Let's break down your **Copper Oxide Layers** question into two steps as requested: --- ### (a) **Bravais Lattice, Primitive Vectors, Unit Cell, and Basis** **Step 1: Sketching the Bravais lattice and primitive vectors (Tetragonal case)** - **Bravais lattice:** The copper oxide layers (CuO₂) have a **tetragonal lattice** in the simplest approximation. This means the lattice has four-fold symmetry in the plane and is stacked identically in the third dimension. - **Primitive lattice vectors:** If the distance between copper atoms is **a** and the stacking distance is **c**, the primitive vectors are: - **a₁ = a x̂** - **a₂ = a ŷ** - **a₃ = c ẑ** - **Unit Cell:** The unit cell is a rectangular prism (a × a × c). - **Basis:** For the CuO₂ layer, the **basis** consists of: - 1 Cu atom at (0, 0, 0) - 2 O atoms at (½, 0, 0) and (0, ½, 0) (in units of a) **Explanation:** In the tetragonal approximation, the Cu atoms sit at the corners of the square lattice, and the O atoms are centered on the edges between Cu atoms, forming the CuO₂ layer. --- ### (b) **La₂CuO₄ - Distorted Structure, Primitive Cell, and Reciprocal Lattice** **Step 2: Distorted structure (Orthorhombic-like), primitive cell, and reciprocal lattice** - **Distorted structure:** In La₂CuO₄, the O atoms are displaced **alternately up and down** out of the CuO₂ plane, so the lattice is **no longer perfectly tetragonal**; it becomes **orthorhombic** (unit cell parameters: a ≠ b ≠ c). - **Primitive cell:** The primitive cell must now be **doubled along one direction** to account for the up/down alternation of O atoms. So, the primitive cell size becomes **a × a × 2c** if the alternation is along the c-axis. - **Lattice spacing:** The spacing between repeating units along c is now **2c** because of the up/down alternation. - **Basis:** The basis now includes: - 1 Cu atom at (0, 0, 0) - 2 O atoms at (½, 0, δ) and (0, ½, –δ) where δ is the displacement above and below the plane. - **Reciprocal lattice:** For the tetragonal cell, the reciprocal lattice vectors are: - **b₁ = (2π/a) x̂** - **b₂ = (2π/a) ŷ** - **b₃ = (2π/c) ẑ** For the doubled cell (along c), the z-component becomes: - **b₃' = (2π)/(2c) = π/c** **Explanation:** The doubling of the unit cell along c due to oxygen displacement halves the periodicity in reciprocal space along z, so the Brillouin zone is smaller in that direction. --- ## **Final Answer:** 1. **(a) In the simple tetragonal CuO₂ structure, the primitive lattice vectors are a₁ = a x̂, a₂ = a ŷ, a₃ = c ẑ; the unit cell is a × a × c, and the basis is 1 Cu at (0,0,0), and 2 O at (½,0,0), (0,½,0).** 2. **(b) In La₂CuO₄, with alternating O displacement, the primitive cell is doubled along c (a × a × 2c), the basis includes displaced O atoms at (½,0,δ) and (0,½,–δ), and the reciprocal lattice vectors are (2π/a, 2π/a, π/c).**

Let's proceed step-by-step to understand the effects of oxygen atom displacement in La₂CuO₄ on its lattice and reciprocal lattice. --- ### **Step 1: Understanding the Structural Distortion** **Explanation:** In the ideal CuO₂ layer, oxygen atoms lie perfectly in the plane, resulting in a simple tetragonal lattice with lattice vectors: - **Primitive vectors:** \[ \mathbf{a}_1 = a \hat{\mathbf{x}}, \quad \mathbf{a}_2 = a \hat{\mathbf{y}}, \quad \mathbf{a}_3 = c \hat{\mathbf{z}} \] - **Basis:** - Cu at \((0, 0, 0)\) - O at \((\frac{1}{2}, 0, 0)\) and \((0, \frac{1}{2}, 0)\) (fractional coordinates relative to the lattice) In La₂CuO₄, the oxygen atoms are displaced **alternately up and down** along the z-axis (out of the plane). This **breaks the perfect tetragonal symmetry**, leading to a **doubling of the unit cell** along the c-axis to accommodate the alternating displacement pattern. --- ### **Step 2: Primitive Cell and Lattice Spacing** **Explanation:** - The **original** unit cell has size \(a \times a \times c\). - Due to the **alternating up/down** displacement of oxygen atoms, the **primitive cell** must now contain **two layers** along the c-axis to include both oxygen positions. **Result:** - The **new primitive cell** in the **distorted structure** has: \[ \boxed{ \text{Size}: \quad a \times a \times 2c } \] - The **lattice points** are now separated by **\(2c\)** along the z-direction, because the pattern repeats every two layers. **Basis positions:** - Cu at \((0,0,0)\) - Oxygen atoms at \(\left(\frac{1}{2}, 0, \delta\right)\) and \(\left(0, \frac{1}{2}, -\delta\right)\), where \(\delta\) is the displacement magnitude along z (up or down). This **doubling** captures the **alternation of oxygen positions**. --- ### **Final Summary:** - The **primitive cell** in La₂CuO₄, accounting for oxygen displacement, is **doubled along the c-axis**: \[ \boxed{ \text{Primitive cell dimensions}: \quad a \times a \times 2c } \] - The **basis atoms** include Cu at \((0,0,0)\) and O atoms at \(\left(\frac{1}{2}, 0, \pm \delta\right)\), \(\left(0, \frac{1}{2}, \pm \delta\right)\). - The **reciprocal lattice vectors** for this **doubled cell** are: \[ \mathbf{b}_1 = \frac{2\pi}{a} \hat{\mathbf{x}}, \quad \mathbf{b}_2 = \frac{2\pi}{a} \hat{\mathbf{y}}, \quad \mathbf{b}_3 = \frac{\pi}{c} \hat{\mathbf{z}} \] This reduction in the Brillouin zone along the z-direction reflects the physical doubling of the real-space periodicity due to oxygen displacement. --- **In brief:** - **Primitive cell in La₂CuO₄:** \(a \times a \times 2c\) - **Lattice spacing:** doubled along c - **Reciprocal lattice vectors:** \(\mathbf{b}_1 = \frac{2\pi}{a} \hat{\mathbf{x}}\), \(\mathbf{b}_2 = \frac{2\pi}{a} \hat{\mathbf{y}}\), \(\mathbf{b}_3 = \frac{\pi}{c} \hat{\mathbf{z}}\) --- If you need further details or clarification, feel free to ask!