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A wavelength of 0.129 nm characterizes \(\mathrm{K}_{\alpha}\) x-rays from zinc. When a beam of these x-rays is incident on the surface of a crystal whose structure is similar to that of NaCl, a first-order maximum is observed at \(8.15^{\circ} .\) Calculate the interplanar spacing based on this information.

Short Answer

Expert verified
The interplanar spacing (d) is approximately 0.457 nm.

Step by step solution

01

Understanding the Given Information

Given that the wavelength \(\lambda\) = 0.129 nm and the angle of the first-order maximum (\( \theta_1 \)) = \( 8.15^{\circ} \). The order of diffraction (n) is 1. We need to find the interplanar spacing (d) for the NaCl-like crystal.
02

Use Bragg's Law

Recall Bragg's Law, which is given as \(n\lambda = 2d\sin\theta\) where n is the order of diffraction, \(\lambda\) is the wavelength, d is the interplanar spacing and \( \theta \) is the angle of incidence. We need to rearrange Bragg's Law to solve for the interplanar spacing, d.
03

Solve for Interplanar Spacing (d)

Rearrange Bragg's Law to solve for d: \(d = (n\lambda) / (2\sin\theta)\). Substituting the known values, \(d = (1 * 0.129) / (2*\sin(8.15))\). Using a calculator, you can solve for interplanar spacing (d).

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Key Concepts

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

Understanding Interplanar Spacing Calculation
Interplanar spacing is a fundamental concept in crystallography that refers to the distance between two parallel planes of atoms within a crystal. This spacing is crucial when it comes to identifying and characterizing crystal structures using X-ray diffraction techniques.

Calculating interplanar spacing involves understanding Bragg's Law, a principle that describes the relationship between X-ray wavelength, the diffraction angle, and the interplanar spacing. The law is mathematically expressed as: \[n\boldsymbol{\rm{\text{λ}\text{}}} = 2d\boldsymbol{\rm{\text{sin}(\rm{\theta})}}\text{,}\] where \(n\) is the order of diffraction, \(\boldsymbol{\rm{\text{λ}\text{}}}\) is the wavelength of the X-rays, \(d\) is the interplanar spacing, and \(\boldsymbol{\rm{\text{θ}\text{}}}\) is the angle of incidence. To find the interplanar spacing, you rearrange the formula to solve for \(d\):\[d = \frac{n\boldsymbol{\rm{\text{λ}\text{}}}}{2\boldsymbol{\rm{\text{sin}(\rm{\theta})}}}\]Using the given X-ray wavelength and angle of diffraction, you can calculate the interplanar spacing by inserting these values into the equation. It's essential to use the correct units when performing these calculations to ensure an accurate result. Computing the spacing allows researchers to infer the crystal structure and determine its properties.
X-ray Diffraction Explained
X-ray diffraction (XRD) is a powerful tool used in material science and crystallography to investigate the atomic and molecular structure of materials. When X-rays are directed at a crystal, they are scattered by the crystal's atoms, producing a diffraction pattern that can be analyzed to gain information about the crystal's structure.

The diffraction pattern results from constructive interference of the X-rays, which occurs at specific angles for a given wavelength according to Bragg's Law. This is the fundamental basis for XRD analysis. The technique can identify unknown substances, determine lattice parameters, and provide insights into the crystal orientation and size of unit cells. XRD is instrumental in fields such as mineralogy, metallurgy, pharmacology, and nanotechnology, where precise knowledge of a material's crystal structure is critical.
The Science of Crystallography
Crystallography is the scientific study of crystals and their structures. It involves analyzing the arrangement of atoms within a crystalline solid, which can be quite complex. Crystals are ordered structures with a periodic arrangement of atoms or molecules, and understanding this structure is vital for a variety of scientific and industrial applications.

The principles of crystallography are based on symmetry, lattice structures, and the idea that a crystal's external form reflects its internal structure. Advanced techniques such as X-ray diffraction are used to detect and describe the three-dimensional structure of the crystals. By doing so, scientists can determine the positions of atoms in a crystal, the chemical bonds, and the disorder within the structure.

Crystallography has led to significant advancements in fields ranging from chemistry to biology, as it has been crucial in determining the structures of complex molecules like proteins and DNA. This knowledge helps in the development of new pharmaceuticals, the design of materials with specific properties, and the advancement of nanotechnology.

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Most popular questions from this chapter

Three discrete spectral lines occur at angles of \(10.09^{\circ}\) \(13.71^{\circ},\) and \(14.77^{\circ}\) in the first-order spectrum of a grating spectrometer. (a) If the grating has 3660 slits/cm, what are the wavelengths of the light? (b) At what angles are these lines found in the second-order spectrum?

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A beam of laser light of wavelength \(632.8 \mathrm{nm}\) has a circular cross section \(2.00 \mathrm{mm}\) in diameter. A rectangular aperture is to be placed in the center of the beam so that, when the light falls perpendicularly on a wall \(4.50 \mathrm{m}\) away, the central maximum fills a rectangle \(110 \mathrm{mm}\) wide and \(6.00 \mathrm{mm}\) high. The dimensions are measured between the minima bracketing the central maximum. Find the required width and height of the aperture.

How much diffraction spreading does a light beam undergo? One quantitative answer is the full width at half maximum of the central maximum of the single- slit Fraunhofer diffraction pattern. You can evaluate this angle of spreading in this problem and in the next. (a) In Equation \(38.4,\) define \(\beta / 2=\phi\) and show that, at the point where \(I=0.5 I_{\max },\) we must have \(\sin \phi=\phi / \sqrt{2} .\) (b) Let \(y_{1}=\sin \phi\) and \(y_{2}=\phi / \sqrt{2} .\) Plot \(y_{1}\) and \(y_{2}\) on the same set of axes over a range from \(\phi=1\) rad to \(\phi=\pi / 2\) rad. Determine \(\phi\) from the point of intersection of the two curves. (c) Then show that, if the fraction \(\lambda / a\) is not large, the angular full width at half maximum of the central diffraction maximum is \(\Delta \theta=0.886 \lambda / a\)

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