Solution:
Solution
Given interatomic spacing = 2r = 2.54 $$A^o$$
In FCC lattice, $$4r = \sqrt 2 a$$ ,where a is lattice constant.
$$2r = \dfrac{\sqrt2 a }{2} = \dfrac{a}{\sqrt2}$$
$$a= 2r\times \sqrt2 = \left ( 2.54 A^o \right )\left ( 1.414 \right ) = 3.59A^o $$
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