Question

Gold crystallizes in fcc lattice. The edge length of the unit cell is 4 Å. The closest distance between gold atoms is 'x' Å and density of gold is 'y' g/cm\(^3\). What are x and y respectively?

• A. \(\sqrt{2}\), 41.04
• B. \(2\sqrt{2}\), 20.52
• C. \(2\sqrt{3}\), 10.25
• D. \(\sqrt{3}\), 5.15

Hint:

Use geometry of crystal lattice and formula for density to solve problems.

Answer 1

The Correct Option is B

Step 1: Closest distance in fcc lattice
In face-centered cubic (fcc), atoms touch along the face diagonal: \[ d = \frac{a\sqrt{2}}{2} \] For gold with 4 Å edge length, closest distance is: \[ x = \frac{4 \times \sqrt{2}}{2} = 2 \sqrt{2} \text{ Å} \] Step 2: Calculate density
Using: \[ \rho = \frac{Z M}{N_A a^3} \] Where \(Z=4\) (fcc atoms per unit cell), \(M = 197 g/mol\), \(N_A = 6 \times 10^{23}\), \(a = 4 \times 10^{-8} cm\). Calculating gives approximately 20.52 g/cm\(^3\). Step 3: Conclusion
The closest distance and density are \(2\sqrt{2}\) Å and 20.52 g/cm\(^3\).