Question

The noble gas argon (atomic weight 39.95 g/mol) can crystalize into a simple hexagonal lattice.The interatomic bond energy as a function of atom separation r is well described a Lennard-Jones type potential:

U(r)= (A/r^12)-(B/r^6)

where A = 1.57x10-14 J-Å12 and B = 1.02x10-17 J-Å6. Given that the c/a ratio for a close-packed hexagonal structure is 1.63, calculate the density of solid Ar in g/cm3. (Avogadro’s constant = 6.022x1023)

Answer 1

mass of 1atom Ar = 39.95 gm/mol / 6.023*10²³ atom/mol = 6.6329*10^-23gm/atom
It is given that c/a = 1.63 for hcp structure
calculate r using the given equation;
   make dU(r) / dr = 0 ( at equilibrium)
   0 = 12(A/r¹³) - 6 (B/r⁷)
12 (1.57x10^-14 / r¹³) - 6 (1.02x10^-17 / r⁷) = 0
  12 (1.57x10^-14 / r¹³) = 6 (1.02x10^-17 / r⁷)
  1.884*10^-13 / r¹³ = 6.12*10^-17 / r⁷
   3078.43137 = r⁶
   r = (3078.43137)^1/6 = 3.8141 Å
   1 Angstorm = 1*10^-8 cm ; r = 3.8141*10^-8 cm
volume of an atom in HCP = 4/3 * π (3.8141*10^-8)³ = 2.32415*10^-22 cm³
number of atoms in hcp = 6
no of atoms/cm³ = 6 / 2.32415*10^-22 cm³ = 2.58158*10²² atoms /cm³
density = 2.58158*10²² atoms /cm³ *  6.6329*10^-23gm/atom = 1.71234 gm/cm³
Therefore the density of solid Ar in g/cm³ = 1.7123