Question

An element ( atomic mass =60 )having fcc crystal has a density of 6.23 g per cm3 . What is the edge length of unit cell

Answer 1

$\textsf{\large{\underline {Concept Used - Solid State }}}$ :

$\textsf{\large{\underline {Solution}}}$ :

Mass of the unit cell, M = $\mathsf{60 g}$

Density of the unit cell, d = $\mathsf{6.23\:g\:{cm}^{-3}}$

We know that,

$\boxed{\mathsf{Density\:=\:{\dfrac{Mass} {Volume}}}}$

$\mathsf{Density\:=\:{\dfrac{M*z} {N_A*V}}}$

where, $\mathsf{M\:=\:Molar\:mass}$

$\mathsf{z\:=\: Effective \:number \:of\: particles\: in\: a \:unit\: cell}$

$\mathsf{N_A\:=\:Avagadro\:Number}$

$\mathsf{V\:=\:Volume \:of\:Unit\:Cell}$

So, Now,

For fcc unit cell, $\mathsf{z\:=\:4}$,

Putting values in the above formula,

$\mathsf{6.23\:=\:{\dfrac{60*4}{(\:6.022*{10}^{23}\:)(\:V\:)}}}$

$\mathsf{V\:=\:{\dfrac{60*4}{(\:6.022*{10}^{23}\:)(\:6.23\:)}}}$

$\mathsf{V\:=\:{\dfrac{240}{(\:6.022*{10}^{23}\:)(\:6.23\:)}}}$

$\mathsf{V\:=\:{\dfrac{240}{(\:37.517*{10}^{23}\:)}}}$

$\mathsf{V\:=\:6.397*{10}^{-23}{cm} ^{3}}$

Since, Volume of a cubic cell, V = $\mathsf{{side}^{3}\:=\:{a} ^{3}}$

➡️ Here, a = Side Length /Edge length of the unit cell .

$\mathsf{{a}^{3}\:=\:6.397*{10}^{-23}}$

$\mathsf{{a}^{3}\:=\:63.97*{10}^{-24}}$

$\boxed{\mathsf{a\:=\:3.99*{10}^{-8}\:cm}}$

➡️ $\textsf{Edge length of the unit cell}$ = $\mathsf{a\:{\approx{\:4*{10}^{-8}\:cm}}}$