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Determine the lattice constant for FCC lead crystal of radius 1.746 Å and also find the spacing of (2 2 0) plane.
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Atomic radius(r) =  2a4

a = 16 r22

a = 16 (1.746)22

| a = 24.388 Aº | | --- |

Given  :   (hkl) = (2 2 0)

Formula     :      

d = al2+m2+n2 

d =  24.38822+22+02

(Spacing)    d = 17.245 Aº

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