Answer:
The density is 10.25 [tex]\frac{atom}{nm^{2} }[/tex]
Explanation:
From the question we are given that the lattice spacing nm
V = [tex](0.546)^{3} }[/tex] × [tex]nm^{3}[/tex]
= 0.1628 [tex]nm^{3}[/tex]
zinc blende lattice has 4 atom per unit cell.
This means that 4 atoms is contained in 0.15056 [tex]nm^{3}[/tex]
Density of the cubic unit in terms of atom is given as =
=26.565 [tex]\frac{atoms}{nm^{3} }[/tex]
For plane (110) the spacing between the cubic unit in the crystal denoted by [tex]d_{hkl}[/tex] is given as = [tex]\frac{a}{\sqrt{1^{2} + 1^{2} +0^{2} } }[/tex] Where a is the lattic spacing = 0.546
= [tex]\frac{0.546}{\sqrt{2} }[/tex] = 0.376 nm
The density of the lattice in (110) plane =26.565 [tex]\frac{atoms}{nm^{3} }[/tex] × 0.386 nm
= 10.25 [tex]nm^{2}[/tex]
