An element has a body-centered cubic structur | vedclass

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(D) For a body-centered cubic $(bcc)$ structure, the number of atoms per unit cell $(Z)$ is $2$.Given edge length $(a)$ = $400 \ pm = 400 	imes 10^{-

Vedclass · Accepted Answer

$1.25$

Vedclass · Answer

(D) For a body-centered cubic $(bcc)$ structure, the number of atoms per unit cell $(Z)$ is $2$.Given edge length $(a)$ = $400 \ pm = 400 	imes 10^{-10} \ cm = 4 	imes 10^{-8} \ cm$.Atomic mass $(M)$ = $24 \ g \ mol^{-1}$.Avogadro's number $(N_A)$ = $6 	imes 10^{23} \ mol^{-1}$.The formula for density $(d)$ is:$d = \frac{Z 	imes M}{a^3 	imes N_A}$Substituting the values:$d = \frac{2 	imes 24}{(4 	imes 10^{-8})^3 	imes 6 	imes 10^{23}}$$d = \frac{48}{64 	imes 10^{-24} 	imes 6 	imes 10^{23}}$$d = \frac{48}{384 	imes 10^{-1}}$$d = \frac{48}{38.4} = 1.25 \ g \ cm^{-3}$.

An element has a body-centered cubic structure with a unit cell edge length of $400 \ pm$. The atomic mass of the element is $24 \ g \ mol^{-1}$. What is the density of the element (in $g \ cm^{-3}$)? $(N_{A} = 6 \times 10^{23} \ mol^{-1})$

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