Question 705068

Given the parameters for the body-centered tetragonal (BCT) lattice: $a = b = a_0$ $c = \frac{a_0}{2}$

Let's place a reference atom at the origin $(0, 0, 0)$ and calculate the distances to all surrounding atoms in the lattice.

1. Distance to adjacent atoms along the c-axis (Top and Bottom): The coordinates of these atoms are $(0, 0, \pm c) \rightarrow (0, 0, \pm \frac{a_0}{2})$. Distance $d_1 = \sqrt{0^2 + 0^2 + (\frac{a_0}{2})^2} = \frac{a_0}{2} = 0.5 a_0$ The total number of such atoms is $2$.

2. Distance to body-centered atoms (diagonally placed in adjacent cells): The coordinates of these atoms are $(\pm \frac{a}{2}, \pm \frac{b}{2}, \pm \frac{c}{2}) \rightarrow (\pm \frac{a_0}{2}, \pm \frac{a_0}{2}, \pm \frac{a_0}{4})$. Distance $d_2 = \sqrt{(\frac{a_0}{2})^2 + (\frac{a_0}{2})^2 + (\frac{a_0}{4})^2}$ $d_2 = \sqrt{\frac{a_0^2}{4} + \frac{a_0^2}{4} + \frac{a_0^2}{16}}$ $d_2 = \sqrt{\frac{4a_0^2 + 4a_0^2 + a_0^2}{16}}$ $d_2 = \sqrt{\frac{9a_0^2}{16}} = \frac{3a_0}{4} = 0.75 a_0$ The total number of such atoms is $8$.

3. Distance to adjacent corner atoms along the a and b axes: The coordinates of these atoms are $(\pm a_0, 0, 0)$ and $(0, \pm a_0, 0)$. Distance $d_3 = a_0 = 1.0 a_0$ The total number of such atoms is $4$.

Conclusion: Comparing the calculated distances, we get: $0.5 a_0 < 0.75 a_0 < 1.0 a_0$.

• Nearest Neighbours: The shortest distance is $d_1$. Therefore, the nearest neighbour distance is $\frac{a_0}{2}$ and the number of nearest neighbours is $2$.
• Next Nearest Neighbours: The second shortest distance is $d_2$. Therefore, the next nearest neighbour distance is $\frac{3a_0}{4}$ and the number of next nearest neighbours is $8$.

Correct Values: $2, \frac{a_0}{2}, 8, \frac{3a_0}{4}$