Q27 / 90 · JEE Main 2017 (02 Apr) · Chemistry

A metal crystallizes in a face centred cubic structure. If the edge length of its unit cell is $' a'$ , the closest approach between two atoms in the metallic crystal will be

A
$2 \sqrt{2} a$
B
$\sqrt{2} a$
Correct answer
C
$\frac{a}{\sqrt{2}}$
D
$2 a$

Solution

$FCC$ unit cell: eight corners with six faces

$r + 2 r + r = \sqrt{2} a 4 r = \sqrt{2} a Closest distance in fcc 2 r = \frac{\sqrt{2} a}{2} = \frac{\sqrt{2} a}{\sqrt{2} \times \sqrt{2}} = \frac{a}{\sqrt{2}}$

Hence, the correct answer is option: 'C'.

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A metal crystallizes in a face centred cubic structure. If the edge length of its unit cell is 'a', the closest approach between two atoms in the metallic crystal will be

Solution

Relevant Questions

A metal crystallizes in a face centred cubic structure. If the edge length of its unit cell is $' a'$ , the closest approach between two atoms in the metallic crystal will be