/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 35 X rays of wavelength \(0.0850 \m... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 36: Problem 35

X rays of wavelength \(0.0850 \mathrm{nm}\) are scattered from the atoms of a crystal. The second-order maximum in the Bragg reflection occurs when the angle \(\theta\) in Fig. 36.22 is \(21.5^{\circ} .\) What is the spacing between adjacent atomic planes in the crystal?

Short Answer

Expert verified
The spacing \(d\) between adjacent atomic planes in the crystal can be calculated using values of \(n\), \(\lambda\) and \(\theta\) in the rearranged Bragg's Law. Ensure the calculator is in degree mode while getting sine of the angle.

Step by step solution

01

Understand Bragg's Law

Bragg's Law is given by \[ n\lambda = 2d\sin\theta \] where \(\lambda\) is the wavelength of the X-rays, \(\theta\) is the scattering angle, \(n\) is the order of the reflection, and \(d\) is the distance between the planes in the crystal.
02

Convert units

Before doing calculations, it's important to ensure all units are compatible. However, in this case, we're dealing with nanometers (nm) and this unit is commonly used in Bragg's Law, so no conversion is necessary. Remember, always check your units first.
03

Insert known values into Bragg's Law

We know that \(\lambda = 0.0850 \, nm, \, n = 2, \, and \, \theta = 21.5^\circ\). We can then substitute these values into Bragg's Law.
04

Rearrange the equation to solve for unknown \(d\)

Now, we rearrange Bragg's Law to isolate \(d\) on one side: \[ d = \frac{n\lambda}{2\sin\theta} \] and substitute the values into it.
05

Calculate the spacing \(d\) between atomic planes

Finally, substitute in values for \(n\), \(\lambda\), and \(\theta\) and calculate the value of \(d\) (crystal spacing). Thus, we get the required expression. Be cautious with the trigonometric function sin, ensure the calculator is in degree mode for calculation.

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

X-ray diffraction
When X-rays interact with a crystal, they undergo a process known as X-ray diffraction. This phenomenon occurs due to the wave-like nature of X-rays, which are a form of electromagnetic radiation with very short wavelengths.
X-ray diffraction is essential for studying the internal structure of crystals. Because the X-rays are scattered by the electron cloud of atoms in the crystal, the pattern of scattered X-rays reveals information about the atomic arrangement inside the crystal.
  • The angles and intensities of these scattered X-rays provide a "fingerprint" for identifying the molecular structure.
  • It helps in determining the spacing between atomic planes.
  • It can reveal defects or imperfections in the crystal structure.
This technique is a cornerstone in fields like materials science, chemistry, and biology, enabling scientists to identify the arrangement of atoms within a crystal lattice.
crystal lattice spacing
In a crystal, atoms are arranged in a repeated and ordered pattern, forming what is known as a crystal lattice. "Lattice spacing" refers to the distance between planes of atoms in the crystal.
Understanding this spacing is vital for multiple reasons:
  • The lattice spacing affects the physical properties of the material, such as its density and thermal expansion.
  • It plays a role in how the material interacts with light and other forms of electromagnetic radiation.
  • In experiments involving X-ray diffraction, calculating the lattice spacing allows us to understand how atoms are arranged within the crystal. This is because the planes of atoms act like a three-dimensional diffraction grating for X-ray waves, scattering them in specific directions.
Bragg's Law provides a way to calculate this spacing by using the wavelength of the X-rays and the angle at which they are diffracted, thus linking the physical property of the lattice spacing with observable data from diffraction experiments.
atomic planes
Atomic planes in a crystal are imaginary layers that extend through the structure, intersecting one another in a systematic way. These planes are crucial in understanding how crystals grow, how they fracture, and how their internal stress distributes.
Every set of planes is defined by the density of atoms it contains and the distance between each plane in the set, known as the plane spacing. This is vital when considering how light, particularly X-rays, interacts with these structures.
  • When X-rays hit a crystal, they can reflect off these atomic planes, creating a pattern dependent on the spacing between the planes and their orientation.
  • Bragg's Law uses this concept to relate the angle of reflection to the wavelength of the X-rays and the spacing between these planes.
  • Understanding atomic planes is not only critical for crystallography but also for engineering applications where the durability of materials under stress is considered.
Thus, atomic planes act as a blueprint for how a crystal can react to various external forces, playing a key role in the field of material science.

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

Monochromatic x rays are incident on a crystal for which the spacing of the atomic planes is \(0.440 \mathrm{nm}\). The first-order maximum in the Bragg reflection occurs when the incident and reflected \(x\) rays make an angle of \(39.4^{\circ}\) with the crystal planes. What is the wavelength of the x rays?

A glass sheet is covered by a very thin opaque coating. In the middle of this sheet there is a thin scratch \(0.00125 \mathrm{~mm}\) thick. The sheet is totally immersed beneath the surface of a liquid. Parallel rays of monochromatic coherent light with wavelength \(612 \mathrm{nm}\) in air strike the sheet perpendicular to its surface and pass through the scratch. A screen is placed in the liquid a distance of \(30.0 \mathrm{~cm}\) away from the sheet and parallel to it. You observe that the first dark fringes on either side of the central bright fringe on the screen are \(22.4 \mathrm{~cm}\) apart. What is the refractive index of the liquid?

Your physics study partner tells you that the width of the central bright band in a single-slit diffraction pattern is inversely proportional to the width of the slit. This means that the width of the central maximum increases when the width of the slit decreases. The claim seems counterintuitive to you, so you make measurements to test it. You shine monochromatic laser light with wavelength \(\lambda\) onto a very narrow slit of width \(a\) and measure the width \(w\) of the central maximum in the diffraction pattern that is produced on a screen \(1.50 \mathrm{~m}\) from the slit. (By "width," you mean the distance on the screen between the two minima on either side of the central maximum.) Your measurements are given in the table. $$ \begin{array}{l|llllllllr} a(\mu \mathrm{m}) & 0.78 & 0.91 & 1.04 & 1.82 & 3.12 & 5.20 & 7.80 & 10.40 & 15.60 \\ \hline w(\mathrm{~m}) & 2.68 & 2.09 & 1.73 & 0.89 & 0.51 & 0.30 & 0.20 & 0.15 & 0.10 \end{array} $$ (a) If \(w\) is inversely proportional to \(a\), then the product \(a w\) is constant, independent of \(a\). For the data in the table, graph \(a w\) versus \(a\). Explain why \(a w\) is not constant for smaller values of \(a\). (b) Use your graph in part (a) to calculate the wavelength \(\lambda\) of the laser light. (c) What is the angular position of the first minimum in the diffraction pattern for (i) \(a=0.78 \mu \mathrm{m}\) and (ii) \(a=15.60 \mu \mathrm{m} ?\)

A diffraction grating has 650 slits \(/ \mathrm{mm}\). What is the highest order that contains the entire visible spectrum? (The wavelength range of the visible spectrum is approximately \(380-750 \mathrm{nm}\).)

Two satellites at an altitude of \(1200 \mathrm{~km}\) are separated by \(28 \mathrm{~km} .\) If they broadcast \(3.6 \mathrm{~cm}\) microwaves, what minimum receiving-dish diameter is needed to resolve (by Rayleigh's criterion) the two transmissions?
See all solutions

Recommended explanations on Physics Textbooks

Energy Physics

Read Explanation

Atoms and Radioactivity

Read Explanation

Engineering Physics

Read Explanation

Measurements

Read Explanation

Torque and Rotational Motion

Read Explanation

Thermodynamics

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.