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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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written 3.3 years ago by
teamques10
★ 66k
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Atomic radius(r) = $\dfrac{\ \sqrt{2a} }{4}$
a = $\dfrac{16 \ {r^2}}{2}$
a = $\dfrac{16\ {(1.746)^2} }{2}$
| a = 24.388 Aº | | --- |
Given : (hkl) = (2 2 0)
Formula :
d = $\dfrac{a}{\sqrt{l^2+m^2+n^2} }$
d = $\dfrac{24.388}{ \sqrt{2^2+2^2+0^2} }$
(Spacing) d = 17.245 Aº
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