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Q.
Total volume of atoms present in a fcc unit cell of a metal with radius r is
NTA AbhyasNTA Abhyas 2022
Solution:
$a=2\sqrt{2}r$
Volume of the cell $=a^{3}=\left(\right.2\sqrt{2}r\left(\left.\right)^{3}=16\sqrt{2}r^{3}$
Number of spheres in fcc $=8\times \frac{1}{8}+6\times \frac{1}{2}=4$
Volume of 4 spheres $=4\times \frac{4}{3}\pi r^{3}$
$=\frac{16}{3}\pi r^{3}$