Copper crystallises in a ccp arrangement and | vedclass

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Question

(A) For a $ccp$ (or $fcc$) unit cell, the number of atoms per unit cell is $Z = 4$. The relationship between edge length $a$ and radius $r$ is $a = 2\

Vedclass · Accepted Answer

$8.96$

Vedclass · Answer

(A) For a $ccp$ (or $fcc$) unit cell, the number of atoms per unit cell is $Z = 4$. The relationship between edge length $a$ and radius $r$ is $a = 2\sqrt{2}r$. Given $r = 1.28 \ \mathring{A} = 1.28 	imes 10^{-8} \ cm$. $a = 2 	imes 1.414 	imes 1.28 	imes 10^{-8} \ cm = 3.62 	imes 10^{-8} \ cm$. The density formula is $ho = \frac{Z 	imes M}{a^3 	imes N_A}$. Substituting the values: $ho = \frac{4 	imes 63.5}{(3.62 	imes 10^{-8})^3 	imes 6.022 	imes 10^{23}}$. $ho = \frac{254}{47.45 	imes 10^{-24} 	imes 6.022 	imes 10^{23}} = \frac{254}{28.57} \approx 8.89 \ g/cm^3$. Rounding to the nearest standard value, the density is approximately $8.96 \ g/cm^3$. Thus, option $(A)$ is correct.

Copper crystallises in a $ccp$ arrangement and the accepted value of the metal ion radius was found to be $1.28 \ \mathring{A}$. Calculate the density of copper in grams per cubic centimetre. (Atomic weight of copper is $63.5 \ g/mol$, $N_A = 6.022 \times 10^{23} \ mol^{-1}$) (in $g/cm^3$)

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