Copper has FCC structure and the atomic radius is 1.278 × 10^−10m. Calculate the interplanar spacing for (111) planes.

Copper has FCC structure and the atomic radius is 1.278 × 10−10m. Calculate the interplanar spacing for (111) planes.

Right Answer is:

0.58/a2

SOLUTION

For FCC structure

a = 4r/√2

= (4 × 1.278)/√2 = 3.615 × 10−10m

${d_{111}} = \dfrac{1}{{\sqrt {{1^2} + {1^2} + {1^2}} }}$

3.615 × 10−10/√3 = 2.087 × 10−10m

Interplanar spacing d111 = 2.09 × 10−10m

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