Question

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

Answer 1

`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)`