If each edge of a cubic unit cell of an eleme | vedclass

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(B) Given: Atomic mass $(M) = 120 \ g \ mol^{-1}$,Density $(d) = 6.25 \ g \ cm^{-3}$,Edge length $(a) = 400 \ pm = 400 	imes 10^{-10} \ cm$,Avogadro

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body centered

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(B) Given: Atomic mass $(M) = 120 \ g \ mol^{-1}$,Density $(d) = 6.25 \ g \ cm^{-3}$,Edge length $(a) = 400 \ pm = 400 	imes 10^{-10} \ cm$,Avogadro number $(N_A) = 6.022 	imes 10^{23} \ mol^{-1}$.Using the formula for density: $d = \frac{Z 	imes M}{N_A 	imes a^3}$.Substituting the values: $6.25 = \frac{Z 	imes 120}{6.022 	imes 10^{23} 	imes (400 	imes 10^{-10})^3}$.$6.25 = \frac{Z 	imes 120}{6.022 	imes 10^{23} 	imes 64 	imes 10^{-24}}$.$6.25 = \frac{Z 	imes 120}{6.022 	imes 64 	imes 10^{-1}}$.$6.25 = \frac{Z 	imes 120}{38.54}$.$Z = \frac{6.25 	imes 38.54}{120} \approx 2$.Since the number of atoms per unit cell $(Z)$ is $2$,the crystal lattice is body centered.

If each edge of a cubic unit cell of an element having atomic mass $120$ and density $6.25 \ g \ cm^{-3}$ measures $400 \ pm$,then the crystal lattice is

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