Question

What is the LD of the [111] direction for silver (in nm-1)?

Answer 1

Answer:

$LD=1.41\ nm^{-1}$

Explanation:

We know that unit cell of silver is FCC.

In FCC

$a=2\sqrt2 \ R$

Plane [111] is passing through the center of cell diagonally.In this plane half part of atoms comes under that plane.

So Z=  2 x 1/2 =1

Z = 1 atom (in plane (1,1,1))

The length of vector $d_{[111]}$ given as

$d_{[111]=2R\sqrt6$

We know that linear density(LD) given as

$LD=\dfrac{Number\ of\ atoms\ in\ the\ direction\ vector}{d_{[111]}}$

$LD=\dfrac{1}{2R\sqrt6}$

We know that for silver

R=0.144 nm

$LD=\dfrac{1}{2\times 0.144\times \sqrt6}$

$LD=1.41\ nm^{-1}$