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Xenon is crystallized in the fcc lattice and the edge of unit length is 620 pm. What is the nearest neighbour distance and radius of xenon atom ? |
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Answer» Solution : EDGE length, a = 620 PM. For fcc lattice, nearest NEIGHBOUR distance, `d = a sqrt2` `= (620)/(sqrt2) =(620)/(1.414) = 438.3` pm For fcc lattice, radius of atom, `r=(a)/(2SQRT2)=(620)/(2 xx 1.414)` = 219.25 pm |
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