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Chapter 18: Problem 9

Briefly tell what is meant by the drift velocity and mobility of a free electron.

Short Answer

Expert verified
Answer: The relationship between drift velocity and mobility of a free electron can be described by the equation \(v_d = \mu E\), where \(v_d\) is the drift velocity, \(\mu\) is the mobility, and \(E\) is the electric field. Drift velocity is directly proportional to the mobility of an electron, which means that a higher mobility indicates that the electron can move more easily in response to the electric field, resulting in a higher drift velocity. The mobility of an electron depends on factors such as the nature of the medium, temperature, and impurity levels present in the material.

Step by step solution

01

Defining Drift Velocity of a Free Electron

Drift velocity is the average velocity of free charge carriers, such as electrons, when they are subjected to an external electric field inside a conducting medium like a metal. The drift velocity depends on the strength of the applied electric field, the charge and the mass of the electron, and the interaction between the electron and the lattice of the conducting medium. The drift velocity can be calculated using the formula: \(v_d = \frac{I}{nAe}\) where \(v_d\) is the drift velocity, \(I\) is the current, \(n\) is the number density of charge carriers (electrons), \(A\) is the cross-sectional area of the conductor, and \(e\) is the elementary charge.
02

Defining Mobility of a Free Electron

Mobility is a property that characterizes how easily electrons can move in response to an electric field. Mathematically, it's defined as the ratio of the drift velocity of the charged particle to the strength of the electric field acting on it. The mobility of an electron can be represented as: \(\mu = \frac{v_d}{E}\) where \(\mu\) is the mobility, \(v_d\) is the drift velocity, and \(E\) is the electric field.
03

Relationship between Drift Velocity and Mobility

There is a direct relationship between drift velocity and mobility. As introduced earlier, mobility can be expressed as the ratio of drift velocity to the electric field: \(\mu = \frac{v_d}{E}\) By rearranging the formula and solving for drift velocity, we can find that: \(v_d = \mu E\) This equation shows that the drift velocity is directly proportional to the mobility of a free electron in an electric field. Higher mobility indicates that the electron can move more easily in response to the electric field, resulting in a higher drift velocity. The mobility of an electron depends on factors such as the nature of the medium (conductor, insulator or semiconductor), temperature, and impurity levels present in the material.

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Key Concepts

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

Electron Mobility
Electron mobility is a fundamental concept that describes how easily an electron can travel through a conducting medium under the influence of an electric field. In simpler terms, it measures the electron's ability to move through a material when a force, such as an electric field, is applied. This property is crucial for determining the efficiency of materials in conducting electricity.
  • Mathematical Definition: Electron mobility is denoted by \( \mu \) and is mathematically expressed as: \[ \mu = \frac{v_d}{E} \]where \( v_d \) is the drift velocity and \( E \) is the electric field.
  • Factors Affecting Mobility: Mobility can vary depending on several factors such as the type of material (metal, semiconductor), temperature, and impurities present.
Knowing the electron mobility helps engineers design better electronic devices by selecting materials that allow electrons to move swiftly, thereby improving performance.
Electric Field
An electric field is a region around a charged particle where the particle exerts a force on other charged particles. Imagine it like an invisible aura that pushes or pulls other charges in its vicinity. Electric fields are fundamental to understanding how electric forces act at a distance.
  • Creation of Electric Fields: Electric fields are created by charged objects. If you've ever rubbed a balloon on your hair and stuck it to a wall, you've experienced the effects of static electricity and the charged field it creates.
  • Representation and Effects: They are often represented by lines pointing in the direction of the force they exert on a positive charge. The strength of the electric field is measured in volts per meter (V/m).
Electric fields are essential for calculating force interactions and understanding how electrons move in a circuit, which directly ties into concepts like drift velocity and electron mobility.
Conducting Medium
A conducting medium is any material that allows the free movement of charge carriers, such as electrons. These materials can easily transport electric current because they contain charge carriers that move freely when an electrical force is applied. Metals like copper and aluminum are typical examples.
  • Characteristics of Conductors: Conductors have low resistance to electricity and high electron mobility, making them ideal for electrical wiring and components.
  • Interaction with Electric Fields: When an electric field is applied, electrons move through these materials, creating an electric current.
The choice of conducting medium plays a vital role in electrical engineering because it impacts the efficiency and reliability of electronic devices.
Charge Carriers
Charge carriers are the particles responsible for carrying electric charge in a medium. In conductors, these are typically electrons, but in certain conditions like semiconductors, they can also be holes (positive charge carriers).
  • Types of Charge Carriers: Depending on the material, charge carriers can be either free electrons or holes. In metals, electrons are the primary charge carriers.
  • Role in Conducting Electricity: The movement of charge carriers under an electric field's influence results in a flow of electric current.
Understanding charge carriers helps explain how electricity moves in different materials, providing insights into designing more efficient electronic components and understanding concepts like drift velocity and electron mobility.

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Most popular questions from this chapter

In terms of electron energy band structure, discuss reasons for the difference in electrical conductivity between metals, semiconductors, and insulators.

(a) Using the data in Table \(18.1\), compute the resistance of a copper wire \(3 \mathrm{~mm}(0.12\) in.) in diameter and \(2 \mathrm{~m}(78.7 \mathrm{in}\).) long. (b) What would be the current flow if the potential drop across the ends of the wire is \(0.05 \mathrm{~V}\) ? (c) What is the current density? (d) What is the magnitude of the electric field across the ends of the wire?

A charge of \(3.5 \times 10^{-11} \mathrm{C}\) is to be stored on each plate of a parallel-plate capacitor having an area of \(160 \mathrm{~mm}^{2}\left(0.25 \mathrm{in} .^{2}\right)\) and a plate separation of \(3.5 \mathrm{~mm}(0.14 \mathrm{in}\).). (a) What voltage is required if a material having a dielectric constant of \(5.0\) is positioned within the plates? (b) What voltage would be required if a vacuum were used? (c) What are the capacitances for parts (a) and (b)? (d) Compute the dielectric displacement for part (a). (e) Compute the polarization for part (a).

The polarization \(P\) of a dielectric material positioned within a parallel- plate capacitor is to be \(1.0 \times 10^{-6} \mathrm{C} / \mathrm{m}^{2}\) (a) What must be the dielectric constant if an electric field of \(5 \times 10^{4} \mathrm{~V} / \mathrm{m}\) is applied? (b) What will be the dielectric displacement \(D ?\)

Germanium to which \(5 \times 10^{22} \mathrm{~m}^{-3} \mathrm{Sb}\) atoms have been added is an extrinsic semiconductor at room temperature, and virtually all the Sb atoms may be thought of as being ionized (i.e., one charge carrier exists for each Sb atom). (a) Is this material \(n\)-type or \(p\)-type? (b) Calculate the electrical conductivity of this material, assuming electron and hole mobilities of \(0.1\) and \(0.05 \mathrm{~m}^{2} / \mathrm{V} \cdot \mathrm{s}\), respectively.
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