Answer to Question #175036 in Mechanical Engineering for Varun

Question #175036

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


1
Expert's answer
2021-04-12T01:45:22-0400

= 2d sin

1/d2 = (h2 + k2 + l2)/a2

a2 = 2(h2 + k2 + l2)/4sin2

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 Mw / (Vc x Na)

Mw = density x Vc x Na / n = 8.96 g/cm3 x 5.832 x 10-24 cm3 x 6.02 x 1023 / 4 = 7.86 g/ mol


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