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X-rays of wavelength 0.103 nm reflects off a crystal and a second-order maximum is recorded at a Bragg angle of \(25.5^{\circ}\). What is the spacing between the scattering planes in this crystal?

Short Answer

Expert verified
The spacing between the scattering planes in the crystal is approximately 0.182 nm.

Step by step solution

01

List down the given parameters

We are given: - Wavelength (\(\lambda\)) = 0.103 nm - Bragg angle (\(\theta\)) = 25.5° - Order of maximum reflection (n) = 2
02

Convert the Bragg angle to radians

To work with trigonometric functions in calculations, we need to convert the angle from degrees to radians. The conversion between degrees and radians is \(1 radian = \frac{180}{\pi}°\). Bragg angle in radians: \(\theta = 25.5° \times \frac{\pi}{180} \approx 0.445 rad\)
03

Apply Bragg's Law to find the spacing between the scattering planes

Now we have all the required variables to calculate the spacing, we will use Bragg's Law: \(n\lambda = 2d\sin{\theta}\) and rearrange the equation to solve for \(d\): \(d = \frac{n\lambda}{2\sin{\theta}}\)
04

Calculate the spacing between the scattering planes

Substitute the given values into the equation: \(d = \frac{(2)(0.103 nm)}{2\sin{(0.445)}}\) Now, calculate the value of \(d\): \(d \approx 0.182 nm\)
05

Report the result

The spacing between the scattering planes in the crystal is approximately 0.182 nm.

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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 technique used to study the atomic structure of crystals. When X-rays encounter a crystal, they are diffracted, or bent, in specific directions. This occurs because the X-ray wavelengths are on the same order of magnitude as the spacing between atoms in a crystal lattice. As these waves interfere with each other, they create a pattern that can reveal details about the crystal's internal structure.

In this process, the X-rays penetrate the crystal and scatter off the planes of atoms. The angles and intensities of these scattered waves provide scientists with information regarding the dimensions and shape of the crystal's unit cell. By analyzing these diffraction patterns, we can deduce crucial information about the arrangement of atoms, which is fundamental in fields like chemistry and material science.

Bragg's Law plays a key role here, as it describes the condition for the constructive interference of X-ray waves scattered by a crystal. It tells us that the path difference between rays reflected off consecutive planes must be an integer multiple of the X-ray's wavelength for constructive interference to occur.
crystal lattice spacing
Crystal lattice spacing refers to the distance between closely packed layers of atoms within a crystal structure. This spacing is a critical parameter because it impacts the material's properties, such as its strength, density, and optical characteristics.

Spacing can be accurately calculated using Bragg's Law, given by the equation \( n\lambda = 2d\sin\theta \), where \( n \) is the order of reflection, \( \lambda \) is the wavelength of the X-rays, \( d \) is the lattice spacing, and \( \theta \) is the Bragg angle. Rearranging this formula allows us to find \( d \) when other variables are known, as in the solution to the original exercise.

Understanding crystal lattice spacing is vital for many applications. For example, semiconductors rely heavily on precise lattice structures to function correctly. In materials science, accurate knowledge of lattice spacing can lead to innovations in creating harder alloys or more efficient photovoltaic cells.
trigonometry in physics
Trigonometry plays a significant role in physics, especially when it comes to analyzing oscillatory or wave phenomena, like X-ray diffraction. In the context of Bragg's Law, trigonometry helps relate angles with distances.

Specifically, the sine function, \( \sin\theta \), is crucial here. This function allows us to translate the angular measurements of scattered X-ray beams into linear distances between the planes in a crystal. Since physics often deals with periodic phenomena, trigonometry provides the tools to handle these periodic or wave-like processes effectively.

Also, converting angles from degrees to radians, as shown in the solution steps, is a common practice when working with trigonometric functions. Radians provide a natural way of describing angles based on their arc length, which is particularly useful in theoretical and applied physics.
  • Radians simplify many physics formulas.
  • Sine and cosine functions describe wave properties.
  • Trigonometry enhances our understanding of spatial relationships in physics.
Understanding how trigonometry applies can provide deeper insights into the principles governing wave interactions in materials and the design of experiments.

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Most popular questions from this chapter

Two slits of width \(2 \mu \mathrm{m},\) each in an opaque material, are separated by a center-to-center distance of \(6 \mu \mathrm{m}\). A monochromatic light of wavelength \(450 \mathrm{nm}\) is incident on the double- slit. One finds a combined interference and diffraction pattern on the screen. (a) How many peaks of the interference will be observed in the central maximum of the diffraction pattem? (b) How many peaks of the interference will be observed if the slit width is doubled while keeping the distance between the slits same? (c) How many peaks of interference will be observed if the slits are separated by twice the distance, that is, \(12 \mu \mathrm{m}\), while keeping the widths of the slits same? (d) What will happen in (a) if instead of 450-nm light another light of wavelength \(680 \mathrm{nm}\) is used? (e) What is the value of the ratio of the intensity of the central peak to the intensity of the next bright peak in (a)? (f) Does this ratio depend on the wavelength of the light? (g) Does this ratio depend on the width or separation of the slits?

Find the wavelength of light that has its third minimum at an angle of \(48.6^{\circ}\) when it falls on a single slit of width \(3.00 \mu \mathrm{m}\)

When dots are placed on a page from a laser printer, they must be close enough so that you do not see the individual dots of ink. To do this, the separation of the dots must be less than Raleigh's criterion. Take the pupil of the eye to be \(3.0 \mathrm{mm}\) and the distance from the paper to the eye of \(35 \mathrm{cm}\); find the minimum separation of two dots such that they cannot be resolved. How many dots per inch (dpi) does this correspond to?

White light falls on two narrow slits separated by \(0.40 \mathrm{mm} .\) The interference pattern is observed on a screen 3.0 \(\mathrm{m}\) away. (a) What is the separation between the first maxima for red light \((\lambda=700 \mathrm{nm})\) and violet light \((\lambda=400 \mathrm{nm}) ? \quad\) (b) \(\mathrm{At}\) what point nearest the central maximum will a maximum for yellow light \((\lambda=600 \mathrm{nm})\) coincide with a maximum for violet light? Identify the order for each maximum.

A microwave of an unknown wavelength is incident on a single slit of width \(6 \mathrm{cm} .\) The angular width of the central peak is found to be \(25^{\circ} .\) Find the wavelength.

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