Question

The total volume of atoms present in an FCC unit cell of a metal radius r is:

• A. $\frac{12}{3}{\mathrm{\pi r}}^3$
• B. $\frac{20}{3}{\mathrm{\pi r}}^3$
• C. $\frac{16}{3}{\mathrm{\pi r}}^3$
• D. $\frac{24}{3}{\mathrm{\pi r}}^3$

Answer 1

The correct option is C

$\frac{16}{3}{\mathrm{\pi r}}^3$

The explanation for the correct option

(c) $\frac{16}{3}{\mathrm{\pi r}}^3$

• The volume occupied by one atom of radius r$=\frac{4}{3}{\mathrm{\pi r}}^3$
• No. of atoms in an FCC[Face Centered Cubic] unit cell$=4$
• So, the volume occupied by an FCC unit cell$=4\times \frac{4}{3}{\mathrm{\pi r}}^3$
• volume occupied$=\frac{16}{3}{\mathrm{\pi r}}^3$

So, option (c) is the correct answer.

The explanation for the incorrect options

(a)$\frac{12}{3}{\mathrm{\pi r}}^3$

• It is the volume occupied by $3$atoms as$\frac{12}{3}{\mathrm{\pi r}}^3=3\times \frac{4}{3}{\mathrm{\pi r}}^3$
• But, an FCC[Face Centered Cubic] unit cell has$4$ atoms.

So, option [a] is incorrect.

[b]$\frac{20}{3}{\mathrm{\pi r}}^3$

• It is the volume occupied by$5$ atoms as$\frac{20}{3}{\mathrm{\pi r}}^3=5\times \frac{4}{3}{\mathrm{\pi r}}^3$.
• But, an FCC[Face Centered Cubic] unit cell has$4$ atoms.

So, option [b] is incorrect.

[d]$\frac{24}{3}{\mathrm{\pi r}}^3$

• It is the volume occupied by $6$atoms as $\frac{24}{3}{\mathrm{\pi r}}^3=6\times \frac{4}{3}{\mathrm{\pi r}}^3$
• But, an FCC[Face Centered Cubic] unit cell has$4$ atoms.

So, option [d] is incorrect.

Hence, option (C) is the correct answer.