Question
Atomic radius of fcc is
✓
Answer
(b)For the $ fcc $ structure
$4r = {({a^2} + {a^2})^{1/2}}$$ = a\sqrt 2 $
==> $r = \frac{{a\sqrt 2 }}{4} = \frac{a}{{2\sqrt 2 }}$
$4r = {({a^2} + {a^2})^{1/2}}$$ = a\sqrt 2 $
==> $r = \frac{{a\sqrt 2 }}{4} = \frac{a}{{2\sqrt 2 }}$
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