If the edge length of the unit cell of an ele | vedclass

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(B) For a $bcc$ structure, the number of atoms per unit cell $Z = 2$. Given: Edge length $a = 400 \, pm = 4 	imes 10^{-8} \, cm$, Atomic weight $M =

Vedclass · Accepted Answer

$5.188$

Vedclass · Answer

(B) For a $bcc$ structure, the number of atoms per unit cell $Z = 2$. Given: Edge length $a = 400 \, pm = 4 	imes 10^{-8} \, cm$, Atomic weight $M = 100 \, 	ext{g/mol}$, Avogadro number $N_A \approx 6.022 	imes 10^{23} \, 	ext{mol}^{-1}$. The formula for density $d$ is: $d = \frac{Z 	imes M}{N_A 	imes a^3}$. Substituting the values: $d = \frac{2 	imes 100}{6.022 	imes 10^{23} 	imes (4 	imes 10^{-8})^3} \approx \frac{200}{38.54} \approx 5.189 \, 	ext{g/cm}^3$. Thus, the density is approximately $5.188 \, 	ext{g/cm}^3$.

If the edge length of the unit cell of an element crystallizing in a $bcc$ structure is $400 \, pm$, then the density of the crystal will be ............ $\text{g/cm}^3$. (Atomic weight $= 100 \, \text{g/mol}$)

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