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Chapter 29: Problem 21

Calculate the separation between the (a) 100 planes, (b) 111 planes, and (c) \(12 \overline{1}\) planes in a cubic lattice whose unit cell length is \(529.8 \mathrm{pm}\)

Short Answer

Expert verified
(a) 529.8 pm, (b) 305.8 pm, (c) 216.2 pm

Step by step solution

01

Understanding the Problem

The task is to calculate the separation between specific crystal planes 鈥� labeled as 100, 111, and 12 \(\overline{1}\) 鈥� in a cubic lattice. The unit cell length (\(a\)) is given as 529.8 pm, or picometers.
02

Formula for Plane Separation

In a cubic lattice, the separation \(d\) between planes labeled as \((hkl)\) is given by the formula: \[ d = \frac{a}{\sqrt{h^2 + k^2 + l^2}} \] where \(a\) is the lattice parameter, and \(h\), \(k\), and \(l\) are the Miller indices of the planes.
03

Calculate Separation for 100 Planes

For the 100 planes, the Miller indices \(h = 1\), \(k = 0\), and \(l = 0\). Substitute these values and \(a = 529.8\,[\text{pm}]\) into the formula: \[ d_{100} = \frac{529.8}{\sqrt{1^2 + 0^2 + 0^2}} = \frac{529.8}{1} = 529.8\,[\text{pm}] \]
04

Calculate Separation for 111 Planes

For the 111 planes, the Miller indices \(h = 1\), \(k = 1\), and \(l = 1\). Substitute these values into the formula: \[ d_{111} = \frac{529.8}{\sqrt{1^2 + 1^2 + 1^2}} = \frac{529.8}{\sqrt{3}} \approx 305.8\,[\text{pm}] \]
05

Calculate Separation for 12 *1 Planes

For the \(12 \overline{1}\) planes, the Miller indices \(h = 1\), \(k = 2\), and \(l = -1\). Substitute these into the formula: \[ d_{12\overline{1}} = \frac{529.8}{\sqrt{1^2 + 2^2 + (-1)^2}} = \frac{529.8}{\sqrt{6}} \approx 216.2\,[\text{pm}] \]

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Key Concepts

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

Miller indices
Miller indices are a set of three numbers used to represent the orientation of a plane or a set of parallel planes in a crystal lattice. They are denoted by h , k, and l, and are determined by the intercepts of the plane with the crystallographic axes:
  • The plane intercepts the x-axis at 1, the y-axis at k, and the z-axis at l.
  • The indices are found by taking the reciprocal of these intercepts and clearing any fractions.
  • This method ensures that planes parallel to each other have identical Miller indices.
Miller indices are central in crystallography because they provide a crystal's researchers with a standardized way of describing planes. This practice helps in various calculations, such as determining the angles between planes and understanding crystal symmetry.
Understanding Miller indices is crucial for calculating distances between planes in a crystal lattice, a fundamental aspect of determining a crystal's structure.
cubic lattice
A cubic lattice is a type of crystal structure where the unit cell is a perfect cube. It is one of the simplest and most symmetrical lattice types found in nature. In a cubic lattice, all three axes are of equal length, and the angles between them are 90 degrees.
There are three types of cubic lattices. They differ in the arrangement of atoms within the unit cell:
  • Simple cubic (SC): Atoms are located at each corner of the cube.
  • Body-centered cubic (BCC): An atom is present at each cube corner and one in the center of the cube.
  • Face-centered cubic (FCC): Atoms are located at each corner and the center of each face of the cube.
Understanding the structure of a cubic lattice is essential in crystallography and materials science. It helps to determine various properties of materials, such as density, atomic packing factor, and coordination number. These properties influence the material's mechanical characteristics, making cubic lattices a fundamental study area in solid-state physics.
unit cell length
The unit cell length, often represented by the symbol a, is the length of the edges of the unit cell in a crystal lattice. It is a key parameter for understanding a material's structure because it dictates the dimensions of the repeating pattern in the lattice.
For cubic lattices, the unit cell length a is the same for all edges, reflecting the cube-shaped symmetry and uniformity.
  • Unit cell length is measured typically in picometers (pm) or Angstroms ( 脜), with 1 脜 = 100 pm.
  • The unit cell dimension is critical for computing various structural properties, including plane separation.
Knowing the unit cell length allows researchers to calculate distances between planes using the formula for plane separation. This calculation is critical for determining the material's properties and understanding interactions within the crystal structure. The unit cell length directly impacts how atoms are packed in the lattice, affecting material density and applied science.
crystallography computation
Crystallography computation involves using mathematical formulas and techniques to understand and describe the geometric structure of crystals. It aids in determining distances, angles, and interactions between different crystal components. In context, the computation focuses on the separation between planes in a crystal lattice.
  • The formula for the separation (d) between planes labeled by Miller indices (hkl) is: \[ d = \frac{a}{\sqrt{h^2 + k^2 + l^2}} \]
  • This formula requires the Miller indices and the unit cell length a.
  • For real-life application, precise measurement of distances guides the analysis of crystal properties, material strength, and electronic behavior.
Engaging in crystallography computation strengthens one's grasp of solid materials and their intrinsic properties. By applying formulas to practical problems, one sees how the microscopic arrangement of atoms translates into macroscopic material characteristics. This computation is indispensable in fields like materials science, chemistry, and condensed matter physics.

