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

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written 9.7 years ago by

teamques10 ★ 70k

modified 4.1 years ago by

chetana1375 • 20

Mumbai university > FE > SEM 1 > Applied Physics 1

Marks: 3M

Year: May 2015

engineering physics

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2.7kviews

written 9.7 years ago by

teamques10 ★ 70k

Given:

It is an FCC structure

r = 1,746 A

Plane (2 2 0)

To find: a = ? and d = ?

Solution:

In FCC structure,

$a= \frac{4r}{\sqrt2}= \frac{4 × 1746 × 10^{-10}}{\sqrt2}$

$a = 4938.43 × 10^{-10}m$

$d = \frac{a}{\sqrt{2^2 + 2^2 + 0^2}}$

$d = \frac{4938.43 × 10^{-10}}{\sqrt8}$

$d = 1745.99 × 10^{-10}m$

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