/*! 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} Q9Q Consider a copper wire that is c... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Consider a copper wire that is carrying, say, a few amperes of current. Is the drift speed vdof the conduction electrons that form that current about equal to, much greater than, or much less than the Fermi speed vFfor copper (the speed associated with the Fermi energy for copper)?

Short Answer

Expert verified

The drift speed of the conduction electrons is much less than the Fermi speed for copper.

Step by step solution

01

The given data

A copper wire is considered.

02

Understanding the concept of drift speed and Fermi speed

A drift speed of conduction electron is the average velocity attained by the charged particles, such as electrons, in a material due to an electric field. In general, an electron propagates randomly at the Fermi velocity, resulting in an average velocity of zero.

Formulae:

The drift speed in a material,vd=inAq (i)

The Fermi speed of the given material,vF=2EFm (ii)

The equation of Fermi energy isE=3162Ï€2/3h2mn2/3 (iii)

where n is the number of conduction electrons per unit volume, m is the mass of an electron and h is Planck’s constant.

03

Calculation of the drift speed and the Fermi speed

When a potential difference is applied across a conductor, free electrons gain velocity in the direction opposite to the electric field. During the motion, they collide with other electrons, ions, and the walls of the conductor. Due to these collisions, the energy is lost and the speed of electrons is reduced. Thus, there is a definite small drift velocity of free electrons.

From equation (i), we can see that the drift speed is inversely proportional to the value of charge carrier density, n.

While using equation (iii) in equation (ii), we can see that the Fermi speed is proportional to the value of the conduction electrons density.

Thus, comparing both the cases, it can be stated that the drift speed is much less than the Fermi speed.

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 91Ó°ÊÓ!

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

A potassium chloride crystal has an energy band gap of 7.6eV above the topmost occupied band, which is full. Is this crystal opaque or transparent to light of wavelength 140 nm?

When a photon enters the depletion zone of a p-njunction, the photon can scatter from the valence electrons there, transferring part of its energy to each electron, which then jumps to the conduction band. Thus, the photon creates electron–hole pairs. For this reason, the junctions are often used as light detectors, especially in the x-ray and gamma-ray regions of the electromagnetic spectrum. Suppose a single 662keV gamma-ray photon transfers its energy to electrons in multiple scattering events inside a semiconductor with an energy gap of 1.1eV, until all the energy is transferred. Assuming that each electron jumps the gap from the top of the valence band to the bottom of the conduction band, find the number of electron – hole pairs created by the process.

The compound gallium arsenide is a commonly used semiconductor, has an energy gap Egof1.43eV. Its crystal structure is like that of silicon, except that half the silicon atoms are replaced by gallium atoms and half by arsenic atoms. Draw a flattened-out sketch of the gallium arsenide lattice, following the pattern of Fig. 41-10a.What is the net charge of the (a) gallium and (b) arsenic ion core? (c) How many electrons per bond are there? (Hint:Consult the periodic table in Appendix G.)

If the temperature of a piece of a metal is increased, does the probability of occupancy 0.1 eV above the Fermi level increase, decrease, or remain the same?

The occupancy probability function (Eq. 41-6) can be applied to semiconductors as well as to metals. In semiconductors the Fermi energy is close to the midpoint of the gap between the valence band and the conduction band. For germanium, the gap width is 0.67eV. What is the probability that (a) a state at the bottom of the conduction band is occupied and (b) a state at the top of the valence band is not occupied? Assume that T = 290K. (Note:In a pure semiconductor, the Fermi energy lies symmetrically between the population of conduction electrons and the population of holes and thus is at the center of the gap. There need not be an available state at the location of the Fermi energy.)

See all solutions

Recommended explanations on Physics Textbooks

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.