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Ch.12 - Solids and Modern Materials
All textbooksBrown 14th EditionCh.12 - Solids and Modern MaterialsProblem 119
Chapter 12, Problem 119

Explain why X-rays can be used to measure atomic distances in crystals but visible light cannot be used for this purpose.

Verified step by step guidance
1
Step 1: Understand the concept of wavelength and its relation to diffraction. The wavelength of light is a crucial factor in determining its ability to resolve small structures. Diffraction occurs when waves encounter obstacles or openings, and the pattern of diffraction can be used to measure distances.
Step 2: Compare the wavelengths of X-rays and visible light. X-rays have much shorter wavelengths (on the order of 0.01 to 10 nanometers) compared to visible light (which ranges from about 400 to 700 nanometers).
Step 3: Relate the wavelength to the size of atomic distances. Atomic distances in crystals are typically on the order of a few angstroms (1 angstrom = 0.1 nanometers). To resolve these distances, the wavelength of the probing radiation must be on the same order of magnitude or smaller.
Step 4: Explain why X-rays are suitable for measuring atomic distances. Because X-rays have wavelengths comparable to the distances between atoms in a crystal, they can be diffracted by the crystal lattice, allowing for the measurement of atomic distances through techniques like X-ray crystallography.
Step 5: Discuss why visible light is not suitable for this purpose. The longer wavelengths of visible light are much larger than atomic distances, making it impossible for visible light to be diffracted by the crystal lattice in a way that would allow for the resolution of atomic-scale features.

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Wavelength and Atomic Scale

The ability to measure atomic distances relies on the wavelength of the radiation used. X-rays have wavelengths on the order of 0.01 to 10 nanometers, which are comparable to the distances between atoms in a crystal lattice. In contrast, visible light has much longer wavelengths (approximately 400 to 700 nanometers), making it unsuitable for probing atomic-scale structures.
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Diffraction and Crystal Structure

X-ray diffraction is a technique that exploits the wave nature of X-rays to study crystal structures. When X-rays interact with the periodic arrangement of atoms in a crystal, they are scattered in specific directions, creating a diffraction pattern. This pattern can be analyzed to determine the arrangement of atoms within the crystal, a process that is not possible with visible light due to its longer wavelength.
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Energy and Interaction with Matter

X-rays possess higher energy than visible light, allowing them to penetrate materials and interact with electrons in atoms more effectively. This interaction is crucial for obtaining detailed information about atomic distances and arrangements. Visible light, having lower energy, is less effective in probing the internal structure of materials, as it is more likely to be absorbed or scattered without providing useful structural information.
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Related Practice
Textbook Question

Sodium oxide (Na2O) adopts a cubic structure with Na atoms represented by green spheres and O atoms by red spheres.

(c) The unit cell edge length is 5.550 Å. Determine the density of Na2O.

Textbook Question

In their study of X-ray diffraction, William and Lawrence Bragg determined that the relationship among the wavelength of the radiation 1l2, the angle at which the radiation is diffracted 1u2, and the distance between planes of atoms in the crystal that cause the diffraction (d) is given by nl = 2d sin u. X rays from a copper X-ray tube that have a wavelength of 1.54 Å are diffracted at an angle of 14.22 degrees by crystalline silicon. Using the Bragg equation, calculate the distance between the planes of atoms responsible for diffraction in this crystal, assuming n = 1 (first-order diffraction).

Textbook Question

Germanium has the same structure as silicon, but the unit cell size is different because Ge and Si atoms are not the same size. If you were to repeat the experiment described in the previous problem but replace the Si crystal with a Ge crystal, would you expect the X rays to be diffracted at a larger or smaller angle u?