Question

A metal crystallizes with a face-centered cubic lattice. The edge length of the unit cell is 408 pm. The diameter of the metal atom is :
(a) 288 pm
(b) 408 pm
(c) 144 pm
(d) 204 pm

Answer 1

Answer:

Answer:

r - radius of atom

4r = √2 a

2d = √2a

Now,

       d = a/√2

    = 408 / √2

    = 408 / 1.414

    = 288.54 pm

Answer 2

The diameter of the metal atom is (a) 288 pm.

Given:

A metal crystallizes with a face-centred cubic lattice and the edge length of the unit cell is 408 pm.

To Find:

The diameter of the metal atom.

Solution:

To the diameter of the metal atom we will follow the following steps:

As we know, in the Fcc unit cell radius and edge length relationship is given by the formula

$a = 2 \sqrt{2}r$

$r = \frac{a}{2 \sqrt{2} }$

Now,

Putting values in the above equation we get,

$r = \frac{408}{2 \sqrt{2} } = \frac{204}{ \sqrt{2} } = \frac{204}{1.414} =144.27pm$

Diameter = 2r = 288.5pm

Henceforth the diameter of the metal atom is (a) 288 pm.

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