A metal has a fcc structure. If the edge leng | vedclass

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(B) $1$. For an $fcc$ unit cell, the number of particles per unit cell $(Z)$ is $4$.$2$. The volume of one particle (assuming spherical shape) is $V_p

Vedclass · Accepted Answer

$5.92 	imes 10^{-24} 	ext{ cm}^3$

Vedclass · Answer

(B) $1$. For an $fcc$ unit cell, the number of particles per unit cell $(Z)$ is $4$.$2$. The volume of one particle (assuming spherical shape) is $V_p = \frac{4}{3} \pi r^3$.$3$. For $fcc$, the relation between edge length $(a)$ and radius $(r)$ is $a = 2\sqrt{2}r$, so $r = \frac{a}{2\sqrt{2}}$.$4$. Volume of $4$ particles = $4 	imes \frac{4}{3} \pi (\frac{a}{2\sqrt{2}})^3 = \frac{16}{3} \pi \frac{a^3}{16\sqrt{2}} = \frac{\pi a^3}{3\sqrt{2}}$.$5$. Given $a = 200 	ext{ pm} = 2 	imes 10^{-8} 	ext{ cm}$.$6$. Volume = $\frac{3.14159 	imes (2 	imes 10^{-8})^3}{3 	imes 1.414} = \frac{3.14159 	imes 8 	imes 10^{-24}}{4.242} \approx 5.92 	imes 10^{-24} 	ext{ cm}^3$.

$A$ metal has a $fcc$ structure. If the edge length of the unit cell is $200 \text{ pm}$, calculate the volume occupied by the particles in a unit cell.

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