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The crystalline structure of \(\mathrm{CuSO}_{4}(\mathrm{s})\) is orthorhombic with unit cell dimensions of \(a=488.2 \mathrm{pm}, b=665.7 \mathrm{pm},\) and \(c=831.6 \mathrm{pm} .\) Calculate the value of \(\theta,\) the first-order Bragg diffraction angle, from the 100 planes, the 110 planes, and the 111 planes if \(\mathrm{CuSO}_{4}(\mathrm{s})\) is irradiated with X-rays with \(\lambda=154.433 \mathrm{pm}\).

Short Answer

Expert verified
θ is approximately 9.07° for (100), 15.6° for (110), and 19.3° for (111) planes.

Step by step solution

01

Understanding Bragg's Law

Bragg's Law is used to determine the diffraction angle \( \theta \) for X-rays reflecting from planes in a crystal lattice. It is given by the formula: \[ n\lambda = 2d\sin\theta \] where \( n \) is the order of reflection (here, \( n = 1 \)), \( \lambda \) is the wavelength of the incident X-rays, \( d \) is the distance between crystal planes, and \( \theta \) is the angle of incidence.
02

Calculate d for (100) Plane

The interplanar spacing \( d \) for a set of crystallographic planes in an orthorhombic lattice is given by:\[ \frac{1}{d^2} = \frac{h^2}{a^2} + \frac{k^2}{b^2} + \frac{l^2}{c^2} \]For the (100) plane, \( h = 1 \), \( k = 0 \), \( l = 0 \). Thus:\[ \frac{1}{d^2} = \frac{1}{a^2} \]\[ d_{100} = a = 488.2 \text{ pm} \].
03

Calculate θ for (100) Plane

Using Bragg's Law for the (100) plane:\[ \lambda = 2d\sin\theta \]\[ 154.433 = 2(488.2)\sin\theta \]\[ \sin\theta = \frac{154.433}{2 \times 488.2} \]\[ \theta \approx \arcsin\left(\frac{154.433}{976.4}\right) \approx 9.07^\circ \].
04

Calculate d for (110) Plane

For the (110) plane, \( h = 1 \), \( k = 1 \), \( l = 0 \):\[ \frac{1}{d^2} = \frac{1^2}{a^2} + \frac{1^2}{b^2} \]\[ \frac{1}{d^2} = \frac{1}{488.2^2} + \frac{1}{665.7^2} \]Calculate \( d_{110} \) using this equation:\[ d_{110} \approx 286.7 \text{ pm} \].
05

Calculate θ for (110) Plane

Using Bragg's Law for the (110) plane:\[ 154.433 = 2\times 286.7 \sin\theta \]\[ \sin\theta = \frac{154.433}{573.4} \]\[ \theta \approx \arcsin\left(\frac{154.433}{573.4}\right) \approx 15.6^\circ \].
06

Calculate d for (111) Plane

For the (111) plane, \( h = 1 \), \( k = 1 \), \( l = 1 \):\[ \frac{1}{d^2} = \frac{1^2}{a^2} + \frac{1^2}{b^2} + \frac{1^2}{c^2} \]\[ \frac{1}{d^2} = \frac{1}{488.2^2} + \frac{1}{665.7^2} + \frac{1}{831.6^2} \]Calculate \( d_{111} \) using this equation:\[ d_{111} \approx 235.5 \text{ pm} \].
07

Calculate θ for (111) Plane

Using Bragg's Law for the (111) plane:\[ 154.433 = 2\times 235.5 \sin\theta \]\[ \sin\theta = \frac{154.433}{471} \]\[ \theta \approx \arcsin\left(\frac{154.433}{471}\right) \approx 19.3^\circ \].

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Key Concepts

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

Orthorhombic Crystal Structure
The orthorhombic crystal structure is one of the seven crystal systems in crystallography. It is characterized by three mutually perpendicular axes that are all of different lengths. This system forms a three-dimensional grid, or lattice, that defines the positions of atoms within a crystal.

In an orthorhombic lattice, the unit cell dimensions are denoted by the parameters \( a \), \( b \), and \( c \). Each of these parameters represents the length of the unit cell along the respective axis. For example:
  • \( a = 488.2 \, \text{pm} \)
  • \( b = 665.7 \, \text{pm} \)
  • \( c = 831.6 \, \text{pm} \)
In our example, the orthorhombic structure describes how the copper sulfate atoms are arranged. Such a structure can affect the material's properties, including its optical, electrical, and mechanical characteristics. Determining the precise arrangement of atoms with this crystal structure can help predict how the material will interact with other substances and react under various conditions.
X-ray Diffraction
X-ray diffraction (XRD) is a technique used to study the atomic structure of crystals through the scattering of X-ray beams. When X-rays are directed at a crystal, they are scattered by the electrons surrounding the atoms, producing a diffraction pattern specific to the arrangement of atoms in the crystal.

This scattering occurs because the crystal's characteristic lattice forms a series of parallel planes, each capable of reflecting the X-rays. This is where Bragg's Law comes into play, which is given by:
  • \( n\lambda = 2d\sin\theta \)
  • \( n \) is the order of reflection
  • \( \lambda \) is the wavelength of the incident X-rays
  • \( d \) is the interplanar spacing of the crystal
  • \( \theta \) is the angle of diffraction
X-ray diffraction provides critical information about the crystal structure, such as:
  • Identifying unknown materials by comparing diffraction patterns
  • Evaluating purity and crystallinity
  • Measuring lattice parameters
By analyzing XRD patterns, scientists can make inferences about the arrangement of atoms within the crystal, which is crucial for applications in material science, chemistry, and physics.
Interplanar Spacing
Interplanar spacing, denoted as \( d \), is a crucial parameter in crystallography and X-ray diffraction analysis. It refers to the distance between two adjacent planes of atoms in a crystal lattice. This concept is critical because it determines how X-rays interact with the crystal, which is essential for applying Bragg's Law.

