Question

Calculate the distance between 111 planes in a crystal of Ca. Repeat the calculation for the 222 planes. (a=0.556nm)

• A. 016.1 nm
• B. 01.61 nm
• C. 0.610 nm
• D. None of the above

Answer 1

Answer: (B)

01.61 nm

Answer 2

We have, $d = \frac{a}{\sqrt{h^{2} + k^{2} + \mathcal{l}^{2}}}$;

$d_{111} = \frac{0.556}{\sqrt{1^{2} + 1^{2} + 1^{2}}} = 0.321nm$ and

$d_{222} = \frac{0.556}{\sqrt{2^{2} + 2^{2} + 2^{2}}} = 0.161nm$

The separation of the 111 planes is twice as great as that of 222 planes.