# cha6_1 - Problem A M 6.1 First make sure you understand...

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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

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## This note was uploaded on 06/07/2010 for the course MECH 1122 taught by Professor Dont during the Spring '10 term at A.T. Still University.

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

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