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Determine the energy required to give an electron a speed equal to \(0.90\) that of light, starting from rest.

Short Answer

Expert verified
The energy required is approximately 0.661 MeV.

Step by step solution

01

Understanding the Problem

We need to find the energy required to accelerate an electron to a speed that is 0.90 times the speed of light. The electron starts from rest, so its initial kinetic energy is zero.
02

Identify the Required Formula

The relativistic kinetic energy of an object is given by the formula \( K = (\gamma - 1)m_0c^2 \), where \( \gamma \) is the Lorentz factor, \( m_0 \) is the rest mass of the electron, and \( c \) is the speed of light.
03

Calculate the Lorentz Factor \( \gamma \)

The Lorentz factor \( \gamma \) is calculated using the formula \( \gamma = \frac{1}{\sqrt{1 - \left( \frac{v}{c} \right)^2}} \). Here, \( v = 0.90c \). Thus,\[ \gamma = \frac{1}{\sqrt{1 - (0.90)^2}} = \frac{1}{\sqrt{1 - 0.81}} = \frac{1}{\sqrt{0.19}} \approx 2.294 \].
04

Insert the Values into the Kinetic Energy Formula

Now we substitute the calculated \( \gamma \) value in the kinetic energy formula: \( K = (\gamma - 1)m_0c^2 \). The rest mass energy \( m_0c^2 \) for an electron is \( 0.511 \) MeV (mega-electronvolts). So \[ K = (2.294 - 1) \times 0.511 \text{ MeV} \approx 0.661 \text{ MeV} \].

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Key Concepts

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

Lorentz Factor
The Lorentz Factor, often denoted by the Greek letter \( \gamma \), is a critical concept in relativistic physics. It provides a measure of how much an object's time, length, and relativistic mass increase as it approaches the speed of light. It is calculated with the formula:\[ \gamma = \frac{1}{\sqrt{1 - \left( \frac{v}{c} \right)^2}} \]where \( v \) is the velocity of the object, and \( c \) is the speed of light. This equation highlights how velocities approaching the speed of light result in very large values of \( \gamma \), indicating significant relativistic effects.
  • At rest (\( v=0 \)): \( \gamma =1 \), meaning no relativistic effects.
  • As \( v \) approaches \( c \): \( \gamma \) grows large, affecting time dilation and length contraction.
In the problem, an electron is accelerated to 0.90 times the speed of light, resulting in a Lorentz factor of approximately 2.294. This factor signifies how much the electron's relativistic kinetic energy will diverge from classical predictions.
Electron Acceleration
When accelerating an electron to high speeds, particularly close to the speed of light, relativistic effects become significant. The electron starts from rest, so its initial velocity is zero, and hence, its initial kinetic energy is also zero.
  • Relativistic acceleration involves significant energy requirements compared to classical physics.
  • As speed increases, the energy needed to continue accelerating the electron increases due to the increase in mass according to relativity.
In this exercise, calculating the energy required to reach 0.90c illustrates the necessity of accounting for relativistic mass and kinetic energy. Unlike in classical physics, energy increases non-linearly in relativistic contexts due to the increased Lorentz factor.
Rest Mass Energy
Rest mass energy is a fundamental concept in Einstein's theory of relativity. It is the intrinsic energy of a particle at rest and is denoted as \( m_0c^2 \), where \( m_0 \) is the rest mass, and \( c \) is the speed of light.
  • For an electron, \( m_0c^2 \) is approximately 0.511 MeV.
  • This energy remains constant regardless of the velocity of the particle.
Even when a particle is at rest, it possesses this energy due to its mass. This concept is crucial in understanding energy-mass equivalence in relativity. When accelerating an electron in the solution, the rest mass energy is utilized as a baseline to calculate additional kinetic energy. Thus, the total energy reflects the combination of rest mass energy and the relativistic kinetic energy.
Speed of Light
The speed of light, denoted by \( c \), is a universal physical constant, significant in both physics and cosmology. It has a value of approximately \( 3 \times 10^8 \) meters per second. It serves as the ultimate speed limit for information and matter in the universe.
  • \( c \) is the denominator in the Lorentz factor, signifying the threshold where relativistic effects are maximized.
  • At speeds close to \( c \), time dilation and length contraction become profound.
In the provided exercise, the speed of light serves as the reference point for determining an electron's relativistic kinetic energy. The electron reaches 0.90c, illustrating the significance of \( c \) in defining and calculating relativistic effects. Understanding \( c \) helps resolve how substantial kinetic energy is required as velocities approach this cosmic speed limit.

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Most popular questions from this chapter

How fast must an object be moving if its corresponding value of \(y\) is to be \(1.0\) percent larger than \(y\) is when the object is at rest? Give your answer to two significant figures. Use the definition \(\gamma=1 / \sqrt{1-(v / \mathrm{c})^{2}}\) to find that at \(v=0, y=1.0\). Hence, the new value of \(y=1.01(1.0)\), and so $$ 1-\left(\frac{v}{\mathrm{c}}\right)^{2}=\left(\frac{1}{1.01}\right)^{2}=0.980 $$ Solving yields \(v=0.14 \mathrm{c}=4.2 \times 10^{7} \mathrm{~m} / \mathrm{s}\)

Compute the rest energy of an electron-that is, the energy equivalent of its mass, \(9.11 \times 10^{-31} \mathrm{~kg}\).

The fastest vehicle leaving Earth so far was NASA's New Horizon Pluto mission. The craft attained a speed of \(16.26 \mathrm{~km} / \mathrm{s}\). Determine the corresponding value of \(y\) using \(\gamma=\left(1-\beta^{2}\right)^{-\frac{1}{3}}\) first; then use Eq. (41.2). [ Hint: Work in \(\mathrm{m} / \mathrm{s}\) and use \(\mathrm{c}=2.998 \times 10^{8} \mathrm{~m} / \mathrm{s}\).

The clock on a spaceship shows that a robot on board took \(10.0 \mathrm{~s}\) to do some job. The ship flies passed a station, and someone watching the robot also notes how long it took to do the job using her own wristwatch. If she computes \(y\) for the ship to be \(1.08\), how long will she say the robot took to do the job? [Hint: Time dilation means time slows and durations appear longer.]

A proton \(\left(\mathrm{m}=1.67 \times 10^{-27} \mathrm{~kg}\right)\) is accelerated to a kinetic energy of \(200 \mathrm{MeV}\). What is its speed at this energy?

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