In fcc lattice, edge length is 400 pm. Find the diameter of the greatest sphere which can be fitted into the interstitial void without distortion of lattice:

Question

In fcc lattice, edge length is 400 pm. Find the diameter of the greatest sphere which can be fitted into the interstitial void without distortion of lattice: (A) 141.4 pm (B) 58.54 pm (C) 117.08 pm (D) None of these

Tardigrade Admin · Accepted Answer

For fcc, r=(a/2 √2)=(400/2 √2)=141.4 pm Radius of Octahedral void ( r )=0.414 R =0.414 × 141.4 =58.54 pm Diameter =2 × 58.54=117.08 pm

Q. In fcc lattice, edge length is $400 p m$ . Find the diameter of the greatest sphere which can be fitted into the interstitial void without distortion of lattice:

141.4 pm

58.54 pm

117.08 pm

None of these

For fcc, $r = \frac{a}{2 2} = \frac{400}{2 2} = 141.4 p m$

Radius of Octahedral void $(r) = 0.414 R$

$= 0.414 \times 141.4$

$= 58.54 p m$

Diameter $= 2 \times 58.54 = 117.08 p m$

Q. In fcc lattice, edge length is 400pm. Find the diameter of the greatest sphere which can be fitted into the interstitial void without distortion of lattice: