Logo
Select Quick Pages:
Home
About
Contact Us
Disclaimer
Privacy policies
Terms and conditions

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

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