/*! 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 65 An \(x\) -ray beam of wavelength... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An \(x\) -ray beam of wavelength \(A\) undergoes first-order reflection (Bragg law diffraction) from a crystal when its angle of incidence to a crystal face is \(23^{\circ}\), and an x-ray beam of wavelength 97 pm undergoes third-order reflection when its angle of incidence to that face is \(60^{\circ} .\) Assuming that the two beams reflect from the same family of reflecting planes, find (a) the interplanar spacing and (b) the wavelength \(A\).

Short Answer

Expert verified
The interplanar spacing is 111.8 pm, and the wavelength \( A \) is 87.3 pm.

Step by step solution

01

Understanding Bragg's Law

Bragg's Law relates the wavelength of x-rays, the angle of incidence, and the interplanar spacing of the crystal. It is written as: \( n \lambda = 2d \sin(\theta) \), where \( n \) is the order of reflection, \( \lambda \) is the wavelength, \( d \) is the interplanar spacing, and \( \theta \) is the angle of incidence.
02

Apply Bragg's Law for First-Order Reflection

For the first x-ray beam with unknown wavelength \( A \), use \( n = 1 \), so the equation becomes \( A = 2d \sin(23^\circ) \).
03

Apply Bragg's Law for Third-Order Reflection

For the second x-ray beam with a wavelength of 97 pm (or \( 97 \times 10^{-12} \) m), use \( n = 3 \), so the equation becomes \( 3(97 \times 10^{-12}) = 2d \sin(60^\circ) \).
04

Solve for Interplanar Spacing \(d\)

From the second equation, calculate \( d \):\[ 2d = 3 \times 97 \times 10^{-12} \div \sin(60^\circ) \].So, using \( \sin(60^\circ) = \sqrt{3}/2 \), calculate:\[ d = \frac{3 \times 97 \times 10^{-12}}{2 \times ( \sqrt{3}/2 )} \approx 111.8 \times 10^{-12} \text{ m} \] or 111.8 pm.
05

Solve for Wavelength \(A\)

Use the calculated \( d \) from the previous step in the first equation:\[ A = 2 \times 111.8 \times 10^{-12} \times \sin(23^\circ) \].Now calculate \( A \), using \( \sin(23^\circ) \approx 0.3907 \), yielding:\[ A \approx 87.3 \times 10^{-12} \text{ m} \] or 87.3 pm.

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 91Ó°ÊÓ!

Key Concepts

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

X-ray diffraction
X-ray diffraction is a phenomenon that occurs when x-rays strike a crystalline material. As the rays interact with the material, they are scattered by the atoms in the crystal lattice. When the angles of the scattered x-rays satisfy certain conditions, they undergo constructive interference. This results in notable patterns of elevated intensity known as diffraction patterns. X-rays are an ideal tool for probing the atomic structure of materials because their wavelengths are comparable to the spacing between atoms in solids. These diffraction patterns are instrumental in determining the internal structures of crystals, leading to insights into material properties and behaviors.
Interplanar spacing
Interplanar spacing, denoted by the symbol \( d \), is the perpendicular distance between adjacent planes of atoms in a crystal lattice. It is a critical factor in Bragg's Law, which uses it to relate the wavelength of incident x-rays to the angle at which they are reflected. The value of \( d \) directly affects the angles and orders at which x-ray diffraction occurs. By calculating \( d \), scientists can gain insights into the crystalline structure and even identify unknown materials. Each type of crystal has its own unique set of interplanar spacings, so measuring these can effectively differentiate between different materials.
Order of reflection
The order of reflection, represented by \( n \), is an integer that signifies the sequence of diffraction events in a crystal. When Bragg's Law (\( n \lambda = 2d \sin(\theta) \)) is applied, \( n \) describes the number of wavelengths that fit into the path difference between x-rays reflected from adjacent planes. For example, a first-order reflection (\( n = 1 \)) involves a single wavelength fitting into the path difference. As \( n \) increases, higher orders like the second-order or third-order reflections occur, involving larger multiples of the wavelength. Choosing the correct order is crucial for accurately applying Bragg's Law to find interplanar spacing or the wavelength of x-rays.
Wavelength calculation
The wavelength of x-rays is a key factor in probing crystal structures. Bragg's Law provides a straightforward method to calculate this wavelength, given known values for reflection order, interplanar spacing, and angle of reflection. In mathematical terms, the basic formula is \( \lambda = \frac{2d\sin(\theta)}{n} \). To calculate the wavelength first involves accurately measuring the angle \( \theta \) and the reflected order \( n \), and determining the interplanar spacing \( d \) through diffraction experiments. This approach not only helps determine the wavelength but also enhances understanding of the crystalline structure and behavior. By using known variables, students can solve for unknown wavelengths, providing a clearer picture of the material's internal arrangements.

