Let's solve each part step by step: --- ## 1. (a)-(d) Identify the Miller indices of the crystallographic planes shown. ### **General Method for Finding Miller Indices (hkl):** 1. **Find the intercepts of the plane with the axes** in terms of the lattice constants (a1, a2, a3). 2. **Take reciprocals** of the intercepts. 3. **Clear fractions** to get the smallest set of integers. 4. **Enclose in parentheses:** (h k l). --- ### **(a)** - The plane cuts the **a3** axis at 1, and is parallel to the **a1** and **a2** axes (never intersects). - Intercepts: (∞, ∞, 1) - Reciprocals: (0, 0, 1) **Miller index:** **(001)** --- ### **(b)** - The plane cuts all axes at 1: intercepts at (1, 1, 1). - Reciprocals: (1, 1, 1) **Miller index:** **(111)** --- ### **(c)** - The plane cuts the **a1** axis at 1, is parallel to the **a2** and **a3** axes (never intersects). - Intercepts: (1, ∞, ∞) - Reciprocals: (1, 0, 0) **Miller index:** **(100)** --- ### **(d)** - The plane cuts the **a2** axis at 1, is parallel to the **a1** and **a3** axes. - Intercepts: (∞, 1, ∞) - Reciprocals: (0, 1, 0) **Miller index:** **(010)** --- ### **(e) Which direction is perpendicular to the plane in (b)?** - The normal to the (111) plane is the [111] direction, i.e., the direction with direction cosines (1,1,1). **Draw an arrow from the origin to the point (1,1,1), and label it as the [111] direction.** --- ## **Summary Table:** | Part | Intercepts | Reciprocals | Miller Index | |------|--------------|-------------|-------------| | (a) | (∞, ∞, 1) | (0, 0, 1) | (001) | | (b) | (1, 1, 1) | (1, 1, 1) | (111) | | (c) | (1, ∞, ∞) | (1, 0, 0) | (100) | | (d) | (∞, 1, ∞) | (0, 1, 0) | (010) | **(e) The direction perpendicular to (b) is [111].** --- ### **2. Nearest Neighbor Distance in Si (diamond cubic), a = 5.43×10⁻⁸ cm** In diamond cubic structure: - Nearest neighbor distance, \( d = \frac{a\sqrt{3}}{4} \) **Calculation:** \[ d = \frac{5.43 \times 10^{-8} \times \sqrt{3}}{4} \] \[ d = \frac{5.43 \times 10^{-8} \times 1.732}{4} \] \[ d = \frac{9.406 \times 10^{-8}}{4} \] \[ d = 2.35 \times 10^{-8} \text{ cm} \] --- ### **3. Germanium: Diamond lattice, 4.42×10²² atoms/cm³. Find unit cell length \(a\).** - Each unit cell contains 8 atoms (for diamond structure). - Number of unit cells per cm³: \( N_{uc} = \frac{4.42 \times 10^{22}}{8} = 5.525 \times 10^{21} \) unit cells/cm³ - Volume of each unit cell: \( V_{uc} = \frac{1}{N_{uc}} = \frac{1}{5.525 \times 10^{21}} = 1.81 \times 10^{-22} \) cm³ - \( a^3 = V_{uc} \) - \( a = (V_{uc})^{1/3} = (1.81 \times 10^{-22})^{1/3} \) Calculate: \[ a = 1.21 \times 10^{-7} \text{ cm} \] --- ## **Final Answers:** 1. **Miller indices:** - (a) (001) - (b) (111) - (c) (100) - (d) (010) - (e) [111] 2. **Nearest neighbor distance in Si:** \(2.35 \times 10^{-8}\) cm 3. **Unit cell length for Ge:** \(1.21 \times 10^{-7}\) cm --- If you need the solution with more detail or any diagram, let me know!