Question

The lattice constant of a cubic lattice is 4.12 A0.  Find the inter planer spacing of (1 0 0) plane ​

Answer 1

Answer:

the interplanar spacing is given by

d( h k l) = a / sqr root of ( h^2 + k^2 + l^ 2)

here in the question a = 4.12

h = 1 , k = 0 , l = 0

so putting all the given values we get

d = 4.12 Ao ,