A crystalline solid consists of atoms stackedup in a repeating lattice structure. Con-sider a crystal as shown in Figure P1.53a. The atoms reside at the cor-ners of cubes of side $L=0.200\mathrm{nm}.$ Onepiece of evidence for the regular arrangement of atoms comesfrom the flat surfacesalong which a crystal separates, or cleaves, when it is broken. Suppose this crystal cleaves along a face diagonal as shown in Figure P1.53b. Calculate the spacing $d$ between two adjacent atomic planes that separate when the crystal cleaves.

Key Concepts: Crystalline Solid, Lattice Structure, Atomic Planes, Spacing Between Planes Explanation: The problem describes a crystalline solid with atoms arranged at the corners of cubes with a side-length of L=0.200nm. The crystal cleaves along a face diagonal, and we need to calculate the spacing between two adjacent atomic planes. Step by Step Solution: Step 1. We can begin by drawing a diagram of the crystal and identifying the planes that separate when the crystal cleaves along a diagonal face. In this case, we can see that the planes are the face diagonal of the cubes, which have a distance of L√3. Step 2. In a cubic lattice, the (hkl) spacing between adjacent planes can be calculated using the formula: \n d = \frac{a}{\sqrt{h^2 + k^2 + l^2}}\n Here, the planes are the face diagonal of the cubes, so we have hkl = 111. Substituting L√3 for the lattice constant 'a', we get:\n d = \frac{L\sqrt{3}}{\sqrt{1^2 + 1^2 + 1^2}} = 0.139 \mathrm{nm} Step 3. Therefore, the spacing between two adjacent atomic planes that separate when the crystal cleaves is 0.139 nm. Final Answer: 0.139 nm