cha6_1 - Problem A& M 6.1 First make sure you...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
First make sure you understand Equation 6 . the planes corresponding to the smallest reciprocal lattice vector yield the smallest angle for the ring. Thus we know that the angle at which the j th diffraction ring occurs is determined by the ratio sin( φ j / 2) sin( φ 1 / 2) = d j d 1 where d j is the magnitude of the the j th reciprocal lattice vector arranged according to increasing magnitudes. Since we need the lengths of the reciprocal lattice vectors we do that first. This is easy since the reciprocal lattice of the fcc lattice is bcc and vice versa. So we just need to find the distances in these lattices in real space. bcc lattice: The lattice sites arranged in order from the origin have coordinates (0 , 0 , 0), (1 / 2 , 1 / 2 , 1 / 2), (1 , 0 , 0), (1 , 1 , 0), (3 / 2 , 1 / 2 , 1 / 2),(1 , 1 , 1), (2 , 0 , 0) and their symmetry coun- terparts. The distances from the origin are 3 2 , 1 , 2 , 11 2 , 3 , 2 which yields the ratios
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 2

cha6_1 - Problem A& M 6.1 First make sure you...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online