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Electrons are accelerated through a potential difference of \(750 \mathrm{kV},\) so that their kinetic energy is \(7.50 \times 10^{5} \mathrm{eV}\) . (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 relativistic speed ratio is approximately 0.938. Classically, the speed is much lower.

Step by step solution

01

Understanding the problem

We are given that electrons are accelerated through a potential difference of 750 kV, granting them kinetic energy of \(7.50 \times 10^{5} \mathrm{eV}\). We need to find the electron's speed \(v\) as a ratio to the speed of light \(c\), and compare it to the speed calculated using classical mechanics principles.
02

Calculating speed using relativistic mechanics

In relativity, the total energy \(E\) of the electron is the sum of its rest energy \(E_0 = mc^2\) and its kinetic energy. Here, \(m\) is the rest mass of the electron, \(c\) is the speed of light, and \(K\) is the kinetic energy. We use the equation:\[K = (\gamma - 1)mc^2\]where \(\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}\) is the Lorentz factor. First, calculate \(\gamma\) using the kinetic energy equation.
03

Solving for the relativistic speed

Rearrange the kinetic energy equation to solve for \(v\):\[ \gamma = 1 + \frac{K}{mc^2} \] using \(K = 7.50 \times 10^{5} \mathrm{eV} = 1.20 \times 10^{-13} \mathrm{J} \) and \( mc^2 = 0.51 \times 10^{6} \mathrm{eV} = 8.19 \times 10^{-14} \mathrm{J}\). Substituting gives \( \gamma \approx 2.47 \). Then use \( \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \) to find \(v\).
04

Calculating speed using classical mechanics

Using classical mechanics, the kinetic energy is equal to \(\frac{1}{2}mv^2 = K\). Therefore, \(v\) can be calculated as: \[v = \sqrt{\frac{2K}{m}} \]. Convert the energy from eV to Joules and use this formula to find \(v_{\text{classical}}\).
05

Comparing relativistic and classical speeds

Plugging in the values, the relativistic speed \(v\) is about \(0.938c\). Using classical mechanics, \(v_{\text{classical}}\) comes out to be approximately \(1.19 \times 10^8 \text{ m/s}\). Observe how close \(v\) is to the speed of light compared to the classical speed.

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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 is a key concept in relativistic mechanics. It describes how much time, length, and relativistic mass are affected for an object that is moving at a significant fraction of the speed of light. The formula for the Lorentz factor \( \gamma \) is given by:
  • \( \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \)
Where \( v \) is the velocity of the moving object, and \( c \) is the speed of light.
For fast-moving particles, like electrons in this problem, \( \gamma \) becomes crucial when calculating relativistic effects, such as kinetic energy. Here, when the electron's speed approaches the speed of light, \( \gamma \) increases significantly, affecting its observed properties. In this exercise, \( \gamma \approx 2.47 \), indicating the electron is moving at a velocity where relativistic effects are notable.
Kinetic Energy
Kinetic energy in classical physics is the energy possessed by an object due to its motion, calculated as \( \frac{1}{2}mv^2 \). However, when the speed increases to near light speed, as in this problem, we employ relativistic kinetic energy. This takes into account the object's increased mass due to the Lorentz factor. The formula for relativistic kinetic energy is:
  • \( K = (\gamma - 1)mc^2 \)
Here, \( K \) is the kinetic energy, \( m \) is the rest mass of the electron, and \( c \) is the speed of light.
This expression shows that the kinetic energy is not just dependent on the velocity squared, as in classical mechanics, but also influenced by the relativistic mass increase represented by \( \gamma \).
For the electron in this exercise, the potential difference of 750 kV gives it a kinetic energy of \( 7.50 \times 10^5 \) eV, which is crucial for understanding its relativistic behavior.
Classical Mechanics
Classical mechanics is the branch of physics dealing with the motion of objects based on Newton's laws, which are accurate for objects moving at speeds much less than the speed of light. In classical mechanics, kinetic energy is calculated simply as:
  • \( K = \frac{1}{2}mv^2 \)
Here, when calculating the speed of the electron, we assume no relativistic effects by simply solving for \( v \):
  • \( v = \sqrt{\frac{2K}{m}} \)
Despite its widespread use, classical mechanics fails to predict the correct values for entities moving at significant fractions of the speed of light, as seen in this exercise.
The discrepancy between the speeds calculated through classical and relativistic considerations is significant, exemplifying why classical mechanics cannot be applied to high-speed particles like electrons.
Electron Speed
The speed of an electron is crucial in understanding its energy and relativistic effects. This exercise involves calculating this speed both classically and relativistically. In relativistic mechanics, the speed of an electron is found using the Lorentz factor, essentially setting up the equation:
  • \( \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \)
Given that the kinetic energy \( K \) is \( 7.50 \times 10^5 \) eV, we first find \( \gamma \), then use it to solve for \( v \), eventually getting a ratio to the speed of light \( c \), resulting in a speed of approximately \( 0.938c \).
For the classical approach, we compute \( v \) using \( v = \sqrt{\frac{2K}{m}} \), yielding a much lower result, demonstrating that without relativistic corrections, we significantly underestimate this speed.
Speed of Light
The speed of light \( c \) is a fundamental constant in physics, approximately \( 3 \times 10^8 \) meters per second. It represents the ultimate speed limit of the universe, and it plays a crucial role in both relativistic and classical physics.
In relativistic mechanics, the speed of light acts as a boundary, influencing how time and mass are perceived as an object's velocity approaches it. Changes and corrections in measurements, like time dilation and length contraction, manifest markedly when approaching this constant boundary.
For this exercise, the speed of light serves as a reference point for calculating the relativity-based speed of the electron, emphasizing how different this speed is compared to the classical calculation.

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Most popular questions from this chapter

An unstable particle is created in the upper atmosphere from a cosmic ray and travels straight down toward the surface of the earth with a speed of 0.99540\(c\) relative to the earth. A scientist at rest on the earth's surface measures that the particle is created at an altitude of 45.0 \(\mathrm{km} .\) (a) As measured by the scientist, how much time does it take the particle to travel the 45.0 \(\mathrm{km}\) to the surface of the earth? (b) Use the length-contraction formula to calculate the distance from where the particle is created to the surface of the earth as measured in the particle's frame. (c) In the particle's frame, how much time does it take the particle to travel from where it is created to the surface of the earth? Calculate this time both by the time dilation formula and from the distance calculated in part (b). Do the two results agree?

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