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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

The wavelength range of the visible spectrum is approximately \(380-750 \mathrm{nm} .\) White light falls at normal incidence on a diffraction grating that has 350 slits \(/ \mathrm{mm} .\) Find the angular width of the visible spectrum in (a) the first order and (b) the third order. (Note: An advantage of working in higher orders is the greater angular spread and better resolution. A disadvantage is the overlapping of different orders, as shown in Example \(36.4 .\) )

The maximum resolution of the eye depends on the diameter of the opening of the pupil (a diffraction effect) and the size of the retinal cells. The size of the retinal cells (about \(5.0 \mu \mathrm{m}\) in diameter) limits the size of an object at the near point \((25 \mathrm{~cm})\) of the eye to a height of about \(50 \mu \mathrm{m}\). (To get a reasonable estimate without having to go through complicated calculations, we shall ignore the effect of the fluid in the eye. eter of the human pupil is about \(2.0 \mathrm{~mm}\), does the Rayleigh criterion allow us to resolve a \(50-\mu \mathrm{m}\) -tall object at \(25 \mathrm{~cm}\) from the eye with light of wavelength \(550 \mathrm{nm} ?\) (b) According to the Rayleigh criterion, what is the shortest object we could resolve at the \(25 \mathrm{~cm}\) near point with light of wavelength \(550 \mathrm{nm} ?\) (c) What angle would the object in part (b) subtend at the eye? Express your answer in minutes \(\left(60 \mathrm{~min}=1^{\circ}\right),\) and compare it with the experimental value of about \(1 \mathrm{min.}\) (d) Which effect is more important in limiting the resolution of our eyes: diffraction or the size of the retinal cells?

The Hubble Space Telescope has an aperture of \(2.4 \mathrm{~m}\) and focuses visible light \((380-750 \mathrm{nm})\). The Arecibo radio telescope in Puerto Rico is \(305 \mathrm{~m}(1000 \mathrm{ft})\) in diameter (it is built in a mountain valley) and focuses radio waves of wavelength \(75 \mathrm{~cm}\). (a) Under optimal viewing conditions, what is the smallest crater that each of these telescopes could resolve on our moon? (b) If the Hubble Space Telescope were to be converted to surveillance use, what is the highest orbit above the surface of the earth it could have and still be able to resolve the license plate (not the letters, just the plate) of a car on the ground? Assume optimal viewing conditions, so that the resolution is diffraction limited.

When laser light of wavelength \(632.8 \mathrm{nm}\) passes through a diffraction grating, the first bright spots occur at \(\pm 17.8^{\circ}\) from the central maximum. (a) What is the line density (in lines/cm) of this grating? (b) How many additional bright spots are there beyond the first bright spots, and at what angles do they occur?

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?
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