Question

question 3
a metal crystallizes in a primitive orthorhombic structure (i.e., one atom at each corner of the unit cell) with lattice parameters a = 0.400 nm, b = 0.300 nm, c = 0.500 nm. the planar density on the (001) plane is pd(001) = select
2 pts

Explanation:

Step1: Recall planar - density formula

Planar density (PD) on a plane is the number of atoms per unit area of the plane. For a primitive orthorhombic unit - cell, on the (001) plane, the area of the (001) plane in the unit - cell is $A = a\times b$. The number of atoms on the (001) plane in a primitive orthorhombic unit - cell (one atom at each corner, and each corner atom is shared by 4 unit - cells on the plane) is $n = 4\times\frac{1}{4}=1$.

Step2: Calculate the area of the (001) plane

Given $a = 0.400\ nm$ and $b = 0.300\ nm$, the area of the (001) plane $A=a\times b=(0.400\ nm)\times(0.300\ nm)=0.12\ nm^{2}$.

Step3: Calculate planar density

Planar density $PD_{(001)}=\frac{n}{A}$, where $n = 1$ (number of atoms on the (001) plane) and $A = 0.12\ nm^{2}$. So $PD_{(001)}=\frac{1}{0.12\ nm^{2}} = 8.33\ nm^{-2}$.

Answer:

$8.33\ nm^{-2}$