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

11kviews

written 8.7 years ago by

teamques10 ★ 69k

modified 3.2 years ago by

chinthasaiteja746 • 30

Mumbai university > FE > SEM 1 > Applied Physics 1

Marks: 3M

Year: May 2015

engineering physics

ADD COMMENT EDIT

1 Answer

2.7kviews

written 8.7 years ago by

teamques10 ★ 69k

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 × …

Create a free account to keep reading this post.

and 2 others joined a min ago.

ADD COMMENT EDIT

Please log in to add an answer.