Answer to Question #175036 in Mechanical Engineering for Varun

Question

Monochromatic X-rays of wavelength 1A° incident on a crystal and are

diffracted. The glancing angles for (100), (110) and (111) planes were found to 

be 16.13°, 23.13° and 28.76°. Identify the crystal structure and find its lattice 

constant. Hence, find the atomic weight of crystal, given the density is 8.96 g/cc

Answer 1

$\lambda$ = 2d sin$\theta$

1/d² = (h² + k² + l²)/a²

a² = $\lambda$²(h² + k² + l²)/4sin²$\theta$

a² = (1 A°)²(1² + 0² + 0²) / 4 sin²(16.13°) = 3.24 A°²

a = 1.80 A°

b² = (1 A°)²(1² + 1² + 0²) / 4 sin²(23.13°) = 3.24 A°²

b = 1.80 A°

c² = (1 A°)²(1² + 1² + 1²) / 4 sin²(28.76°) = 3.24 A°²

c = 1.80 A°

FCC crystal (a = b = c)

Volume for cubic a³ = (1.80 A°)³ = 5.832 A°³ = 5.832 x 10^-24 cm³

In a FCC lattice there rer 8 atoms at eight corners and 6 at face centers,

Now each corner contributes to eight cells so per unit cell contribution is 1/8 x 8 =1 atom. Now similarly each face center contributes to two unit cells so contribution per unit cell by six face centers is equal to 1/2 x 6 = 3 atoms. Hence total no. of atoms per unit cell of FCC lattice is n = (1+3) = 4 atoms.

density = n x M_w / (V_c x N_a)

M_w = density x V_c x N_a / n = 8.96 g/cm³ x 5.832 x 10^-24 cm³ x 6.02 x 10²³ / 4 = 7.86 g/ mol