Question

An element has atomic mass 93 g mol^–1 and density 11.5 g cm^–3. If the edge length of its unit cell is 300 pm, identify the type of unit cell

Answer 1

Given;

M = 93 g mol^-1

d = 11.5 g cm^-3

a = 300 pm = 300×10^-10cm = 3×10^-8 cm

We know that

${\displaystyle d=\frac{ZXXM}{N_A\times a^3}}$

${\displaystyle Z=\frac{d\times N_A\times a^3}{M}=\frac{11.5\times 6.022\times {10}^{23}{(3\times {10}^{-8})}^3}{93}}$

= 2.01 (approx)

As the number of atoms present in given unit cells are coming nearly equal to 2, hence the given unit cell is body centred cubic unit cell (BCC).