If a metal crystallises in a face-centred cub | vedclass

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(D) For an $FCC$ unit cell, the relation between edge length $(a)$ and radius $(r)$ is $a = 2\sqrt{2}r$.Given $r = 25 \ pm = 25 	imes 10^{-10} \ cm$.

Vedclass · Accepted Answer

$2.828 	imes 10^{24}$

Vedclass · Answer

(D) For an $FCC$ unit cell, the relation between edge length $(a)$ and radius $(r)$ is $a = 2\sqrt{2}r$.Given $r = 25 \ pm = 25 	imes 10^{-10} \ cm$.$a = 2 	imes 1.414 	imes 25 	imes 10^{-10} \ cm = 70.7 	imes 10^{-10} \ cm = 7.07 	imes 10^{-9} \ cm$.The volume of one unit cell $(V_{cell})$ is $a^3 = (7.07 	imes 10^{-9} \ cm)^3 \approx 353.5 	imes 10^{-27} \ cm^3 = 3.535 	imes 10^{-25} \ cm^3$.The number of unit cells in $1.0 \ cm^3$ is $\frac{1.0 \ cm^3}{V_{cell}} = \frac{1}{3.535 	imes 10^{-25}} \approx 0.2828 	imes 10^{25} = 2.828 	imes 10^{24}$.

If a metal crystallises in a face-centred cubic $(FCC)$ structure with a metallic radius of $25 \ pm$, the number of unit cells in $1.0 \ cm^3$ of the lattice is:

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