The number of atoms in 100 g of an fcc cryst | vedclass

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Question

(A) For an $fcc$ crystal,the number of atoms per unit cell $z = 4$. Density $ho = 10 \ g/cm^3$ and edge length $a = 100 \ pm = 100 	imes 10^{-10} \

Vedclass · Accepted Answer

$4 	imes 10^{25}$

Vedclass · Answer

(A) For an $fcc$ crystal,the number of atoms per unit cell $z = 4$. Density $ho = 10 \ g/cm^3$ and edge length $a = 100 \ pm = 100 	imes 10^{-10} \ cm = 10^{-8} \ cm$. The volume of the unit cell $V = a^3 = (10^{-8} \ cm)^3 = 10^{-24} \ cm^3$. The mass of one unit cell = $ho 	imes V = 10 \ g/cm^3 	imes 10^{-24} \ cm^3 = 10^{-23} \ g$. The number of unit cells in $100 \ g$ of the crystal = $\frac{100 \ g}{10^{-23} \ g/	ext{unit cell}} = 10^{25} \ 	ext{unit cells}$. Since each $fcc$ unit cell contains $4$ atoms,the total number of atoms = $4 	imes 10^{25}$.

The number of atoms in $100 \ g$ of an $fcc$ crystal with density $d = 10 \ g/cm^3$ and cell edge length $a = 100 \ pm$ is:

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