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Most popular questions from this chapter

Multilayer physisorption is often described by the \(B E T\) adsorption isotherm $$ \frac{P}{V\left(P^{*}-P\right)}=\frac{1}{c V_{\mathrm{m}}}+\frac{(c-1) P}{V_{\mathrm{m}} c P^{*}} $$ where \(P^{*}\) is the vapor pressure of the adsorbate at the temperature of the experiment, \(V_{\mathrm{m}}\) is the volume corresponding to a monolayer of coverage on the surface, \(V\) is the total volume adsorbed at pressure \(P\), and \(c\) is a constant. Rewrite the equation for the BET adsorption isotherm in the form $$ \frac{V}{V_{\mathrm{m}}}=f\left(P / P^{*}\right) $$ Plot \(V / V_{m}\) versus \(P / P^{*}\) for \(c=0.1,1.0,10\), and \(100 .\) Discuss the shapes of the curves.

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?

Consider a surface-catalyzed bimolecular reaction between molecules \(A\) and \(B\) that has a rate law of the form \\[ v=k_{3} \theta_{\mathrm{A}} \theta_{\mathrm{B}} \\] where \(\theta_{\mathrm{A}}\) is the fraction of surface sites occupied by reactant \(\mathrm{A}\) and \(\theta_{\mathrm{B}}\) is the fraction of surface sites occupied by reactant \(\mathrm{B}\). A mechanism consistent with this reaction is as follows: Take \(K_{\mathrm{A}}\) and \(K_{\mathrm{B}}\) to be the equilibrium constants for Equations 1 and \(2,\) respectively. Derive expressions for \(\theta_{\mathrm{A}}\) and \(\theta_{\mathrm{B}}\) in terms of \([\mathrm{A}],[\mathrm{B}], K_{\mathrm{A}},\) and \(K_{\mathrm{B}} .\) Use your results to show that the rate law can be written as \\[ v=\frac{k_{3} K_{\mathrm{A}} K_{\mathrm{B}}[\mathrm{A}][\mathrm{B}]}{\left(1+K_{\mathrm{A}}[\mathrm{A}]+K_{\mathrm{B}}[\mathrm{B}]\right)^{2}} \\]

