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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.

written 3.5 years ago by teamques10 ★ 68k

•   modified 3.5 years ago

engineering physics

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written 3.5 years ago by teamques10 ★ 68k

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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