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

(A) How far from grains of red sand must you be to position yourself just at the limit of resolving the grains if your pupil diameter is \(1.5 \mathrm{~mm}\), the grains are spherical with radius \(50 \mu \mathrm{m}\), and the light from the grains has wavelength \(650 \mathrm{nm} ?\) (b) If the grains were blue and the light from them had wavelength \(400 \mathrm{nm}\), would the answer to (a) be larger or smaller?

Light containing a mixture of two wavelengths, 500 and \(600 \mathrm{nm},\) is incident normally on a diffraction grating. It is desired (1) that the first and second maxima for each wavelength appear at \(\theta \leq 30^{\circ},\) (2) that the dispersion be as high as possible, and (3) that the third order for the \(600 \mathrm{nm}\) light be a missing order. (a) What should be the slit separation? (b) What is the smallest individual slit width that can be used? (c) For the values calculated in (a) and (b) and the light of wavelength \(600 \mathrm{nm}\), what is the largest order of maxima produced by the grating?

(A) What is the angular separation of two stars if their images are barely resolved by the Thaw refracting telescope at the Allegheny Observatory in Pittsburgh? The lens diameter is \(76 \mathrm{~cm}\) and its focal length is \(14 \mathrm{~m}\). Assume \(\lambda=550 \mathrm{nm}\). (b) Find the distance between these barely resolved stars if each of them is 10 light-years distant from Earth. (c) For the image of a single star in this telescope, find the diameter of the first dark ring in the diffraction pattern, as measured on a photographic plate placed at the focal plane of the telescope lens. Assume that the structure of the image is associated entirely with diffraction at the lens aperture and not with lens "errors."

A diffraction grating is made up of slits of width \(300 \mathrm{nm}\) with separation \(900 \mathrm{nm}\). The grating is illuminated by monochromatic plane waves of wavelength \(\lambda=600 \mathrm{nm}\) at normal incidence. (a) How many maxima are there in the full diffraction pattern? (b) What is the angular width of a spectral line observed in the first order if the grating has 1000 slits?

The telescopes on some commercial surveillance satellites can resolve objects on the ground as small as \(85 \mathrm{~cm}\) across (see Google Earth), and the telescopes on military surveillance satellites reportedly can resolve objects as small as \(10 \mathrm{~cm}\) across. Assume first that object resolution is determined entirely by Rayleigh's criterion and is not degraded by turbulence in the atmosphere. Also assume that the satellites are at a typical altitude of \(400 \mathrm{~km}\) and that the wavelength of visible light is \(550 \mathrm{nm}\). What would be the required diameter of the telescope aperture for (a) \(85 \mathrm{~cm}\) resolution and (b) \(10 \mathrm{~cm}\) resolution? (c) Now, considering that turbulence is certain to degrade resolution and that the aperture diameter of the Hubble Space Telescope is \(2.4 \mathrm{~m},\) what can you say about the answer to (b) and about how the military surveillance resolutions are accomplished?

See all solutions

Recommended explanations on Physics Textbooks

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.