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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.
1 Answer
written 3.9 years ago by |
Atomic radius(r) = √2a4
a = 16 r22
a = 16 (1.746)22
| a = 24.388 Aº | | --- |
Given : (hkl) = (2 2 0)
Formula :
d = a√l2+m2+n2
d = 24.388√22+22+02
(Spacing) d = 17.245 Aº