Question

If the radius of copper is 0.128 nm, what is the pd of the (100) plane for copper in m-2?

Answer 1

Copper has a FCC i.e. face centered cubic crystal structure. The 100 plane is essentially a planar section of the cubic cell where 4 Cu atoms occupy the 4 corners of the plane along with 1 Cu atom at the center of that plane. Each of the Cu atoms in the corners is shared by 4 adjacent unit cells. Thus, there are 2 Cu atoms present in the 100 plane (4*1/4 + 1 = 2).

Now, the planar density PD along the 100 plane is given as:

PD(100) = # atoms in the 100 plane/Area of 100 plane

            = $\frac{2}{8R^{2} }$

Here R = radius = 0.128 nm = $0.128*10^{-9} m$

PD = $\frac{2}{8*(0.128*10^{-9}) ^{2} } = 1.526*10^{19} m^{-2}$