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Chapter 18: Problem 41

A hypothetical metal is known to have an electrical resistivity of \(4 \times 10^{-8}(\Omega \cdot \mathrm{m})\). Through a specimen of this metal that is \(25 \mathrm{~mm}\) thick is passed a current of \(30 \mathrm{~A}\); when a magnetic field of \(0.75\) tesla is simultaneously imposed in a direction perpendicular to that of the current, a Hall voltage of \(-1.26 \times\) \(10^{-7} \mathrm{~V}\) is measured. Compute (a) the electron mobility for this metal and (b) the number of free electrons per cubic meter.

Short Answer

Expert verified
a) Calculate the electron mobility of the hypothetical metal. Using the formula for electron mobility, we have: \(\mu=\dfrac{R_H}{\rho}=\dfrac{-1.26 \times 10^{-7}\mathrm{~V}}{qvB} \cdot \dfrac{1}{4\times10^{-8}(\Omega\cdot\mathrm{m})}\) First, we need to find the magnetic force on the charges (\(F_B\)). We use the formula \(F_B = qvB\) and assuming the charge of an electron (\(q\)) is \(1.6 \times 10^{-19} \mathrm{~C}\). We don't have the drift velocity (\(v\)) in this case, but we can rewrite the Hall coefficient formula as follows: \(R_H= \dfrac{V_H}{F_B} = \dfrac{-1.26 \times 10^{-7}\mathrm{~V}}{q\cdot v \cdot B}\) Now, we can plug in the values and solve for electron mobility (\(\mu\)): \(\mu=\dfrac{-1.26 \times 10^{-7}\mathrm{~V}}{(1.6 \times 10^{-19}\mathrm{~C})\cdot v \cdot (0.75\, \mathrm{T})} \cdot \dfrac{1}{4\times10^{-8}(\Omega\cdot\mathrm{m})}\) \(\mu = \dfrac{-1.26 \times 10^{-7}\mathrm{~V}}{(1.6 \times 10^{-19}\mathrm{~C})\cdot(0.75\,\mathrm{T}) \cdot v \cdot 4\times10^{-8}(\Omega\cdot\mathrm{m})}\) b) Calculate the number of free electrons per cubic meter. Now, we can find the number of free electrons per cubic meter (\(n\)) using the Hall coefficient (\(R_H\)) and the charge of the electron (\(q\)): \(n = \dfrac{-1}{R_Hq}\) Plugging the values into the formula, we have: \(n = \dfrac{-1}{\dfrac{-1.26 \times 10^{-7}\mathrm{~V}}{(1.6 \times 10^{-19}\mathrm{~C})\cdot(0.75\,\mathrm{T}) \cdot v} \cdot (1.6 \times 10^{-19}\mathrm{~C})}\) \(n = \dfrac{-1}{-1.26 \times 10^{-7}\mathrm{~V} \cdot (1.6 \times 10^{-19}\mathrm{~C}) \cdot(0.75\,\mathrm{T}) \cdot v}\) To find the number of free electrons per cubic meter (\(n\)), we need the drift velocity (\(v\)) of the electrons, which isn't provided in this exercise. Therefore, we cannot calculate a definite value for the number of free electrons per cubic meter without further data.

Step by step solution

01

Convert thickness to meters

We are given that the thickness of the metal specimen is \(25 \mathrm{~mm}\). To convert this value to meters, we can use the following conversion: \(1\mathrm{~mm}=10^{-3}\mathrm{~m}\) So, \(25 \mathrm{~mm}= 25\cdot10^{-3} \mathrm{~m}=0.025 \mathrm{~m}\).
02

Calculate the current density and magnetic force on the charges

The current density (\(J\)) can be calculated using the formula: \(J =\dfrac{I}{A}\) Where \(I\) is the current (\(30 \mathrm{~A}\)) and \(A\) is the cross-sectional area. Considering the thickness of the specimen and assuming a width (\(w\)) and height (\(h\)) of 1 meter, we can find the cross-sectional area as: \(A = wh = 1\cdot0.025 = 0.025\mathrm{~m^2}\) Now, we can calculate the current density: \(J = \dfrac{30\mathrm{~A}}{0.025\mathrm{~m^2}} = 1200 \mathrm{~A/m^2}\) The magnetic force on the charges (\(F_B\)) can be calculated using the formula: \(F_B = qvB\) Where \(q\) is the charge of the electron, \(v\) is the drift velocity of the electrons, and \(B\) is the magnetic field (\(0.75\) tesla).
03

Find the Hall coefficient

From the Hall voltage (\(V_H\)) and magnetic force (\(F_B\)), we can find the Hall coefficient (\(R_H\)) using the formula: \(V_H = R_H \cdot F_B\) \(R_H= \dfrac{V_H}{F_B} = \dfrac{-1.26 \times 10^{-7}\mathrm{~V}}{qvB}\)
04

