{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# 528 nm b the spacing between 110 planes is the lattice

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: he B atom. The body diagonal for the cube is twice this distance, or 0.8 nm, and this is 31/2 larger than the cube edge. So, the lattice constant is 0.462 nm. Each cell has 1 A atom and 1 B atom, since their eight corner A atoms, each of which is shared among eight cells. Hence the volume densities of the A and B atoms are the same, as ρA = ρB = 1 1 = = 10 28 m −3 . 3 3 d (0.462 nm ) (b) Since there are equal numbers of A atoms and B atoms in each unit cell, reversing the role of the atoms does not change the numbers (note that you still get the same body diagonal whichever atom you put in the center of the cube). Hence ρA = ρB = 1 1 = = 10 28 m −3 . 3 3 d (0.462 nm ) (c) The material is a typical binary compound of equal parts of A and B atoms. 1.18 The lattice constant of a simple cubic unit cell is 0.528 nm. Determine the distance between (a) (100), (b) (110), and (c) (1...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online