Learning Materials
Features
Discover
Chapter 10: 6CQ (page 413)
why is covalent bonding directional, while ionic bonding is not?
Because electrons are shared unequally among different elements.
Over 30 million students worldwide already upgrade their learning with Vaia!
All the tools & learning materials you need for study success - in one app.
Get started for free
Exercise 29 notes that more energy is required to ionize sodium that is retrieved by adding that electrons to an isolated chlorine atom, but the NaCl bond represents a lower energy because the attracting ions draw close together. Quantifying the energy-lowering effect of having alternating plus and minus charges can be rather involved for 3D lattice, but a one dimensional calculation is instructive. Consider an infinite line of point charges alternating between +e and –e, with a uniform spacing between adjacent opposite charge of a. (a) The electrostatic potential energy per ion is the same for a given positive ion as for a given negative ion. Why? (b) Calculate the electrostatic potential energy per ion for simplicity. Assume that a positive charge is at the origin. The following power series expansion will be helpful: .
Question: The Fermi velocity VF is defined by , where is the fermi energy. The Fermi energy for silver is 5.5eV.(a) Calculate the Fermi velocity.(b) what would be the wavelength of an electron with this velocity. (c)How does this compare with the lattice spacing of 0.41 nm? Does the order of magnitude makes sence?
In Figure 10.24, the n=1 band ends at , while in Figure 10.27 it ends at
Question: Referring to equations(10-2), lobe I of the hybrid states combines the spherically symmetric s state with the state that is oriented along thez-axis. and thus sticks out in the direction (see Exercises 28 and 33), If Figure is a true picture, then in a coordinate system rotated counterclockwise about they-axis by the tetrahedral angle, lobe II should become lobe. In the new frame. -values are unaffected. but what had been values in the 2x -plane become values in the -plane. according to and , where is , or .
(a) Show that lobe II becomes lobe I. Note that since neither the 2s state nor the radial part of the p states is affected by a rotation. only the angular parts given in equations (10-1) need be considered.
(b) Show that if lobe II is instead rotated about thez-axis by simply shifting . it becomes lobes III and IV.
Explain the dependence of conductivity on temperature for conductors and for semiconductors.
What do you think about this solution?
We value your feedback to improve our textbook solutions.
Explanations 
Textbooks 
Exams 
GCSE 
IGCSE 
AS 
A Level 
International A Level 
University Admissions Tests 
Flashcards 
Vaia AI 
Notes 
Study Plans 
Study Sets 
Find a job 
Find a degree 
Student Deals 
Magazine 
Mobile App