Copper crystallises in fcc unit cell with cel | vedclass

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(D) The formula for density is $d = \frac{Z 	imes M}{N_A 	imes a^3}$.For an $fcc$ unit cell,the number of atoms per unit cell $Z = 4$.The edge lengt

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$63.1$

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(D) The formula for density is $d = \frac{Z 	imes M}{N_A 	imes a^3}$.For an $fcc$ unit cell,the number of atoms per unit cell $Z = 4$.The edge length $a = 3.608 	imes 10^{-8} \, cm$.Avogadro's number $N_A = 6.022 	imes 10^{23} \, mol^{-1}$.Substituting the values: $8.92 = \frac{4 	imes M}{6.022 	imes 10^{23} 	imes (3.608 	imes 10^{-8})^3}$.$8.92 = \frac{4 	imes M}{6.022 	imes 10^{23} 	imes 46.96 	imes 10^{-24}}$.$M = \frac{8.92 	imes 6.022 	imes 10^{23} 	imes 46.96 	imes 10^{-24}}{4}$.$M = 63.1 \, g \, mol^{-1}$.Thus,the atomic mass of copper is $63.1 \, u$.

Copper crystallises in $fcc$ unit cell with cell edge length of $3.608 \times 10^{-8} \, cm$. The density of copper is $8.92 \, g \, cm^{-3}$. Calculate the atomic mass of copper. (In $u$)

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