A metal forms an fcc structure. Calculate the | vedclass

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(D) $1$. In an $fcc$ unit cell, the packing efficiency is $74\%$, which means $74\%$ of the unit cell volume is occupied by atoms.$2$. The remaining v

Vedclass · Accepted Answer

$6.385 	imes 10^{-23} 	ext{ cm}^3$

Vedclass · Answer

(D) $1$. In an $fcc$ unit cell, the packing efficiency is $74\%$, which means $74\%$ of the unit cell volume is occupied by atoms.$2$. The remaining volume is the void volume, which is $100\% - 74\% = 26\%$ of the total unit cell volume $(V_{cell})$.$3$. Given that the void volume is $1.66 	imes 10^{-23} 	ext{ cm}^3$, we can write: $0.26 	imes V_{cell} = 1.66 	imes 10^{-23} 	ext{ cm}^3$.$4$. Solving for $V_{cell}$: $V_{cell} = \frac{1.66 	imes 10^{-23}}{0.26} 	ext{ cm}^3$.$5$. $V_{cell} \approx 6.385 	imes 10^{-23} 	ext{ cm}^3$.

$A$ metal forms an $fcc$ structure. Calculate the volume of the $fcc$ unit cell in $\text{cm}^3$ if the void volume is $1.66 \times 10^{-23} \text{ cm}^3$.

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