Calculate the electron mobility

Electron mobility (\(\mu\)) can be calculated using the formula: \(\mu=\dfrac{R_H}{\rho}=\dfrac{-1.26 \times 10^{-7}\mathrm{~V}}{qvB} \cdot \dfrac{1}{4\times10^{-8}(\Omega\cdot\mathrm{m})}\)
05

Calculate the number of free electrons per cubic meter

We can find the number of free electrons per cubic meter (\(n\)) using the Hall coefficient (\(R_H\)) and the charge of the electron (\(q\)): \(n = \dfrac{-1}{R_Hq}\) By following these steps and using the given values, we can calculate the electron mobility and the number of free electrons per cubic meter for the given hypothetical metal.

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Key Concepts

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

Understanding Hall Voltage
The concept of Hall voltage is fundamental in the study of electromagnetism and plays a crucial role in determining the characteristics of conductive materials. It's named after the physicist Edwin Hall, who discovered the effect in 1879. Hall voltage arises when a magnetic field is applied perpendicular to the direction of electric current in a conductor.
When this happens, the charge carriers (which are usually electrons) experience a force known as the Lorentz force. This force pushes the charge carriers to one side of the conductor, leading to a voltage difference across the material. This voltage difference is what we refer to as Hall voltage. In the provided exercise, the negative sign of the Hall voltage indicates that the charge carriers in the hypothetical metal are negatively charged, which is typical for electrons.
By measuring the Hall voltage, we can gather information about the charge carriers in a material, such as their concentration and mobility, which explains why this phenomenon is instrumental in semiconductor research and technology. In simpler terms, Hall voltage helps us 'see' how electrons behave when they are flowing through a material and how they react to an external magnetic field.
Electron Mobility's Role in Conductivity
When diving into the world of semiconductor physics, electron mobility emerges as a key parameter. It measures how quickly electrons can move through a conductor when subjected to an electric field. This property is crucial because it affects how well the material conducts electricity; higher mobility means that electrons can move more freely, resulting in better conductivity.
In our exercise, mobility is derived from both the Hall voltage and the known electrical resistivity using a formula. To imagine electron mobility, you could think of it similarly to a person running through a crowd. A higher mobility would be akin to an agile runner who can swiftly weave through the crowd (conductor) without much resistance, portraying electrons effortlessly drifting through the lattice structure of the given metal.
Moreover, the electron mobility can reveal impurities and defects in a material. In an ideal, pure crystal, electrons would encounter minimal resistance and thus, exhibit high mobility. Conversely, in a real-world material with impurities and crystal defects, electron mobility tends to drop. This characteristic makes electron mobility a significant factor in designing and manufacturing electronic devices.
Free Electrons Per Cubic Meter and Electrical Properties
A fundamental characteristic of conductive materials is the concentration of free electrons per cubic meter. This is essentially the number of electrons available to conduct electric current within a given volume of the material. Understanding this quantity is critical for characterizing the electrical properties of conductors and semiconductors.
In the context of our exercise, knowing the number of free electrons helps determine the effectiveness of the metal in conducting electricity. The higher the number of free electrons, the more conductive the material is. This is because more charge carriers are available to participate in the flow of electric current. You can liken this to having more workers in a factory; the more workers you have, the more productive the factory can be.
Moreover, the concentration of free electrons is interlinked with the material's resistivity – a high number of free electrons typically means lower resistivity, and therefore, better conductivity. In semiconductors, this number is fine-tuned by introducing small amounts of impurities in a process known as doping, which allows for the precise control of the material's conductivity. The exercise guides us through calculating this number using the Hall voltage and the fundamental charge of an electron, giving us insight into how densely packed the 'electron sea' is in our hypothetical metal.

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Most popular questions from this chapter

Cite the differences in operation and application for junction transistors and MOSFETs.

Briefly describe electron and hole motions in a \(p-n\) junction for forward and reverse biases; then explain how these lead to rectification.

(a) For each of the three types of polarization, briefly describe the mechanism by which dipoles are induced and/or oriented by the action of an applied electric field. (b) For solid lead titanate \(\left(\mathrm{PbT} \mathrm{O}_{3}\right)\), gaseous neon, diamond, solid \(\mathrm{KCl}\), and liquid \(\mathrm{NH}_{3}\), what kind(s) of polarization is (are) possible? Why?

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.

At room temperature the electrical conductivity of \(\mathrm{PbTe}\) is \(500(\Omega \cdot \mathrm{m})^{-1}\), whereas the electron and hole mobilities are \(0.16\) and \(0.075 \mathrm{~m}^{2} / \mathrm{V} \cdot \mathrm{s}\), respectively. Compute the intrinsic carrier concentration for PbTe at room temperature.
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