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In face centered cubic (fcc) crystal lattice, edge length is 400 pm. Find the diameter of greatest sphere which can be fit into the interstitial void without distortion of lattice.

In face centered cubic (fcc) crystal lattice, edge length is 400 pm. Find the diameter of greatest sphere which can be fit into the interstitial void without distortion of lattice.

Grade:upto college level

2 Answers

Deepak Patra
askIITians Faculty 471 Points
9 years ago
Hello Student,
Please find the answer to your question
For an octahedral void a = 2 (r + R) In fcc lattice the largest void present is octahedral void. If the radius of void sphere is R and of lattice sphere is r. Then,
r = √2 *400/5 = 141.12 pm (a = 400 pm)
applying condition for octahedral void, 2 (r + R) = a
∴ 2 R = a 2r = 400 – 2 * 141.12
∴ Diameter of greatest sphere = 117.16 pm

Thanks
Deepak patra
askIITians Faculty
Ayush Bhardwaj
11 Points
6 years ago
In fcc lattice, thr largest void present is octahedral void and its radius(R) is given byR=(a-2r)/2Where,a= edge length,r=radius of the lattice atomBut r=?We know thatr=√2 × a/4r=1.414×400/4r=141.4pmNow diameter of the biggest sphere=2RFrom relation given above,2R=a - 2r 2R = 400 - 2(141.4) 2R = 400 - 282.6 2R = 117.4pmHence the diameter of the biggest sphere that can be inserted is 141.4pm

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