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Chapter 18: Problem 56

A Hall effect experiment uses a silver bar \(3.50 \mu \mathrm{m}\) thick. When the bar carries a current of \(1.42 \mathrm{~A},\) a perpendicular magnetic field of \(0.155 \mathrm{~T}\) results in a Hall potential difference of \(6.70 \mu \mathrm{V}\). (a) Use these data to determine the density of conduction electrons in silver. (b) How many conduction electrons are there per atom of silver? [Note: The density of silver is \(\left.10,490 \mathrm{~kg} / \mathrm{m}^{3} .\right]\)

Short Answer

Expert verified
(a) Electron density is \(5.86 \times 10^{28} \text{ electrons/m}^3\). (b) 1 conduction electron per atom of silver.

Step by step solution

01

Understand the Hall Effect Equation

The Hall effect for a conductor relates the Hall voltage \( V_H \) to the current \( I \), magnetic field \( B \), thickness \( t \), and charge carrier density \( n \) as: \( V_H = \frac{IB}{net} \). Identify known values: \( V_H = 6.70 \times 10^{-6} \text{ V} \), \( I = 1.42 \text{ A} \), \( B = 0.155 \text{ T} \), and thickness \( t = 3.50 \times 10^{-6} \text{ m} \). The charge of an electron \( e \) is \( 1.6 \times 10^{-19} \text{ C} \).
02

Rearrange the Hall Effect Equation to Solve for \( n \)

Rearrange the formula to solve for electron density \( n \): \( n = \frac{IB}{eV_Ht} \).
03

Substitute the Known Values into the Equation

Substitute the given values into the rearranged equation: \( n = \frac{1.42 \times 0.155}{1.6 \times 10^{-19} \times 6.70 \times 10^{-6} \times 3.50 \times 10^{-6}} \). Calculate \( n \).
04

Calculate the Electron Density \( n \)

Perform the calculation: \[ n = \frac{1.42 \times 0.155}{1.6 \times 10^{-19} \times 6.70 \times 10^{-6} \times 3.50 \times 10^{-6}} \approx 5.86 \times 10^{28} \text{ electrons/m}^3 \]. This is the density of conduction electrons in silver.
05

Calculate the Number of Silver Atoms per Cubic Meter

Use the density of silver: \( 10,490 \text{ kg/m}^3 \) and the molar mass of silver \( 107.87 \text{ g/mol} \) to find atoms per cubic meter. Convert \( 10,490 \text{ kg/m}^3 \) to \( 10,490,000 \text{ g/m}^3 \), then use \( \frac{\text{density of silver}}{\text{molar mass}} \times N_A \), where \( N_A = 6.022 \times 10^{23} \text{ mol}^{-1} \).
06

Calculate Atoms per Cubic Meter

Perform the calculation: \( \frac{10,490,000}{107.87} \times 6.022 \times 10^{23} \approx 5.85 \times 10^{28} \text{ atoms/m}^3 \). This is the number of silver atoms per cubic meter.
07

Determine the Number of Conduction Electrons per Silver Atom

Divide the density of conduction electrons by the number of silver atoms: \( \frac{5.86 \times 10^{28}}{5.85 \times 10^{28}} \approx 1.00 \). Thus, there is one conduction electron per silver atom.

