Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 3: Problem 61

(a) Derive planar density expressions for \(\mathrm{BCC}\) (100) and (110) planes in terms of the atomic radius \(R\). (b) Compute and compare planar density values for these same two planes for molybdenum (Mo).

Short Answer

Expert verified
Question: Calculate and compare the planar density values for the (100) and (110) planes in a BCC unit cell for Molybdenum (Mo). Answer: The planar density of the (100) plane for Mo = 1.23 x 10^(-14) atoms / pm^2, and the planar density of the (110) plane for Mo = 1.47 x 10^(-14) atoms / pm^2. The (110) plane has a higher planar density than the (100) plane.

Step by step solution

01

Definition of planar density

Planar density is defined as the number of atoms in a unit cell per unit area of a specific crystallographic plane. In this case, we are dealing with BCC unit cells, and we have to find the planar density for the (100) and (110) planes.
02

Determine atoms on the (100) plane

In a BCC unit cell, we have atoms at each corner and one atom at the center of the cube. The (100) plane cuts through four corner atoms and one atom at the center. We should note that the corner atoms are shared with other unit cells. Therefore, each corner atom contributes 1/4 to the planar density. The center atom contributes fully to the plane. So, we have: Number of atoms on the (100) plane = (4 x 1/4) + 1 = 2 atoms
03

Calculate the area of the (100) plane

The side length of the BCC unit cell (a) can be given by the equation: a = 4R / sqrt(3). The (100) plane is a square with the side length equal to a. So, the area of the (100) plane is given by: Area of the (100) plane = a^2 = (4R / sqrt(3))^2 = (16R^2) / 3
04

Calculate the planar density of the (100) plane

Now, we can find the planar density for the (100) plane using the number of atoms and area of the plane: Planar density of the (100) plane = Number of atoms / Area of the plane = 2 / [(16R^2) / 3] = 3 / (8R^2)
05

Determine atoms on the (110) plane

For the (110) plane, it passes through four corner atoms and two center atoms on the corresponding diagonal face. So, we have: Number of atoms on the (110) plane = (4 x 1/4) + (2 x 1/2) = 2 atoms
06

Calculate the area of the (110) plane

The (110) plane forms a rectangle with a diagonal of the square face (a) and the edge length (a) of the unit cell. Hence, the area of the (110) plane is given by: Area of the (110) plane = a x a/sqrt(2) = (4R / sqrt(3)) x (4R / sqrt(3))/sqrt(2) = (16R^2) / (3 sqrt(2))
07

Calculate the planar density of the (110) plane

Now, we can find the planar density for the (110) plane using the number of atoms and area of the plane: Planar density of the (110) plane = Number of atoms / Area of the plane = 2 / [(16R^2) / (3 sqrt(2))] = sqrt(2) / (4R^2) (b)
08

Atomic radius of Mo

The atomic radius of Molybdenum (Mo) is approximately R = 139 pm.
09

Calculate the planar densities for Mo

Using the derived expressions for the planar density and substituting R = 139 pm, we can find the planar density values: Planar density of the (100) plane for Mo = 3 / (8 * (139 pm)^2) ≈ 1.23 x 10^(-14) atoms / pm^2 Planar density of the (110) plane for Mo = sqrt(2) / (4 * (139 pm)^2) ≈ 1.47 x 10^(-14) atoms / pm^2
10

Compare the planar density values

Comparing the planar density values for the (100) and (110) planes in Mo, we can observe that the (110) plane has a higher planar density (1.47 x 10^(-14) atoms / pm^2) than the (100) plane (1.23 x 10^(-14) atoms / pm^2).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Iron (Fe) undergoes an allotropic transformation at \(912^{\circ} \mathrm{C}:\) upon heating from a \(\mathrm{BCC}\) ( \(\alpha\) phase) to an FCC \((\gamma\) phase). Accompanying this transformation is a change in the atomic radius of \(\mathrm{Fe}-\) from \(R_{\mathrm{BCC}}=\) \(0.12584 \mathrm{~nm}\) to \(R_{\mathrm{FCC}}=0.12894 \mathrm{~nm}-\) and, in addition, a change in density (and volume). Compute the percentage volume change associated with this reaction. Does the volume increase or decrease?

For tetragonal crystals, cite the indices of directions that are equivalent to each of the following directions: (a) \([011]\) (b) \([100]\)

Sketch a tetragonal unit cell, and within that cell indicate locations of the \(1 \frac{1}{2} \frac{1}{2}\) and \(\frac{1}{2} \frac{1}{4} \frac{1}{2}\) point coordinates.

(a) What are the direction indices for a vector that passes from point \(\frac{1}{4} 0 \frac{1}{2}\) to point \(\frac{3}{4} \frac{1}{2}\) in a cubic unit cell? (b) Repeat part (a) for a monoclinic unit cell.

Magnesium (Mg) has an HCP crystal structure, a \(c / a\) ratio of \(1.624\), and a density of \(1.74 \mathrm{~g} / \mathrm{cm}^{3}\). Compute the atomic radius for \(\mathrm{Mg}\).
See all solutions

Recommended explanations on Physics Textbooks

Engineering Physics

Read Explanation

Measurements

Read Explanation

Famous Physicists

Read Explanation

Modern Physics

Read Explanation

Physical Quantities And Units

Read Explanation

Magnetism

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free