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Electrons are accelerated through a potential difference of 750 kV, so that their kinetic energy is 7.50 \(\times\) 10\(^5\) eV. (a) What is the ratio of the speed \(v\) of an electron having this energy to the speed of light, \(c\)? (b) What would the speed be if it were computed from the principles of classical mechanics?

Short Answer

Expert verified
The speed ratio, \(v/c\), is approximately 0.774. The classical speed is much higher.

Step by step solution

01

Identify the Given Data

The potential difference is 750 kV, which corresponds to an energy of 7.50 \(\times\) 10\(^5\) eV for the electron. Thus, we use this energy to determine the speed ratio to the speed of light.
02

Calculate Speed Using Relativistic Kinetic Energy

The relativistic kinetic energy is given by the equation: \( KE = (\gamma - 1) mc^2 \), where \( \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \). Given \(KE = 7.50 \times 10^5 eV\) or \(1.2 \times 10^{-13} J\), solve for speed \(v\) in terms of \(c\).
03

Calculate the Ratio \(v/c\)

Rearrange the formula from the previous step to solve for \(v\) in terms of \(c\), leading to \(\frac{v}{c} = \sqrt{1 - \left(\frac{m c^2}{KE + mc^2}\right)^2}\). Substitute the values for \(m e\) (the mass-energy of an electron), and solve for the ratio \(v/c\).
04

Calculate Speed Using Classical Mechanics

For an electron, use the classical formula \( KE = \frac{1}{2}mv^2 \) to find \(v\). Use \( KE = 7.50 \times 10^5 eV = 1.2 \times 10^{-13} J \), and calculate \(v = \sqrt{\frac{2KE}{m}}\) where \(m = 9.11 \times 10^{-31} kg\).
05

Comparison of Classical and Relativistic Speeds

Calculate and compare both speeds. Note that the classical speed will be notably higher than the relativistic speed because classical mechanics doesn't account for relativistic mass increases at high speeds.

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Key Concepts

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

Electron Speed Calculation
When an electron is accelerated, its speed becomes a crucial aspect in understanding its behavior, especially at high energies. In this scenario, the electron's speed is deduced using principles from relativistic mechanics instead of classical mechanics due to the involvement of high energies. The crux here is to distinguish between the relativistic and classical approaches: - **Relativistic Speed Calculation:** Utilizes the relativistic kinetic energy equation to account for the increase in mass when particles approach the speed of light.- **Classical Speed Calculation:** Involves the standard kinetic energy equation, which does not consider mass changes at high velocities.To compute the speed of the electron using relativistic principles, we start with its kinetic energy, provided as 7.50 \( \times \) 10\(^5\) eV. This energy corresponds to a significant fraction of the electron's rest energy. Thus, we solve for its speed using the formula for relativistic kinetic energy and determine the ratio \( \frac{v}{c} \), where \( v \) is the speed of the electron and \( c \) is the speed of light. The formula itself stems from the time dilation and length contraction effects described by Einstein's theory of relativity.
Relativistic Kinetic Energy
Relativistic Kinetic Energy becomes pivotal when particles are moving close to the speed of light. Unlike classical kinetic energy, this type of energy embraces the effects of special relativity.In the context of our exercise, the equation for relativistic kinetic energy is given by:\[ KE = (\gamma - 1) mc^2 \] This is where: - \( \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \) is the Lorentz factor, which increases as velocity \( v \) approaches \( c \),- \( m \) is the mass of the electron, and- \( c \) is the speed of light.The Lorentz factor introduces corrections to classical mechanics by acknowledging that the mass of a moving object increases with speed. Here, we solve for \( v \) by rearranging the relativistic kinetic energy formula and incorporating known values such as the electron's rest mass energy. This ensures that calculations accurately reflect the faster-than-classical decrease in speed as the electron's energy increases, closely aligning with experimental data.
Potential Difference
Understanding the role of potential difference is key when analyzing scenarios involving charged particles, such as electrons. A potential difference, often measured in volts, provides the energy required to accelerate charged particles, like electrons, along a path. In this exercise, a potential difference of 750 kV accelerates the electron, providing it with the substantial energy of 7.50 \( \times \) 10\(^5\) eV.Here's how the potential difference ties into this:- **Potential Energy Conversion:** When an electron moves through a potential difference, it gains kinetic energy. The potential energy difference gets converted into kinetic energy, enhancing the particle's motion.- **Energy Relationship:** The amount of energy gained by an electron is directly related to the potential difference it travels through. This is vital for understanding phenomena like electron acceleration in electric fields, used in devices like particle accelerators.In effect, potential difference serves as the catalyst enabling the electron to gain sufficient kinetic energy, propelling it towards relativistic speeds. Recognizing this transformation is crucial for comprehending broader concepts in electromagnetism and particle physics.

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Most popular questions from this chapter

Inside a spaceship flying past the earth at three-fourths the speed of light, a pendulum is swinging. (a) If each swing takes 1.80 s as measured by an astronaut performing an experiment inside the spaceship, how long will the swing take as measured by a person at mission control (on earth) who is watching the experiment? (b) If each swing takes 1.80 s as measured by a person at mission control, how long will it take as measured by the astronaut in the spaceship?

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