1\. In this problem, we will derive the structure factor for a sodium chloride-type unit cell. First, show that the coordinates of the cations at the eight corners are (0,0,0),(1,0,0),(0,1,0) \((0,0,1),(1,1,0),(1,0,1),(0,1,1,),\) and (1,1,1) and those at the six faces are \(\left(\frac{1}{2}, \frac{1}{2}, 0\right),\left(\frac{1}{2}, 0, \frac{1}{2}\right)\) \(\left(0, \frac{1}{2}, \frac{1}{2}\right),\left(\frac{1}{2}, \frac{1}{2}, 1\right),\left(\frac{1}{2}, 1, \frac{1}{2}\right),\) and \(\left(1, \frac{1}{2}, \frac{1}{2}\right) .\) Similarly, show that the coordinates of the anions along the 12 edges are \(\left(\frac{1}{2}, 0,0\right),\left(0, \frac{1}{2}, 0\right),\left(0,0, \frac{1}{2}\right),\left(\frac{1}{2}, 1,0\right),\left(1, \frac{1}{2}, 0\right),\left(0, \frac{1}{2}, 1\right),\left(\frac{1}{2}, 0,1\right),\left(1,0, \frac{1}{2}\right)\) \(\left(0,1, \frac{1}{2}\right),\left(\frac{1}{2}, 1,1\right),\left(1, \frac{1}{2}, 1\right),\) and \(\left(1,1, \frac{1}{2}\right)\) and those of the anion at the center of the unit cell are \(\left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right) .\) Now show that \\[ \begin{aligned} F(h k l)=& \frac{f_{+}}{8}\left[1+e^{2 \pi i h}+e^{2 \pi i k}+e^{2 \pi i l}+e^{2 \pi i(h+k)}+e^{2 \pi i(h+l)}+e^{2 \pi i(k+l)}+e^{2 \pi i(h+k+l)}\right] \\ &+\frac{f_{+}}{2}\left[e^{\pi i(h+k)}+e^{\pi i(h+l)}+e^{\pi i(k+l)}+e^{\pi i(h+k+2 l)}+e^{\pi i(h+2 k+l)}+e^{\pi i(2 h+k+l)}\right] \\ &+\frac{f}{4}\left[e^{\pi i h}+e^{\pi i k}+e^{\pi i l}+e^{\pi i(h+2 k)}+e^{\pi i(2 h+k)}+e^{\pi i(k+2 l)}+e^{\pi i(h+2 l)}+e^{\pi i(2 h+l)}+e^{\pi i(2 k+l)}\right. \end{aligned} \\] \\[ \begin{array}{l} \left.\quad+e^{\pi i(h+2 k+2 l)}+e^{\pi i(2 h+k+2 l)}+e^{\pi i(2 h+2 k+l)}\right]+f_{-} e^{\pi i(h+k+l)} \\ =f_{+}\left[1+(-1)^{h+k}+(-1)^{h+l}+(-1)^{k+l}\right] \\ \quad+f_{-}\left[(-1)^{h}+(-1)^{k}+(-1)^{l}+(-1)^{h+k+l}\right] \end{array} \\] Finally, show that \\[ F(h k l)=4\left(f_{+}+f_{-}\right) \\] if \(h, k,\) and \(l\) are all even; that \\[ F(h k l)=4\left(f_{+}-f_{-}\right) \\] if \(h, k,\) and \(l\) are all odd, and that \(F(h k l)=0\) otherwise.

In this problem, we will prove that a crystal lattice can have only one-, two-, threefour-, and six-fold axes of symmetry. Consider the following figure, where \(P_{1}, P_{2},\) and \(P_{3}\) are three lattice points, each separated by the lattice vector a. If the lattice has \(n\) -fold symmetry, then both a clockwise and a counter- clockwise rotation by \(\phi=360^{\circ} / n\) about the point \(P_{2}\) will lead to the points \(P_{1}^{\prime}\) and \(P_{2}^{\prime},\) which must be lattice points (because of the fact that the lattice has an \(n\) -fold axis of symmetry). Show that the vector distance \(P_{1}^{\prime} P_{2}^{\prime}\) must satisfy the relation \\[ 2 \mathbf{a} \cos \phi=N \mathbf{a} \\] where \(N\) is a positive or negative integer. Now show that the only values of \(\phi\) that satsify the above relation are \(360^{\circ}(n=1), 180^{\circ}(n=2), 120^{\circ}(n=3), 90^{\circ}(n=4),\) and \(60^{\circ}(n=6)\) corresponding to \(N=2,-2,-1,0,\) and \(1,\) respectively.
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