Sodium has a body-centered cubic (BCC) packin | vedclass

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Question

(B) For a body-centered cubic $(BCC)$ structure,the nearest neighbor distance $d$ is given by the distance between the corner atom and the body-center

Vedclass · Accepted Answer

$4.3$

Vedclass · Answer

(B) For a body-centered cubic $(BCC)$ structure,the nearest neighbor distance $d$ is given by the distance between the corner atom and the body-centered atom.In a $BCC$ unit cell with lattice parameter $a$,the body diagonal is $\sqrt{3}a$.The distance between the nearest atoms is $d = \frac{\sqrt{3}a}{2}$.Given $d = 3.7 \mathring{A}$,we can solve for $a$:$a = \frac{2d}{\sqrt{3}} = \frac{2 	imes 3.7}{1.732} \approx 4.27 \mathring{A}$.Rounding to one decimal place,we get $a = 4.3 \mathring{A}$.

$LIST$ $I$	$LIST$ $II$
$A$. Intrinsic Semiconductor	$I$. Fermi-level near conduction band
$B$. $n$-type semiconductor	$II$. Fermi-level at middle
$C$. $p$-type semiconductor	$III$. Fermi-level near valence band
$D$. Metals	$IV$. Fermi-level inside conduction band

Sodium has a body-centered cubic $(BCC)$ packing. If the distance between two nearest atoms is $3.7 \mathring{A}$,then its lattice parameter is.....$\mathring{A}$

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