An fcc unit cell of aluminium contains the eq | vedclass

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(D) In an $fcc$ (face-centered cubic) unit cell:$1.$ Atoms at the corners: There are $8$ corners,and each corner atom contributes $\frac{1}{8}$ to the

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$4$

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(D) In an $fcc$ (face-centered cubic) unit cell:$1.$ Atoms at the corners: There are $8$ corners,and each corner atom contributes $\frac{1}{8}$ to the unit cell. So,$\frac{1}{8} 	imes 8 = 1$ atom.$2.$ Atoms at the face centers: There are $6$ faces,and each face-centered atom contributes $\frac{1}{2}$ to the unit cell. So,$\frac{1}{2} 	imes 6 = 3$ atoms.$3.$ Total number of atoms $(Z)$ = $1 + 3 = 4$ atoms.

An $fcc$ unit cell of aluminium contains the equivalent of how many atoms?

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