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\(\mathrm{X}\) 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 between atomic planes is approximately 0.232 nm.

Step by step solution

01

Understanding Bragg's Law

The given problem involves Bragg's Law, which is used to find the spacing between adjacent planes in a crystal. Bragg's Law is expressed as \( n\lambda = 2d\sin\theta \), where \( n \) is the order of the maximum, \( \lambda \) is the wavelength of the X-rays, \( d \) is the plane spacing, and \( \theta \) is the angle of incidence.
02

Identifying Known Values

According to the problem, the wavelength \( \lambda = 0.0850 \, \text{nm} \), the angle \( \theta = 21.5^{\circ} \), and the order of maximum \( n = 2 \). We need to find \( d \), the spacing between atomic planes.
03

Rearranging Bragg’s Equation

Reorganize Bragg's Law for \( d \):\[d = \frac{n\lambda}{2\sin\theta}\]
04

Substituting Values

Now substitute the known values for \( n \), \( \lambda \), and \( \theta \) into the rearranged Bragg's Law:\[d = \frac{2 \times 0.0850 \, \text{nm}}{2 \sin(21.5^{\circ})} = \frac{0.170 \, \text{nm}}{2 \sin(21.5^{\circ})}\]
05

Calculating the Sine and Solving for d

Use a calculator to find \( \sin(21.5^{\circ}) \), which is approximately 0.3665. Now plug this back into the equation to solve for \( d \):\[d = \frac{0.170 \, \text{nm}}{2 \times 0.3665} \approx 0.232 \, \text{nm}\].
06

Final Result

The spacing between adjacent atomic planes in the crystal is approximately 0.232 nm.

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Key Concepts

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

X-ray scattering
X-ray scattering is a fascinating process where X-rays are directed onto a crystal and the pattern formed provides insights into the crystal's structure. When X-rays hit the atoms in a crystal, they are scattered in various directions. Since the atoms in a crystal are arranged in a periodic order, the scattered X-rays can interfere with each other, either constructively or destructively.
This interference pattern is crucial because it can reveal the internal spacing between crystal planes, which are layers of atoms within the crystal. Using Bragg's Law, which relates the angle at which these X-rays are scattered to the spacing within the crystal, scientists can deduce the spacing between these planes. This ability is essential for understanding the material's structural properties. Even tiny differences in the scattering angles and intensities can indicate different arrangements of atoms or even the presence of defects within the crystal.
  • Constructive interference occurs when scattered waves overlap to intensify the pattern, forming sharp maxima in scattering strength.
  • Destructive interference happens when waves cancel each other out, resulting in minima in patterns.
crystal plane spacing
The spacing between crystal planes, denoted as \(d\), defines the distance between parallel planes of atoms within a crystal structure. This is a central element of understanding how X-rays interact with these planes. Crystal plane spacing is linked to the diffraction pattern produced when X-rays interact with the atomic structure. The arrangement and distance between layers significantly impact various physical properties of materials, such as their strength, elasticity, and optical characteristics.
In the context of Bragg's Law, the spacing \(d\) can be determined by measuring the angle at which X-rays are scattered and the known wavelength of the X-rays. This relationship allows scientists to map out how tightly packed the atomic layers are in a crystal. This measurement can be crucial for both physicists and chemists when characterizing new materials or exploring unknown substances. It can also aid in finding the material's purity and detecting any flaws or inconsistencies at a microscopic level.
Understanding crystal plane spacing can help in various applications including:
  • Designing new materials with desired mechanical or optical properties.
  • Analyzing and improving electronic components like semiconductors.
  • Researching geological samples for mineral content and distribution.
second-order maximum
The concept of a second-order maximum arises from the conditions set by Bragg's Law. When X-rays are diffracted by a crystal, multiple maxima can occur at different angles for the same wavelength, depending on the order number \(n\).
A second-order maximum refers to the scenario where \(n = 2\) in Bragg's Law \(n\lambda = 2d\sin\theta\). It essentially means that the path difference between the X-rays reflecting off adjacent planes is equivalent to twice the X-ray wavelength. This creates a condition where constructive interference leads to another strong peak in the diffraction pattern. Returning back to the exercise, this was observed at an angle \(\theta = 21.5^{\circ}\) for X-rays with a wavelength of 0.0850 nm.
The importance of different order maxima lies in their ability to provide more data points, which can strengthen confidence in measuring the plane spacing accurately. These insights can help further unlock information about the internal structure of complex materials and optimize performance in technological applications.
Characteristics of second-order maximum:
  • Signals a higher intensity peak due to constructive interference.
  • Occurs at wider angle \(\theta\) compared to the first-order peak.
  • Provides additional information for validating the plane-spacing calculation.

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Most popular questions from this chapter

The intensity of light in the Fraunhofer diffraction pattern of a single slit is $$I=I_{0}\left(\frac{\sin \gamma}{\gamma}\right)^{2}$$ where $$\gamma=\frac{\pi a \sin \theta}{\lambda}$$ (a) Show that the equation for the values of \(\gamma\) at which \(I\) is a maximum is \(\tan \gamma=\gamma .(\) b) Determine the three 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.)

A converging lens 7.20 \(\mathrm{cm}\) in diameter has a focal length of 300 \(\mathrm{mm} .\) If the resolution is diffraction limited, how far away can an object be if points on it 4.00 \(\mathrm{mm}\) apart are to be resolved (according to Rayleigh's criterion)? Use \(\lambda=550 \mathrm{nm} .\)

An astronaut in the space shuttle can just resolve two point sources on earth that are 65.0 \(\mathrm{m}\) apart. Assume that the resolution is diffraction limited and use Rayleigh's criterion. What is the astronaut's altitude above the earth? Treat his eye as a circular aperture with a diameter of 4.00 \(\mathrm{mm}\) (the diameter of his pupil), and take the wavelength of the light to be 550 \(\mathrm{nm} .\) Ignore the effect of fluid in the eye.

Thickness of Human Hair. 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?

Parallel rays of monochromatic light with wavelength 568 nm illuminate two identical slits and produce an interference pattern on a screen that is 75.0 \(\mathrm{cm}\) from the slits. The centers of the slits are 0.640 \(\mathrm{mm}\) apart and the width of each slit is 0.434 \(\mathrm{mm}\) . If the intensity at the center of the central maximum is \(5.00 \times\) \(10^{-4} \mathrm{W} / \mathrm{m}^{2},\) what is the intensity at a point on the screen that is 0.900 \(\mathrm{mm}\) from the center of the central maximum?

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