Atoms of element X form hcp lattice and those of element Y occupy $$\frac{2}{3}$$ of its tetrahedral voids. The percentage of element X in the lattice is (Nearest integer) ______

Let the number of atoms forming the hcp lattice be $$\mathrm{N}$$.

In a close-packed structure, total tetrahedral voids are $$\mathrm{2N}$$.

Atoms occupy $$\mathrm{\frac{2}{3}}$$ of the tetrahedral voids.

Hence, number of atoms in tetrahedral voids is $$\mathrm{\frac{2}{3} \times 2N = \frac{4N}{3}}$$.

Ratio of lattice atoms to atoms in voids:

$$\mathrm{N : \frac{4N}{3}}$$

Multiplying by 3 gives:

$$\mathrm{3 : 4}$$

Therefore, the formula is $$\mathrm{X_3Y_4}$$.

Total atoms in one formula unit:

$$\mathrm{3 + 4 = 7}$$

Percentage of lattice atoms:

$$\mathrm{\frac{3}{7} \times 100}$$

$$\mathrm{= 42.857\% \approx 43\%}$$