To calculate interplanar spacing \( d \) in an orthorhombic crystal, the formula is:
  • \( \frac{1}{d^2} = \frac{h^2}{a^2} + \frac{k^2}{b^2} + \frac{l^2}{c^2} \)
  • \( h \), \( k \), and \( l \) are Miller indices representing the orientation of the plane
  • \( a \), \( b \), and \( c \) are the unit cell dimensions

The Miller indices \((hkl)\) define the orientation of crystallographic planes. For instance:
  • The \( (100) \) plane has \( h = 1, k = 0, l = 0 \)
  • The \( (110) \) plane has \( h = 1, k = 1, l = 0 \)
  • The \( (111) \) plane has \( h = 1, k = 1, l = 1 \)
Understanding these spacings helps predict how a crystal will diffract X-rays, revealing information about its atomic structure. This knowledge is vital for designing materials with specific properties, whether for industrial applications or scientific research.

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Most popular questions from this chapter

The von Laue equations are often expressed in vector notation. The following figure illustrates the X-ray scattering from two lattice points \(P_{1}\) and \(P_{2}\). Let \(s_{0}\) be a unit vector in the direction of the incident radiation and \(s\) be a unit vector in the direction of the scattered X-radiation. Show that the difference in the path lengths of the waves scattered from \(P_{1}\) and \(P_{2}\) is given by \\[ \delta=P_{1} A-P_{2} B=\mathbf{r} \cdot \mathbf{s}-\mathbf{r} \cdot \mathbf{s}_{0}=\mathbf{r} \cdot \mathbf{S} \\] where \(\mathbf{S}=\mathbf{s}-\mathbf{s}_{0} .\) Because \(P_{1}\) and \(P_{2}\) are lattice points, \(\mathbf{r}\) must be expressible as \(m \mathbf{a}+\) \(n \mathbf{b}+p \mathbf{c},\) where \(m, n,\) and \(p\) are integers, and \(\mathbf{a}, \mathbf{b},\) and \(\mathbf{c}\) are the unit cell axes. Show that the fact that \(\delta\) must be an integral multiple of the wavelength \(\lambda\) leads to the equations \\[ \begin{array}{l} \mathbf{a} \cdot \mathbf{S}=h \lambda \\ \mathbf{b} \cdot \mathbf{S}=k \lambda \\ \mathbf{c} \cdot \mathbf{S}=l \lambda \end{array} \\] where \(h, k,\) and \(l\) are integers. These equations are the von Laue equations in vector notation.

Potassium crystallizes as a body-centered cubic lattice, and the length of a unit cell is \(533.3 \mathrm{pm} .\) Given that the density of potassium is \(0.8560 \mathrm{g} \cdot \mathrm{cm}^{-3},\) calculate the Avogadro constant.

In this problem, we will derive the Bragg equation, Equation 29.12. William and Lawrence Bragg (father and son) assumed that X-rays are scattered by successive planes of atoms within a crystal (see the following figure). Each set of planes reflects the X-rays specularly; that is, the angle of incidence is equal to the angle of reflection, as shown in the figure. The X-radiation reflected from the lower plane in the figure travels a distance \(P Q R\) longer than the X-radiation reflected by the upper layer. Show that \(P Q R=2 d \sin \theta,\) and argue that \(2 d \sin \theta\) must be an integral number of wavelengths for constructive interference and hence a diffraction pattern to be observed.

In this problem we derive Equation \(29.45,\) the rate law for the oxidation reaction \(2 \mathrm{CO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{CO}_{2}(\mathrm{g})\) assuming that the reaction occurs by the Langmuir-Hinshel-wood mechanism. The overall rate law for this mechanism is \\[ v=k_{3} \theta_{\mathrm{CO}} \theta_{\mathrm{O}_{2}} \\] \(\theta_{\mathrm{O}_{2}}=\frac{\left(K_{\mathrm{O}_{2}}\left[\mathrm{O}_{2}\right]\right)^{1 / 2}}{1+\left(K_{\mathrm{O}_{2}}\left[\mathrm{O}_{2}\right]\right)^{1 / 2}+K_{\mathrm{CO}}[\mathrm{CO}]}\) \(\theta_{\mathrm{CO}}=\frac{K_{\mathrm{CO}}[\mathrm{CO}]}{1+\left(K_{\mathrm{O}_{2}}\left[\mathrm{O}_{2}\right]\right)^{1 / 2}+K_{\mathrm{CO}}[\mathrm{CO}]}\)

The hydrogenation of ethene on copper obeys the rate law \\[ v=\frac{k\left[\mathrm{H}_{2}\right]^{1 / 2}\left[\mathrm{C}_{2} \mathrm{H}_{4}\right]}{\left(1+K\left[\mathrm{C}_{2} \mathrm{H}_{4}\right]\right)^{2}} \\] where \(k\) and \(K\) are constants. Mechanistic studies show that the reaction occurs by the Langmuir-Hinshelwood mechanism. How are \(k\) and \(K\) related to the rate constants for the individual steps of the reaction mechanism? What can you conclude about the relative adsorption of \(\mathrm{H}_{2}(\mathrm{g})\) and \(\mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{g})\) to the copper surface from the form of the observed rate law?

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