangement of atoms or particles in a crystal lattice, specifically focusing on the available interstitial spaces and the overall density of the lattice.

a) The maximum diameter associated with the octahedra holes in the FCC super lattices: Let us consider diameter of hole is d = 2r where r is radius of hole Diameter of gold (Au) nanoparticle, D = 2R = 4 nm where R is radius of Au nanoparticle 2 D - D = d 2 x 4 - 4 = d 1.414x4 -4 = d d = 5.656 4 = 1.656 nm Maximum diameter of hole is d = 2r = 1.656 nm or D(2-1) = d D (1.414-1) = d 0.414D = d Similarly for tetrahedral holes, the formula for diameter is 0.225D = 0.225 x 4 = 0.9 nm3) Atomic mass of Gold (Au) = 196.97 amu Mass of one Au atom = atomic mass/Avogadro's number = 196.97/6.023 x 1023 = 32.703x 10-23 g In FCC lattice there are 4 number atoms Mass of 4 Au atom = 4 x 32.703x 10-23 g = 130.812 x 10-23 g Diameter of nanoparticle that crystallize as FCC is 4nm Radius, R = diameter/2 = 4/2 = 2 nm = 2 x 10-9 m = 2 x 10-7 cm (100 cm = 1 m) Volume of unit cell, V = R8)3 = (2 x 10-7 cm8)3 = 8 x 10-21 x 22.617 = 180.936 x 10-21 cm3 Density associated with Au nanoparticles in FCC superlattice= mass/volume = 130.812 x 10-23 g/180.936 x 10-21 cm3 = 0.00723 g/cm3