MCQ
Atomic radius of fcc is
- A$\frac{a}{2}$
- ✓$\frac{a}{{2\sqrt 2 }}$
- C$\frac{{\sqrt 3 }}{4}a$
- D$\frac{{\sqrt 3 }}{2}a$
✓
Answer
Correct option: B.
$\frac{a}{{2\sqrt 2 }}$
b
(b)For the $ fcc $ structure
$4r = {({a^2} + {a^2})^{1/2}}$$ = a\sqrt 2 $
==> $r = \frac{{a\sqrt 2 }}{4} = \frac{a}{{2\sqrt 2 }}$
(b)For the $ fcc $ structure
$4r = {({a^2} + {a^2})^{1/2}}$$ = a\sqrt 2 $
==> $r = \frac{{a\sqrt 2 }}{4} = \frac{a}{{2\sqrt 2 }}$
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