An element X (atomic weight = 24) crystallise | vedclass

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$(A)$ For an $FCC$ lattice,the number of atoms per unit cell $(Z)$ is $4$.Due to $4\%$ Schottky defects,the effective number of atoms per unit cell $(

Vedclass · Accepted Answer

$2.4$

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$(A)$ For an $FCC$ lattice,the number of atoms per unit cell $(Z)$ is $4$.Due to $4\%$ Schottky defects,the effective number of atoms per unit cell $(Z_{	ext{effective}})$ is:$Z_{	ext{effective}} = 4 	imes (1 - 0.04) = 4 	imes 0.96 = 3.84$.The edge length $a = 400 \, pm = 400 	imes 10^{-10} \, cm = 4 	imes 10^{-8} \, cm$.The observed density $(d)$ is given by the formula:$d = \frac{Z_{	ext{effective}} 	imes M}{N_A 	imes a^3}$.Substituting the values:$d = \frac{3.84 	imes 24}{6 	imes 10^{23} 	imes (4 	imes 10^{-8})^3} = \frac{92.16}{6 	imes 10^{23} 	imes 64 	imes 10^{-24}} = \frac{92.16}{38.4} = 2.4 \, g/cm^3$.

An element $X$ (atomic weight = $24$) crystallises in a face-centered cubic $(FCC)$ lattice. The edge length of the unit cell is $400 \, pm$. If this lattice contains $4\%$ Schottky defects,calculate the observed density of this element in $g/cm^3$ $(N_A = 6 \times 10^{23})$.

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