Find the number of equivalent (100), (110), and (111) planes in a cubic crystal.
Take the combinations of 1 and 1bars and plot you will get
6 members-- in (100) as (100),(010),(001),(1bar00),(01bar0 ),(001bar) having interplaner distance d=a
'a' be the lattice parameter
12 members--in (110) as (110),(101) ,(011),(11bar0) ,(101bar), (011bar), (1bar10), (1bar01),(01bar1),(1bar1bar0), ( 1bar01bar),(01bar1bar) having interplaner distance
8 members --in(111) as (111),(111bar),(11bar1),(1bar11),(11bar1bar),(1bar11bar),(1bar1bar1),(1bar1bar1bar)
having interplaner distance
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