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

Question

asked Nov 3, 2020 211k views

1 Answer

Thanasi · Answer 1 · 2020-11-10T00:41:24+0000

Answer:

$LD=1.41\ nm^(-1)$

Step-by-step 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

$LD=(Number\ of\ atoms\ in\ the\ direction\ vector)/(d_([111]))$

$LD=(1)/(2R\sqrt6)$

We know that for silver

R=0.144 nm

$LD=(1)/(2* 0.144* \sqrt6)$

$LD=1.41\ nm^(-1)$