Let's break down the problem into the two required steps for each part, with explanations at each step, and end with the final answer. --- ### (a) **Tetragonal Structure: Bravais Lattice, Primitive Vectors, Unit Cell, and Basis** #### **Step 1: Sketch the Bravais Lattice & Primitive Vectors** In the tetragonal structure, the Cu atoms form a square lattice in the \( ab \)-plane, and the layers are stacked along the \( c \)-axis. - **Bravais lattice:** Simple tetragonal (think of a simple cubic lattice stretched along one axis). - **Primitive lattice vectors:** - \(\vec{a}_1 = a\hat{x}\) - \(\vec{a}_2 = a\hat{y}\) - \(\vec{a}_3 = c\hat{z}\) **Explanation:** These vectors define a parallelepiped box (unit cell) that repeats to fill space. The vectors \(a\) and \(c\) are the lattice spacings in-plane and out-of-plane. --- #### **Step 2: Unit Cell and Basis** - **Unit cell:** The parallelepiped defined by \(\vec{a}_1, \vec{a}_2, \vec{a}_3\). - **Basis:** At each lattice point: - 1 Cu atom at (0, 0, 0) - 2 O atoms at e.g. \((a/2, 0, 0)\) and \((0, a/2, 0)\) within the plane. **Explanation:** The basis tells us the additional atoms associated with each lattice point. For CuO₂, the basis describes the Cu at the origin and two O atoms displaced along the axes. --- #### **Final Answer (a):** - **Bravais lattice:** Simple tetragonal. - **Primitive lattice vectors:** \(\vec{a}_1 = a\hat{x}, \vec{a}_2 = a\hat{y}, \vec{a}_3 = c\hat{z}\). - **Unit cell:** Parallelepiped defined by these vectors. - **Basis:** 1 Cu at (0,0,0) and 2 O at \((a/2, 0, 0)\), \((0, a/2, 0)\). --- ### (b) **La₂CuO₄: Buckled Oxygen Atoms** #### **Step 1: Primitive Cell and Lattice Spacing** When O atoms are alternately shifted up/down out of the CuO₂ plane (like a checkerboard), the unit cell must be enlarged to accommodate this alternation. - The new primitive cell doubles in the in-plane direction: It now contains two Cu atoms (because the pattern alternates every other site). - **Primitive vectors:** - \(\vec{a}'_1 = a\hat{x} + a\hat{y}\) - \(\vec{a}'_2 = -a\hat{x} + a\hat{y}\) - \(\vec{a}'_3 = c\hat{z}\) - **Lattice spacing:** In-plane spacing is now \(\sqrt{2}a\) (along the diagonal), out-of-plane is \(c\). **Explanation:** Because the O atoms alternate up and down on adjacent sites, the smallest repeating unit must include both types (up and down) — so the primitive cell is doubled. --- #### **Step 2: Reciprocal Lattice** - The reciprocal lattice of a body-centered or doubled cell (as in this case) is **smaller** in the in-plane directions than for the original tetragonal cell. - **Primitive reciprocal lattice vectors:** - \(\vec{b}_1 = \frac{2\pi}{a}\left(\frac{1}{2}\hat{x} + \frac{1}{2}\hat{y}\right)\) - \(\vec{b}_2 = \frac{2\pi}{a}\left(-\frac{1}{2}\hat{x} + \frac{1}{2}\hat{y}\right)\) - \(\vec{b}_3 = \frac{2\pi}{c}\hat{z}\) **Explanation:** The reciprocal lattice vectors are defined by the real-space primitive vectors, and the doubling (or enlargement) of the unit cell halves the reciprocal lattice spacing in those directions. --- #### **Final Answer (b):** - **Primitive cell:** Contains two Cu atoms and four O atoms (one up, one down) with vectors \(\vec{a}'_1 = a\hat{x} + a\hat{y}\), \(\vec{a}'_2 = -a\hat{x} + a\hat{y}\), \(\vec{a}'_3 = c\hat{z}\); lattice spacing in-plane is \(\sqrt{2}a\), out-of-plane is \(c\). - **Reciprocal lattice:** Primitive vectors \(\vec{b}_1 = \frac{\pi}{a}(\hat{x} + \hat{y})\), \(\vec{b}_2 = \frac{\pi}{a}(-\hat{x} + \hat{y})\), \(\vec{b}_3 = \frac{2\pi}{c}\hat{z}\).