A metal X crystallises in a face-centred cubic arrangement with the edge length 862 pm. What is the shortest separation of any two nuclei of the atom?
(a) 406 pm
(b) 707 pm
(c) 862 pm
(d) 609.6 pm
full solution plz
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Answer:
Correct option is
D
609.6 pm
The nucleus of the nearest neighbour of an atom in FCC lattice lies at the mid point of a face diagonal.
So, the shortest distance between two nuclei of atom in FCC lattice =
2
Length of Face Diagonal
=
2
a
2
=
2
a
Given a = 862 pm;
So, the shortest distance between two nuclei of atom in FCC lattice =
2
a
=
2
862
= 609.6 pm
Hence, Option "D" is the correct answer.
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