Question

An element (atomic mass $=100$ amu) having a bcc structure has a unit cell edge equal to $400$ pm. The density (in g cm${}^{-3}$) of the element is:

• A. $10.376$
• B. $5.200$
• C. $7.289$
• D. $2.144$

Answer 1

The correct option is B $5.200$

The formula for calculating density is given below:

$d=\frac{z\times M}{N_A\times a^3}$

where,

$d=$ density of the unit cell
$z=$ effective number of atoms in the bcc unit cell $=2$
$M=$ molar mass $=100g/mol$

$N_A=$ Avogadro number

$a^3=$ volume of the unit cell $=(400\times {10}^{-10}{)}^{3}c{m}^3$
Substituting the value, we get

$d=\frac{2\times 100}{{6\times 10}^{23}\times {(400\times {10}^{-10}cm)}^3}=5.2g.c{m}^{-3}$