At T K,copper (atomic mass = 63.5 u) has fc | vedclass

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(B) For an $fcc$ unit cell,the number of atoms per unit cell,$Z = 4$. Edge length $a = x \ \mathring{A} = x 	imes 10^{-8} \ cm$. Atomic mass of $Cu =

Vedclass · Accepted Answer

$\frac{423}{x^3}$

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(B) For an $fcc$ unit cell,the number of atoms per unit cell,$Z = 4$. Edge length $a = x \ \mathring{A} = x 	imes 10^{-8} \ cm$. Atomic mass of $Cu = 63.5 \ g \ mol^{-1}$. Avogadro's number $N_A = 6.0 	imes 10^{23} \ mol^{-1}$. The formula for density $d$ is: $d = \frac{Z 	imes M}{a^3 	imes N_A}$ Substituting the values: $d = \frac{4 	imes 63.5}{(x 	imes 10^{-8})^3 	imes 6.0 	imes 10^{23}}$ $d = \frac{254}{x^3 	imes 10^{-24} 	imes 6.0 	imes 10^{23}}$ $d = \frac{254}{x^3 	imes 0.6} = \frac{423.33}{x^3} \approx \frac{423}{x^3} \ g \ cm^{-3}$. Thus,the correct option is $(B)$ or $(C)$.

At $T \ K$,copper (atomic mass $= 63.5 \ u$) has $fcc$ unit cell structure with edge length of $x \ \mathring{A}$. What is the approximate density of $Cu$ in $g \ cm^{-3}$ at that temperature? $(N_A = 6.0 \times 10^{23} \ mol^{-1})$

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