Xenon crystallizes in the face-centered cubic lattice and the edge of the unit cell is $620 p m$ . What is that nearest neighbour distance and what is the radius of the xenon atom?

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Solution

$1.$ The nearest neighbor distance would be $\frac{a}{\sqrt{2}} = \frac{620}{\sqrt{2}} = 438.4$ $p m$
$2.$ In FCC $4 r = a \sqrt{2}$ (along the face diagonal)
$r = \frac{a}{2 \sqrt{2}} == \frac{620}{2 \sqrt{2}} = 219.2$ $p m$

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Xenon crystallizes in the face-centered cubic lattice and the edge of the unit cell is 620pm. What is that nearest neighbour distance and what is the radius of the xenon atom?

1. The nearest neighbor distance would be a√2=620√2=438.4 pm2. In FCC 4r=a√2 (along the face diagonal)r=a2√2==6202√2=219.2 pm

Xenon crystallizes in the face-centered cubic lattice and the edge of the unit cell is $620 p m$ . What is that nearest neighbour distance and what is the radius of the xenon atom?

$1.$ The nearest neighbor distance would be $\frac{a}{\sqrt{2}} = \frac{620}{\sqrt{2}} = 438.4$ $p m$
$2.$ In FCC $4 r = a \sqrt{2}$ (along the face diagonal)
$r = \frac{a}{2 \sqrt{2}} == \frac{620}{2 \sqrt{2}} = 219.2$ $p m$