A metal (X) of atomic weight M g mol^{-1} c | vedclass

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Question

(A) The density $(d)$ of a unit cell is given by the formula: $d = \frac{Z 	imes M}{N 	imes a^3}$,where $Z$ is the number of atoms per unit cell,$M$

Vedclass · Accepted Answer

$a = \left( \frac{2M}{Nd} \right)^{\frac{1}{3}}$

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(A) The density $(d)$ of a unit cell is given by the formula: $d = \frac{Z 	imes M}{N 	imes a^3}$,where $Z$ is the number of atoms per unit cell,$M$ is the molar mass,$N$ is Avogadro's number,and $a$ is the edge length of the unit cell.For a $bcc$ (body-centered cubic) lattice,the number of atoms per unit cell $(Z)$ is $2$.Substituting $Z = 2$ into the density formula: $d = \frac{2M}{N 	imes a^3}$.Rearranging for $a^3$: $a^3 = \frac{2M}{Nd}$.Taking the cube root on both sides: $a = \left( \frac{2M}{Nd} ight)^{\frac{1}{3}}$.

$A$ metal $(X)$ of atomic weight $M \ g \ mol^{-1}$ crystallizes in a $bcc$ lattice. Its density is $d \ g \ cm^{-3}$. What is the equation for the unit cell edge length $(a)$? ($N=$ Avogadro number)

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