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Chapter 36: Problem 36

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鈥檚 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

Monochromatic light from a distant source is incident on a slit \(0.750 \mathrm{~mm}\) wide. On a screen \(2.00 \mathrm{~m}\) away, the distance from the central maximum of the diffraction pattern to the first minimum is measured to be \(1.35 \mathrm{~mm}\). Calculate the wavelength of the light.

Coherent electromagnetic waves with wavelength \(\lambda\) pass through a narrow slit of width \(a\). The diffraction pattern is observed on a tall screen that is \(2.00 \mathrm{~m}\) from the slit. When \(\lambda=500 \mathrm{nm}\), the width on the screen of the central maximum in the diffraction pattern is \(8.00 \mathrm{~mm}\). For the same slit and screen, what is the width of the central maximum when \(\lambda=0.125 \mathrm{~mm} ?\)

In the 1920s Clinton Davisson and Lester Germer accidentally observed diffraction when electrons with \(54 \mathrm{eV}\) of energy were scattered off crystalline nickel. The diffraction peak occurred when the angle between the incident beam and the scattered beam was \(50^{\circ}\). (a) What is the corresponding angle \(\theta\) relevant for Eq. (36.16)\(?\) (b) The planes in crystalline nickel are separated by \(0.091 \mathrm{nm}\), as determined by x-ray scattering experiments. According to the Bragg condition, what wavelength do the electrons in these experiments have? (c) Given the mass of an electron as \(9.11 \times 10^{-31} \mathrm{~kg},\) what is the corresponding classical speed \(v_{\mathrm{cl}}\) of the diffracted electrons? (d) Assuming the electrons correspond to a wave with speed \(v_{\mathrm{cl}}\) and wavelength \(\lambda,\) what is the frequency \(f\) of the diffracted waves? (e) Quantum mechanics postulates that the energy \(E\) and the frequency \(f\) of a particle are related by \(E=h f,\) where \(h\) is known as Planck's constant. Estimate \(h\) from these observations. (f) Our analysis has a small flaw: The relevant wave velocity, known as a quantum phase velocity, is half the classical particle velocity, for reasons explained by deeper aspects of quantum physics. Re-estimate the value of \(h\) using this modification. (g) The established value of Planck's constant is \(6.6 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}\). Does this agree with your estimate?

A laser beam of wavelength \(\lambda=632.8 \mathrm{nm}\) shines at normal incidence on the reflective side of a compact disc. (a) The tracks of tiny pits in which information is coded onto the CD are \(1.60 \mu \mathrm{m}\) apart. For what angles of reflection (measured from the normal) will the intensity of light be maximum? (b) On a DVD, the tracks are only \(0.740 \mu \mathrm{m}\) apart. Repeat the calculation of part (a) for the DVD.

An interference pattern is produced by light of wavelength \(580 \mathrm{nm}\) from a distant source incident on two identical parallel slits separated by a distance (between centers) of \(0.530 \mathrm{~mm} .\) (a) If the slits are very narrow, what would be the angular positions of the first-order and second- order, two-slit interference maxima? (b) Let the slits have width \(0.320 \mathrm{~mm} .\) In terms of the intensity \(I_{0}\) at the center of the central maximum, what is the intensity at each of the angular positions in part (a)?
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