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Determine the lattice constant for FCC lead crystal of radius 1,746 A* and also find the spacing of (2 2 0) plane.

Mumbai university > FE > SEM 1 > Applied Physics 1

Marks: 3M

Year: May 2015

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Given:

It is an FCC structure

r = 1,746 A

Plane (2 2 0)

To find: a = ? and d = ?

Solution:

In FCC structure,

a=4r2=4×1746×10102

a=4938.43×1010m

d=a22+22+02

$d = \frac{4938.43 × …

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