WBJEE · Chemistry · Solid State

In the face-centred cubic lattice structure of gold the closest distance between gold atoms is ('a' being the edge length of the cubic unit cell)

A \(a \sqrt{2}\)
B \(\frac{a}{\sqrt{2}}\)
C \(\frac{a}{2 \sqrt{2}}\)
D \(2 \sqrt{2}\) a

Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{a}{\sqrt{2}}\)

Step-by-step Solution

Detailed explanation

Hint: Closest distance between two gold atoms in fcc lattice of gold is \(\frac{a}{\sqrt{2}} .\) This is the distance between the corner atom and the closest face centre atom.

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Explore more questions on app

From WBJEE

Explore more questions on app