Question

Hetero atom

(1) calculate radius of atom which forms simple cubic crystal structure. Lattice constant=4.0Å

(2) calculate radius of atom which forms HCP crystal structure. Lattice constant a=3.0Å

(3) calculate nuclear distance of body-centered tetragonal(Lattice constant a=b=5Å, c=6Å), and

atomic packing factor.

Answer 1

(1) For simple cubic structure radius of atom (r) is given by

r = a/2,   a = lattice constant

here a = 4.0 A   ,   r = 4.0/2 = 2 A

(2) For HCP structure radius of atom (r) is given by

r = a/2,   a = lattice constant

here a = 3.0 A   ,   r = 3.0/2 = 1.5 A

(3) nuclear distance , r = 1/4(2a² + c² )^0.5

a= 5A, C = 6A

r = 2.31 A

Atomic packing factor = Volume of spheres/ Volume of unit cell

total Volume of 2 spheres = 2 x (4/3) x 3.14x r³

Volume of unit cell = axbxc =a² x c

F = 103.21/150 = 0.688