Question

If the radius of silver is 0.144 nm, what is the PD of the (100) plane for silver in m-2?

Answer 1

Before going to solve this question first we have to understand planar density of a crystal lattice.

Planar density of a plane is given defined-

                                                 $PD=\frac{N}{V}$

Where N is number of atoms centred in the plane and A is the area of the plane.

As per the question we have been given silver.

We know that silver is a face centered cubic crystal. i.e FCC

The radius of silver [r] is given as 0.144 nm.

$1 nm=10^{-9} m$.

Hence radius $r =0.144*10^{-9} m$

We have to calculate the PD of [100] plane of silver.

The area of [100] FCC is 8r^2 where r is the radius silver.

The number of atoms centered on [100] plane is 2.It is so because four atoms will be shared by four corners of the cube and there will be one central atom.Hence there will be only two atoms in one unit cell of [100] plane of silver.

Now we have to calculate the planar density PD.We know that-

                                                       $PD=\frac{N}{A}$

                                                               $=\frac{2}{8r^2}$

                                                               $=\frac{2}{8*[0.144*10^{-9}]^2 }$

                                                               $=12.05632716*10^{18} m^{-2}$  [ans]