Question

find the distance between (345)planes in a cubic lattice having length 7.7A°​

Answer 1

The distance between (345) planes in a cubic lattice having length 7.7 A°​ is 1.089

• We know that, the distance can be represented as :

        1/ d² = h²/a² + k²/b² + l²/c²               eq(1)

• As per the question, a = b = c =  7.7 A°​  and h = 3, k = 4, l = 5

• Now, putting the values in eq (1), we get,

        1/d² = 3²/7.7² + 4²/7.7² + 5²/7.7²

               = (3² + 4² + 5²) / 7.7²

               = 50/59.29 x 10 ⁻²⁰

               = 0.843 x 10²⁰

• Now, d² = 1/ 0.843 x 10²⁰

                 d² = 1.186

                  d = 1.089