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It is stated in the text that special relativity must be used to calculate the de Broglie wavelength of electrons in an electron microscope. Let us discover how much of an effect relativity has. Consider an electron accelerated through a potential difference of \(1.00 \times 10^{5} \mathrm{V}\) a. Using the Newtonian (nonrelativistic) expressions for kinetic energy and momentum, what is the electron's de Broglie wavelength? b. The de Broglie wavelength is \(\lambda=h / p,\) but the momentum of a relativistic particle is not \(m v .\) Using the relativistic expressions for kinetic energy and momentum, what is the electron's de Broglie wavelength?

Short Answer

Expert verified
The de Broglie wavelength of an electron accelerated through a potential difference of \(1.00 \times 10^{5} \mathrm{V}\) computed using Newtonian physics and relativistic physics differ. The contrast illustrates the impact of relativistic effects on particles moving at higher speeds.

Step by step solution

01

Calculate De Broglie Wavelength Using Newtonian Physics

First, determine the electron's kinetic energy using the Newtonian relation \(KE = qV\), where \(q\) is the charge of the electron (\(1.60 \times 10^{-19} \mathrm{C}\)) and \(V\) is the potential difference (\(1.00 \times 10^{5} \mathrm{V}\)). Next, calculate the momentum from the kinetic energy using \(p=\sqrt{2mE}\), where \(m\) is the electron's mass (\(9.109 \times 10^{-31} \mathrm{kg}\)) and \(E\) is the kinetic energy. Lastly, get the de Broglie wavelength using \(\lambda=h / p\), where \(h\) is Planck's constant (\(6.626 \times 10^{-34} \mathrm{J} \cdot \mathrm{s}\)) and \(p\) is the momentum.
02

Compute De Broglie Wavelength Using Relativistic Physics

With the same potential difference, compute the electron's kinetic energy using the relativistic formula \(KE = mc^2(\gamma - 1)\), where \(m\) is the electron's mass, \(c\) is the speed of light (\(3.00 \times 10^8 \mathrm{m/s}\)), and \(\gamma\) is the Lorentz factor, which can be found via \(\gamma = 1/\sqrt{1 - (v/c)^2}\). Having obtained kinetic energy, compute the relativistic momentum using \(p = \gamma m v\). Lastly, find the de Broglie wavelength using \(\lambda=h / p\), where \(h\) is Planck's constant and \(p\) is the momentum.

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Key Concepts

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

Special Relativity
When studying the behavior of particles moving at speeds comparable to the speed of light, we cannot rely on the familiar formulas from Newtonian mechanics. Instead, we use special relativity, a theory proposed by Albert Einstein in 1905, which revolutionized our understanding of time, space, and mass.

In special relativity, we learn that as an object's velocity increases, its mass appears to increase from the perspective of a stationary observer, resulting in a relativistic mass. This effect is negligible at everyday speeds but becomes significant as the object's speed approaches the speed of light.

A key concept in special relativity is the Lorentz factor \[\gamma = \frac{1}{\sqrt{1 - (\frac{v}{c})^2}}\], where \(v\) is the velocity of the particle and \(c\) is the speed of light. The Lorentz factor is used to calculate time dilation, length contraction, and the relativistic mass of objects in motion. For particles in an electron microscope, this means the traditional Newtonian calculation for the de Broglie wavelength must be modified to include relativistic effects.
Relativistic Physics
Relativistic physics involves the study of phenomena that occur at velocities close to the speed of light. At such high speeds, the equations of motion deviate from those predicted by Newtonian mechanics. For example, an electron accelerated through a potential difference of \(1.00 \times 10^{5} \mathrm{V}\) will have a significant fraction of the speed of light, meaning relativistic formulas must be used to accurately calculate its properties, such as kinetic energy and momentum.

The relativistic expression for kinetic energy is \(KE = mc^2(\gamma - 1)\), and the momentum of a relativistic particle is given by \(p = \gamma m v\). These adjustments are vital for high-energy physics and technologies that rely on the precise behavior of particles at high velocities, like electron microscopes—or when dealing with cosmic rays and particles in accelerators.
Newtonian Mechanics
Newtonian mechanics, the foundation of classical physics, is governed by Isaac Newton's laws of motion and the law of universal gravitation. It works remarkably well for the everyday speeds and masses we encounter in most aspects of life.

The Newtonian formula for kinetic energy is \(KE = \frac{1}{2}mv^2\), where \(m\) is the mass of the particle and \(v\) is its velocity. However, for calculating the momentum of a particle, we use \(p = mv\). When the speed of a particle isn't too high, these formulas are sufficient to calculate variables like the de Broglie wavelength of electrons.

It's important for students to understand that while these Newtonian formulas are simpler and often adequate, they fail to account for relativistic effects that occur at high velocities, highlighting the importance of context when choosing the appropriate physics framework.
Electron Microscopy
Electron microscopy is a powerful technique that allows us to see structures at the nanometer scale by using electrons instead of light to form an image. Unlike photons, electrons have mass and can be described by their de Broglie wavelength, which determines the resolution of the microscope.

The de Broglie wavelength of an electron is inversely proportional to its momentum: \[\lambda = \frac{h}{p}\], where \(h\) is Planck's constant and \(p\) is the momentum of the electron. A lower wavelength (higher momentum) allows for higher resolution imaging.

In electron microscopes, electrons are accelerated to high speeds, necessitating the use of relativistic physics to calculate their de Broglie wavelength correctly. The difference between wavelengths calculated using Newtonian and relativistic physics can significantly affect the accuracy of the images produced by an electron microscope.

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Most popular questions from this chapter

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