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\mathrm{X}\( rays of wavelength \)0.10 \mathrm{~nm}\( are found to undergo secondorder reflection at a Bragg angle of \)23^{\circ}$ from a lithium fluoride crystal. What is the interplanar spacing of the reflecting planes in the crustal?

Short Answer

Expert verified
The interplanar spacing \( d \) is approximately 0.256 nm.

Step by step solution

01

Understand Bragg's Law

Bragg's Law is expressed as \( n\lambda = 2d\sin\theta \), where \( n \) is the order of reflection, \( \lambda \) is the wavelength of the incident wave, \( d \) is the interplanar spacing, and \( \theta \) is the Bragg angle. In this problem, \( n = 2 \), \( \lambda = 0.10 \text{ nm} \), and \( \theta = 23^{\circ} \). We need to find \( d \).
02

Convert Angles to Radians

Angles should often be converted to radians for calculations. However, trigonometric functions in calculators usually accept angles in degrees directly. In this case, \( \theta = 23^{\circ} \) can be used directly in the sine function since the sine of the angle is dimensionless.
03

Rearrange Bragg's Law Formulation

Rearrange Bragg's Law to solve for \( d \) by dividing both sides of the equation by \( 2\sin\theta \): \[ d = \frac{n\lambda}{2\sin\theta} \]
04

Substitute Values Into Rearranged Equation

Substitute the given values into the equation: \[ d = \frac{2 \times 0.10 \text{ nm}}{2\sin(23^{\circ})} \]. Simplifying the equation gives \[ d = \frac{0.20 \text{ nm}}{2\sin(23^{\circ})} \].
05

Calculate Sin of Bragg Angle

Use a calculator to find \( \sin(23^{\circ}) \), which gives approximately 0.3907.
06

Compute Interplanar Spacing

Substitute \( \sin(23^{\circ}) = 0.3907 \) into the formula: \[ d = \frac{0.20 \text{ nm}}{2 \times 0.3907} \]. This calculation simplifies to \[ d \approx \frac{0.20}{0.7814} \approx 0.256 \text{ nm} \].

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Key Concepts

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

Interplanar Spacing
Interplanar spacing refers to the distance between parallel planes of atoms in a crystal lattice. This is a key concept when discussing Bragg's Law, as it helps define how X-rays interact with a crystal. When X-rays strike a crystal, they reflect off these planes of atoms at specific angles. The distance between these planes, known as the interplanar spacing, is critical in determining the angles at which X-ray diffraction occurs.

Bragg's Law, given by the equation \( n\lambda = 2d\sin\theta \), allows us to calculate the interplanar spacing \( d \) when we know the order of reflection \( n \), the wavelength of the X-ray \( \lambda \), and the Bragg angle \( \theta \). By rearranging the equation, we solve for \( d \) as follows:
  • Substitute known values: \( n = 2 \), \( \lambda = 0.10 \text{ nm} \), and \( \theta = 23^\circ \).
  • Rearrange Bragg’s Law: \( d = \frac{n\lambda}{2\sin\theta} \).
  • Using the sine of the Bragg angle, \( \sin(23^\circ) \approx 0.3907 \), we find the interplanar spacing \( d \approx 0.256 \text{ nm} \).
By understanding interplanar spacing, students can gain insights into the structural properties of the crystal being examined.
X-ray Diffraction
X-ray diffraction is a powerful tool used to study the structure of crystals. It involves the scattering of X-rays by the electrons in a crystal lattice. This scattering occurs at specific angles and is used to derive information about the atomic structure of the material being studied.

When X-rays hit a crystal, they can reflect off the planes of atoms within the structure. If these reflections add up constructively, they intensify and are observable as distinct patterns called diffraction peaks. These patterns allow scientists to analyze the arrangement of atoms within the crystal. Such analyses are crucial in many scientific fields, including materials science and chemistry.

The angles at which these reflections occur are not arbitrary; they are dictated by the atomic structure of the crystal and can be calculated using Bragg’s Equation. This way, X-ray diffraction provides direct evidence of how atoms, ions, or molecules are arranged in the crystal structure, enabling researchers to determine sizes, shapes, and bond lengths that would otherwise be challenging to measure.
Wavelength
The wavelength of an X-ray is a fundamental concept in understanding how X-rays interact with matter. Wavelength refers to the distance between successive crests of a wave and is normally denoted by the Greek letter \( \lambda \). In the context of X-ray diffraction, it is the wavelength of the X-rays that determines how they are diffracted by a crystal.

X-rays used in diffraction studies typically have very short wavelengths, often in the range of 0.01 to 10 nanometers (nm). The specific wavelength is crucial because it needs to be on the same order of magnitude as the interplanar spacing in the crystal. This condition must be fulfilled for X-rays to effectively interact with the crystal lattice and produce a clear diffraction pattern.
  • In the given exercise, the X-ray wavelength is 0.10 nm.
  • This wavelength allows for the examination of internal crystal structures due to its short distance.
  • Short wavelengths, like those of X-rays, enable the analysis of small-scale structures not visible with longer-wavelength light.
Understanding the wavelength helps make crucial connections between the properties of X-rays and their practical applications in analyzing crystal structures.

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Most popular questions from this chapter

In conventional television, signals are broadcast from towers to home receivers. Even when a receiver is not in direct view of a tower because of a hill or building, it can still intercept a signal if the signal diffracts enough around the obstacle, into the obstacle's "shadow region." Previously, television signals had a wavelength of about \(50 \mathrm{~cm}\), but digital television signals that are transmitted from towers have a wavelength of about \(10 \mathrm{~mm}\). (a) Did this change in wavelength increase or decrease the diffraction of the signals into the shadow regions of obstacles? Assume that a signal passes through an opening of \(6.0 \mathrm{~m}\) width between two adjacent buildings. What is the angular spread of the central diffraction maximum (out to the first minima) for wavelengths of (b) \(50 \mathrm{~cm}\) and (c) \(10 \mathrm{~mm}\) ?

A grating has 600 rulings/mm and is \(6.0 \mathrm{~mm}\) wide. (a) What is the smallest wavelength interval it can resolve in the third order at \(\lambda=500 \mathrm{~nm}\) ? (b) How many higher orders of maxima can be seen?

In the single-slit diffraction experiment of Fig. \(36-4\), let the wavelength of the light be \(500 \mathrm{~nm}\), the slit width be \(6.00 \mu \mathrm{m}\), and the viewing screen be at distance \(D=4.00 \mathrm{~m}\). Let a \(y\) axis extend upward along the viewing screen, with its origin at the center of the diffraction pattern. Also let \(I_{p}\) represent the intensity of the diffracted light at point \(P\) at \(y=15.0 \mathrm{~cm}\). (a) What is the ratio of \(I_{P}\) to the intensity \(I_{m}\) at the center of the pattern? (b) Determine where point \(P\) is in the diffraction pattern by giving the maximum and minimum between which it lies, or the two minima between which it lies.

Visible light is incident perpendicularly on a grating with 325 rulings/mm. What is the longest wavelength that can be seen in the fifth-order diffraction?

Light at wavelength \(589 \mathrm{~nm}\) from a sodium lamp is incident perpendicularly on a grating with 40000 rulings over width \(96 \mathrm{~mm}\). What are the first-order (a) dispersion \(D\) and (b) resolving power \(R\), the second-order (c) \(D\) and (d) \(R\), and the third-order (e) \(D\) and (f) \(R\) ?

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