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A metal crystallizes with a face-centered cubic lattice. The edge of the unit cell is 408 pm. The diameter of the metal atom is
  • a)
    144 pm
  • b)
    204 pm
  • c)
    288 pm
  • d)
    408 pm
Correct answer is option 'C'. Can you explain this answer?
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A metal crystallizes with a face-centered cubic lattice. The edge of t...
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A metal crystallizes with a face-centered cubic lattice. The edge of t...
Given information:
- Metal crystallizes with a face-centered cubic lattice
- Edge of the unit cell = 408 pm

To find:
- Diameter of the metal atom

Solution:
Face-centered cubic (fcc) lattice:

In an fcc lattice, there are atoms at the corners and in the center of each face of the cube. The unit cell of an fcc lattice is shown below:

![image.png](https://www.askiitians.com/iit-jee-solid-state/fcc-lattice.png)

The edge length of the unit cell (a) can be related to the radius of the atom (r) by the following formula:

a = 4r/√2

The diameter of the atom (d) is twice the radius (r), so we can rewrite the above formula as:

a = 2d√2/4

d = a√2/2

Substituting the given value of edge length (a = 408 pm) in the above formula, we get:

d = 408 pm × √2/2

d = 408 pm × 1.414/2

d = 288 pm

Therefore, the diameter of the metal atom is 288 pm.

Answer: c) 288 pm
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Community Answer
A metal crystallizes with a face-centered cubic lattice. The edge of t...
For fcc lattice
4r=√2a
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A metal crystallizes with a face-centered cubic lattice. The edge of the unit cell is 408 pm. The diameter of the metal atom isa)144 pmb)204 pmc)288 pmd)408 pmCorrect answer is option 'C'. Can you explain this answer?
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