Question

Copper has face-centered cubic (fcc) lattice with interatomic spacing equal to 2.54 Å.The value of the lattice constant for this lattice is:

• A. 2.29 Å
• B. 3.29 Å
• C. 2.59 Å
• D. 3.59 Å

Answer 1

The Correct Option is D

To find the lattice constant of a face-centered cubic (fcc) lattice, we need to understand the relationship between the lattice parameter (lattice constant) and the interatomic spacing in such a crystal structure.

In an fcc lattice, the atoms are located at each corner and the centers of all the cube faces of the unit cell. The face diagonal can be related to the lattice constant a and the interatomic spacing d.

For an fcc structure, the face diagonal is \sqrt{2}a, and since it consists of four atomic radii (or two interatomic distances), we have:

\sqrt{2}a = 2d

Given that the interatomic spacing d = 2.54 \, \text{Å}, we plug this value into the equation:

\sqrt{2}a = 2 \times 2.54 \, \text{Å}

\sqrt{2}a = 5.08 \, \text{Å}

Solving for a, we obtain:

a = \frac{5.08}{\sqrt{2}} \approx 3.59 \, \text{Å}

Thus, the value of the lattice constant for the fcc lattice of copper is 3.59 Å.

Let's confirm our calculation:

The face diagonal formula is derived from considering a right-angle triangle along the face of the unit cell. This reaffirms that the choice of methodology is correct.

Therefore, among the given options, 3.59 Å is the correct answer.