Log out Go to App
Go to App

Learning Materials

Features

Discover

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.

Short Answer

Expert verified
X-rays can be used to measure atomic distances in crystals because their shorter wavelengths (0.01 to 10 nm) are more suitable for constructive interference with the crystal lattice, as described by Bragg's Law (\(n\lambda = 2d \sin{\theta}\)). In contrast, visible light has longer wavelengths (400 to 700 nm) which are too large compared to the atomic distances (on the order of angstroms), resulting in limited resolution and inability to accurately measure atomic distances in crystals.

Step by step solution

01

Wavelengths of X-rays and Visible Light

To understand the difference between X-rays and visible light when it comes to analyzing crystal structures, we should first note the difference in their wavelengths. X-rays have much shorter wavelengths, typically in the range of 0.01 to 10 nanometers (nm). In contrast, visible light has wavelengths between 400 and 700 nm.
02

Crystal Structure and Atomic Distances

In a crystal, atoms are arranged in a regularly repeating pattern, with a fixed distance between adjacent atoms called the atomic distance or lattice spacing. This atomic distance is typically on the order of angstroms (Å, 10^(-10) meters) – which is significantly smaller than the wavelengths of visible light.
03

Bragg's Law

Bragg's Law is a relationship that describes how X-rays (or other types of electromagnetic radiation) scatter off a crystalline structure. The law states that constructive interference between scattered waves occurs when \(n\lambda = 2d \sin{\theta}\), where: - \(n\) is an integer (1, 2, 3, ...) - \(\lambda\) is the wavelength of the incident wave - \(d\) is the atomic distance (lattice spacing) - \(\theta\) is the angle between the incident wave and the scattered wave
04

Constructive Interference and Resolution

For us to measure atomic distances in a crystal structure, we need constructive interference between the scattered waves. This constructive interference will create a detectable pattern that can be analyzed to determine the atomic spacing. Since the atomic spacing is on the order of angstroms, X-rays with shorter wavelengths in the range of 0.01 to 10 nm (1 to 100 Å) are better suited for this task than visible light.
05

Visible Light and Atomic Distances

When it comes to visible light, its wavelengths are much larger than the atomic distances in a crystal. This means that visible light cannot resolve the features within a crystal lattice due to the limited resolution caused by its longer wavelengths. Consequently, visible light is unable to provide enough information to accurately measure atomic distances in crystals. In conclusion, X-rays can be used to measure atomic distances in crystals because their shorter wavelengths are more suited to constructively interfere with the crystalline structure, while visible light cannot be used for this purpose due to its longer wavelengths that limit its resolution capabilities.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Unlike metals, semiconductors increase their conductivity as you heat them (up to a point). Suggest an explanation.

Which would you expect to be the more ductile element, (a) \(\mathrm{Ag}\) or \(\mathrm{Cr}\) (b) Zn or Ge? In each case explain your reasoning.

Amorphous silica, \(\mathrm{SiO}_{2}\), has a density of about $2.2 \mathrm{~g} / \mathrm{cm}^{3}$, whereas the density of crystalline quartz, another form of \(\mathrm{SiO}_{2}\), is \(2.65 \mathrm{~g} / \mathrm{cm}^{3}\). Which of the following statements is the best explanation for the difference in density? (a) Amorphous silica is a network-covalent solid, but quartz is metallic. (b) Amorphous silica crystallizes in a primitive cubic lattice. (c) Quartz is harder than amorphous silica. (d) Quartz must have a larger unit cell than amorphous silica. (e) The atoms in amorphous silica do not pack as efficiently in three dimensions as compared to the atoms in quartz.

You are given a gray substance that melts at \(700^{\circ} \mathrm{C} ;\) the solid is a conductor of electricity and is insoluble in water. Which type of solid (molecular, metallic, covalent-network, or ionic) might this substance be?

If you want to make a polymer for plastic wrap, should you strive to make a polymer that has a high or low degree of crystallinity?
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Inorganic Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Making Measurements

Read Explanation

Kinetics

Read Explanation

Chemical Reactions

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free