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Key Concepts

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

Hall voltage
The Hall voltage arises in the context of the Hall effect. This phenomenon occurs when a current-carrying conductor is placed in a perpendicular magnetic field. As electrons move through the conductor's lattice, they experience the Lorentz force due to the magnetic field. This force pushes them to one side of the conductor, building up an electric field that balances the force due to the current. The potential difference across the conductor, created by this electric field, is known as the Hall voltage.
  • In our example, a Hall voltage of 6.70 碌V is observed.
  • This voltage reflects the interaction between the current of 1.42 A and the magnetic field of 0.155 T.
This voltage is essential to compute the electron density in the conductor.
Charge carrier density
Charge carrier density, often denoted as \( n \), refers to the number of charge carriers, such as electrons, per unit volume in a material. It is a crucial factor in determining the electrical properties of conductors and semiconductors. In the context of the Hall effect, the formula connecting the Hall voltage \( V_H \), magnetic field \( B \), current \( I \), thickness \( t \), and charge carrier density is:\[ V_H = \frac{IB}{net} \]Rearranging this to solve for \( n \) gives:\[ n = \frac{IB}{eV_Ht} \]
  • For silver, using the given values, the computed conduction electron density is approximately \(5.86 \times 10^{28} \text{ electrons/m}^3\).
  • This density is pivotal to understanding how efficiently a material can conduct electricity.
Conduction electrons
Conduction electrons are free electrons in a metal that facilitate electrical conduction. Unlike bound electrons, conduction electrons are delocalized and move freely within the metal lattice. They respond to external electric fields, thus moving charges and creating electric currents.
  • In silver, these electrons are responsible for its excellent conductivity.
  • During the Hall effect experiment, conduction electrons are deflected, creating the measured Hall voltage.
The density of conduction electrons is a key variable in these analyses.
Electron density
Electron density in the Hall effect context refers to how closely packed conduction electrons are in a material. It is expressed as the number of electrons per unit volume. To calculate it from the Hall effect, you must identify factors like current, magnetic field strength, and geometric parameters of the material in question.
  • The derived formula: \( n = \frac{IB}{eV_Ht} \) helps compute this density efficiently.
  • For silver, the computed electron density matches closely the atom density, suggesting one conduction electron per atom.
This correlation is vital for validating models of electrical conductivity in materials.
Magnetic field
A magnetic field, denoted as \( B \), is central to inducing the Hall effect in a conductor. The direction and strength of this field influence how conduction electrons are deflected and, consequently, the magnitude of the Hall voltage. In this exercise:
  • A magnetic field of 0.155 T is applied perpendicularly to the silver bar.
  • This field, interacting with the current and the structure of the metal, facilitates the measurable Hall voltage of 6.70 碌V.
Understanding the magnetic field's role helps explain how and why electrons move within a conductor under specific conditions.
Current calculation
Current calculation in the context of the Hall effect relates directly to the resulting Hall voltage and the computed charge carrier density. The current, \( I \), flows along the conductor, perpendicular to the applied magnetic field, and is critical in the calculation of the Hall voltage:
  • With a given current of 1.42 A and known parameters, we determine the density of conduction electrons.
  • These calculations lend insight into the physical processes underlying electrical conduction.
Thus, understanding current and its interaction with magnetic fields and material properties is essential for comprehensive Hall effect analyses.

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Most popular questions from this chapter

A 0.45 -T magnetic field points due east. (a) Find the magnitude and direction of the force on a \(+0.10-\mu\) C charge moving southward at \(3.5 \mathrm{~m} / \mathrm{s}\). (b) Repeat for a \(-0.10-\mu \mathrm{C}\) charge with the same velocity.

A proton moving at \(4.50 \mathrm{~km} / \mathrm{s}\) in the \(+y\) -direction enters a region containing a uniform magnetic field of \(450 \mathrm{mT}\) in the \(+z\) -direction. Find the electric field that must be applied to give zero net force on the proton.

The conduction electron density in germanium (a semiconductor) is \(2.01 \times 10^{24} \mathrm{~m}^{-3}\). The Hall effect is observed in germanium using a \(0.150-\mathrm{mm}\) -thick bar in a magnetic field of \(1.25 \mathrm{~T}\). Find the current required in the bar in order to produce a Hall potential difference of \(1.0 \mathrm{mV}\).

Two long parallel wires each carry \(2.5 \mathrm{~A}\) in the same direction, with their centers \(1.5 \mathrm{~cm}\) apart. (a) Find the magnetic field halfway between the wires. (b) Find the magnetic field at a point in the same plane as the wires, \(1.5 \mathrm{~cm}\) from one wire and \(3.0 \mathrm{~cm}\) from the other. (c) Find the force of interaction between the wires, and tell whether it's attractive or repulsive.

A straight conducting wire carries \(0.25 \mathrm{~A}\) in the \(+y\) -direction. Find the magnetic field that results in a force per unit length on the wire of \(0.20 \mathrm{~N} / \mathrm{m}\) in the \(+z\) -direction. (a) \(1.25 \mathrm{~T},-x\) -direction; (b) \(1.25 \mathrm{~T},+x\) -direction; (c) \(0.80 \mathrm{~T},-x\) -direction; (d) \(0.80 \mathrm{~T}\) \(+x\) -direction.
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