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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?

Short Answer

Expert verified
The wavelength of the x rays is \(0.283 \: nm\).

Step by step solution

01

Understand and write down the givens

From the problem, we understand that the first-order maximum (which means \(n = 1\)) occurs with a diffraction angle of \(39.4^{\circ}\). To use this angle in calculations, convert it to radians by multiplying by \(\pi/180\). Also, we are given the spacing of the atomic planes in the crystal is \(0.440 \mathrm{nm}\) or \(0.440 \times 10^{-9} m\). We need to find the wavelength \(\lambda\).
02

Apply Bragg’s Law

We substitute the given values into Bragg's law to solve for the wavelength \(\lambda\).
03

Substitute the givens into the equation

Substitute \(n = 1\), \(d = 0.440 \times 10^{-9} \: m\), and \(\theta = 39.4 \times (\pi/180) \: rad \) into the equation to get \(\lambda = 2d \sin \theta / n\). So, \(\lambda = 2 \times 0.440 \times 10^{-9}m \times \sin(39.4 \times \pi/180) / 1\).
04

Compute the wavelength

After computation, you will find the wavelength \(\lambda\) to be approximately \(0.283 \times 10^{-9} \: m\) or \(0.283 \: nm\). This is the wavelength of the x rays.

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Key Concepts

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

X-ray diffraction
X-ray diffraction is a crucial technique used in the study of materials' atomic structures. It occurs when X-rays, which are a form of electromagnetic radiation, interact with the material's crystalline structure. When these rays hit a crystal, they scatter in various directions.
A key concept in X-ray diffraction is the interference of waves: when the scattered waves are in sync or in phase, they amplify the signal, forming what are known as "diffraction peaks." This phenomenon is the basis of Bragg's Law, which helps determine the distances between atomic planes within a crystal.
X-ray diffraction is used for several applications:
  • Identifying the composition and arrangement of atoms in a material
  • Determining the size and the shape of the unit cell in the crystal lattice
  • Characterizing the size, shape, and distribution of materials at the molecular level
Understanding X-ray diffraction can deeply enhance one's grasp of the microscopic world, influencing fields from material science to chemistry.
Crystal lattice spacing
Crystal lattice spacing refers to the distance between adjacent planes of atoms in a crystal. This distance is crucial because it affects how X-rays diffract when they hit the crystal. In the context of Bragg's Law, this spacing is denoted as \(d\).
Crystal lattice spacing determines the conditions under which constructive interference of X-rays will occur, producing measurable diffraction peaks. The term "spacing" can also reveal insights into the material's density and structure.
In practical applications, knowing the crystal lattice spacing helps:
  • Identify crystalline phases and structures
  • Predict material properties like strength and conductivity
  • Enable the synthesis of new materials with desired attributes
Understanding this concept is essential for fields like crystallography, where the arrangement of atoms within a crystal is studied.
Wavelength calculation
The calculation of X-ray wavelength in the context of crystal diffraction involves using Bragg's Law, formulated by the equation \( n\lambda = 2d\sin\theta \), where \(n\) is the order of reflection, \(\lambda\) is the wavelength, \(d\) is the spacing between crystal planes, and \(\theta\) is the angle of incidence that causes reflection.
This formula allows scientists to calculate the wavelength of incident X-rays when the other variables are known. Accurate measurements of wavelength are crucial in identifying the type of X-ray being employed, which in turn affects the resolution and quality of the diffraction data.
Calculating wavelengths is critical for several reasons:
  • It helps in the identification of substances
  • It aids in the design of experiments, ensuring appropriate resolution
  • It provides insights into the electronic structure of materials
Mastering wavelength calculations is beneficial for those delving into materials science or physics, serving as a foundational tool in X-ray crystallography.

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Most popular questions from this chapter

The intensity of light in the Fraunhofer diffraction pattern of a single slit is given by Eq. (36.5). Let \(\gamma=\beta / 2\). (a) Show that the equation for the values of \(\gamma\) at which \(I\) is a maximum is \(\tan \gamma=\gamma\). (b) Determine the two smallest positive values of \(\gamma\) that are solutions of this equation. (Hint: You can use a trial-and-error procedure. Guess a value of \(\gamma\) and adjust your guess to bring \(\tan \gamma\) closer to \(\gamma\). A graphical solution of the equation is very helpful in locating the solutions approximately, to get good initial guesses.) (c) What are the positive values of \(\gamma\) for the first, second, and third minima on one side of the central maximum? Are the \(\gamma\) values in part (b) precisely halfway between the \(\gamma\) values for adjacent minima? (d) If \(a=12 \lambda,\) what are the angles \(\theta\) (in degrees) that locate the first minimum, the first maximum beyond the central maximum, and the second minimum?

A wildlife photographer uses a moderate telephoto lens of focal length \(135 \mathrm{~mm}\) and maximum aperture \(f / 4.00\) to photograph a bear that is \(11.5 \mathrm{~m}\) away. Assume the wavelength is \(550 \mathrm{nm}\). (a) What is the width of the smallest feature on the bear that this lens can resolve if it is opened to its maximum aperture (b) If, to gain depth of field, the photographer stops the lens down to \(f / 22.0\), what would be the width of the smallest resolvable feature on the bear?

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?

Although we have discussed single-slit diffraction only for a slit, a similar result holds when light bends around a straight, thin object, such as a strand of hair. In that case, \(a\) is the width of the strand. From actual laboratory measurements on a human hair, it was found that when a beam of light of wavelength \(632.8 \mathrm{nm}\) was shone on a single strand of hair, and the diffracted light was viewed on a screen \(1.25 \mathrm{~m}\) away, the first dark fringes on either side of the central bright spot were \(5.22 \mathrm{~cm}\) apart. How thick was this strand of hair?

Red light of wavelength \(633 \mathrm{nm}\) from a helium-neon laser passes through a slit \(0.350 \mathrm{~mm}\) wide. The diffraction pattern is observed on a screen \(3.00 \mathrm{~m}\) away. Define the width of a bright fringe as the distance between the minima on either side. (a) What is the width of the central bright fringe? (b) What is the width of the first bright fringe on either side of the central one?

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