/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 4 For the HCP crystal structure, s... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 4

For the HCP crystal structure, show that the ideal \(c / a\) ratio is \(1.633\).

Short Answer

Expert verified
Answer: The ideal c/a ratio for a Hexagonal Close Packed (HCP) crystal structure is 1.633. It is determined using the geometry of the unit cell and trigonometry, primarily focusing on the relationship between the edge lengths of the unit cell and the height of the tetrahedron that forms the HCP structure.

Step by step solution

01

Understand the Hexagonal Close Packed Structure

The HCP crystal structure is formed by stacking layers of atoms in a close-packed arrangement. The atoms in one layer are situated on top of the triangular voids in the next layer, resulting in a repeating ABABAB... pattern. The unit cell of the HCP structure has edges a, b, and c, with angles alpha=90掳, beta=90掳, and gamma=120掳. The c-axis is perpendicular to the hexagonal layers, thus the c/a ratio will represent the relative height of the HCP unit cell.
02

Tetrahedron in HCP Structure

Considering the HCP unit cell, we can focus on one tetrahedron formed by the layers A and B, where one vertex is at the center of the unit cell and the other three vertices are at three adjacent atoms in one of the layers. We will denote the side length of this tetrahedron as 'd' and the height of the tetrahedron from the vertex at the center of the unit cell as 'h'.
03

Calculate 'd' using edge length 'a'

The side length 'd' is equal to the distance between two adjacent atoms in the same hexagonal layer. This is equal to the edge length 'a' of the unit cell. Therefore, we have \(d = a\).
04

Calculate the height 'h' of the tetrahedron

From the tetrahedron, we can form a right triangle with base 'd/2' and height 'h', where the hypotenuse is the distance between the vertex at the center of the unit cell and the midpoint of the opposite edge. As we found in Step 3, 'd' is equal to 'a', so we can calculate 'h' using the Pythagorean theorem and the fact that we have \(120^{\circ}\) angles in the hexagonal layer: \(h^2 + (\dfrac{d}{2})^2 = d^2\) \(h^2 + (\dfrac{a}{2})^2 = a^2\) \(h^2 = a^2 - (\dfrac{a^2}{4})\) \(h^2 = \dfrac{3}{4}a^2\) \(h = \dfrac{\sqrt{3}}{2}a\)
05

Relate 'h' to 'c'

Now, we need to relate the height of the tetrahedron to the height of the unit cell (c). In HCP structure, 3 layers A, B and A align on top of each other, hence the height 'c' of the HCP structure will be 2 times the height of the tetrahedron 'h'. \(c = 2h\) \(c = 2 * \dfrac{\sqrt{3}}{2}a\) \(c = \sqrt{3}a\)
06

Calculate the ideal c/a ratio

Finally, to get the ideal c/a ratio, we will divide the value of 'c' derived in Step 5 by 'a': \(c/a = \dfrac{\sqrt{3}a}{a}\) \(c/a = \sqrt{3} \approx 1.633\) Thus, the ideal c/a ratio for a Hexagonal Close Packed (HCP) crystal structure is 1.633.

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

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

Show that the atomic packing factor for HCP is \(0.74\).

Convert the (111) and (012) planes into the fourindex Miller-Bravais scheme for hexagonal unit cells.

Determine the expected diffraction angle for the first-order reflection from the (310) set of planes for BCC chromium (Cr) when monochromatic radiation of wavelength \(0.0711 \mathrm{~nm}\) is used.

Molybdenum (Mo) has a BCC crystal structure, an atomic radius of \(0.1363 \mathrm{~nm}\), and an atomic weight of \(95.94 \mathrm{~g} / \mathrm{mol}\). Compute and compare its theoretical density with the experimental value found inside the front cover of the book.

(a) Derive planar density expressions for FCC (100) and (111) planes in terms of the atomic radius \(R\). (b) Compute and compare planar density values for these same two planes for aluminum (Al).
See all solutions

Recommended explanations on Physics Textbooks

Famous Physicists

Read Explanation

Electric Charge Field and Potential

Read Explanation

Oscillations

Read Explanation

Modern Physics

Read Explanation

Particle Model of Matter

Read Explanation

Classical Mechanics

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.