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 $(z)$ is $4$.The volume of one unit cell is $V = a^3 = (100 \ pm)^3 = (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 $(z)$ is $4$.The volume of one unit cell is $V = a^3 = (100 \ pm)^3 = (100 	imes 10^{-10} \ cm)^3 = 10^{-24} \ cm^3$.The mass of one unit cell is $m = 	ext{density} 	imes 	ext{volume} = 10 \ g/cm^3 	imes 10^{-24} \ cm^3 = 10^{-23} \ g$.The number of unit cells in $100 \ g$ is $N_{cells} = \frac{100 \ g}{10^{-23} \ g/unit cell} = 10^{25} \ unit cells$.Since each $fcc$ unit cell contains $4$ atoms, the total number of atoms is $4 	imes 10^{25}$.

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

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