The edge length of the unit cell of a metal ( | vedclass

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(B) The formula for density is $d = \frac{Z 	imes M_W}{N_A 	imes a^3}$.Given: $d = 1.74 \, g \, cm^{-3}$,$M_W = 24 \, g \, mol^{-1}$,$a = 4.53 \, \m

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$4$

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(B) The formula for density is $d = \frac{Z 	imes M_W}{N_A 	imes a^3}$.Given: $d = 1.74 \, g \, cm^{-3}$,$M_W = 24 \, g \, mol^{-1}$,$a = 4.53 \, \mathring{A} = 4.53 	imes 10^{-8} \, cm$,and $N_A = 6 	imes 10^{23} \, mol^{-1}$.Rearranging for $Z$: $Z = \frac{d 	imes N_A 	imes a^3}{M_W}$.$Z = \frac{1.74 	imes (6 	imes 10^{23}) 	imes (4.53 	imes 10^{-8})^3}{24}$.$Z = \frac{1.74 	imes 6 	imes 10^{23} 	imes 93.01 	imes 10^{-24}}{24}$.$Z = \frac{1.74 	imes 6 	imes 93.01 	imes 10^{-1}}{24} \approx \frac{971.08}{240} \approx 4.04$.Thus,the effective number of atoms $Z$ is $4$.

The edge length of the unit cell of a metal $(M_W = 24 \, g \, mol^{-1})$ having a cubic structure is $4.53 \, \mathring{A}$. If the density of the metal is $1.74 \, g \, cm^{-3}$,then the effective number of atoms in the unit cell is :- $(N_A = 6 \times 10^{23} \, mol^{-1})$

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