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(a) Through what potential difference does an electron have to be accelerated, starting from rest, to achieve a speed of \(0.980 c ?\) (b) What is the kinetic energy of the electron at this speed? Express your answer in joules and in electronvolts.

Short Answer

Expert verified
Potential difference is approximately 5.81 x 10^5 V; kinetic energy is about 5.81 x 10^5 eV or 9.31 x 10^-14 J.

Step by step solution

01

Understanding the electron's velocity and energy context

The problem asks us to find the potential difference required to accelerate an electron to a velocity of 0.980 times the speed of light, denoted as \( c \). We also need to calculate the kinetic energy of the electron at this speed in both joules and electronvolts.
02

Calculate the relativistic mass of the electron

The relativistic mass of an electron increases with speed and is calculated using Lorentz factor \( \gamma \), defined as \( \gamma = \frac{1}{\sqrt{1 - \left( \frac{v}{c} \right)^2}} \). Substituting \( v = 0.980c \), calculate \( \gamma \).
03

Find relativistic kinetic energy of the electron

The relativistic kinetic energy is computed as \( KE = (\gamma - 1) m_0 c^2 \), where \( m_0 \) is the rest mass energy of an electron \( 9.11 \times 10^{-31} \text{ kg} \) and \( c \) is the speed of light \( 3 \times 10^8 \text{ m/s} \). Calculate \( KE \).
04

Convert kinetic energy in joules to electronvolts

Convert the kinetic energy from Step 3 from joules to electronvolts using the conversion factor \( 1 \text{ eV} = 1.602 \times 10^{-19} \text{ J} \).
05

Relate kinetic energy to potential difference

In an electric field, potential difference \( \Delta V \) is related to change in kinetic energy \( \Delta KE \) by \( \Delta KE = e \Delta V \), where \( e \) is the electron charge \( 1.602 \times 10^{-19} \text{ C} \). Solve for \( \Delta V \).

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Key Concepts

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

Lorentz Factor
When an object moves close to the speed of light, its dynamics are described by relativity, not classical mechanics. The Lorentz factor, denoted as \( \gamma \), helps us understand the adjustments needed due to such high speeds.
The Lorentz factor is given by the formula:
  • \( \gamma = \frac{1}{\sqrt{1 - \left( \frac{v}{c} \right)^2}} \)
Here, \( v \) is the velocity of the object and \( c \) is the speed of light. The Lorentz factor increases as an object's speed approaches the speed of light, indicating a significant increase in properties like mass and time relative to an observer. This concept is crucial in calculating values like the relativistic kinetic energy.For an electron moving at \( 0.980c \), the speed is very high. One must use the Lorentz factor to accurately assess its relativistic properties. This factor significantly impacts calculations of kinetic energy and other aspects as the object's speed nears light speed.
Potential Difference
Potential difference, also known as voltage, represents the work needed to move a charge from one point to another in an electric field. It's measured in volts (V).
  • Expressed as \( \Delta V \), this difference provides the energy required to accelerate charged particles like electrons.
In the context of the problem, understanding the potential difference helps determine how much voltage is necessary to achieve a desired speed for an electron initially at rest.

The relationship between kinetic energy change \( \Delta KE \) and potential difference \( \Delta V \) is expressed by:
  • \( \Delta KE = e \Delta V \), where \( e \) is the charge of the electron (\( 1.602 \times 10^{-19} \text{ C} \)).
Using this equation allows us to calculate precisely how much potential difference is needed to achieve and sustain high-speed motion as described in the problem.
Kinetic Energy
Kinetic energy refers to the energy an object possesses due to its motion. In relativistic physics, kinetic energy calculations differ significantly from classical kinetic energy computations.
  • The classical formula \( KE = \frac{1}{2} mv^2 \) is replaced by the relativistic formula:
\[ KE = (\gamma - 1) m_0 c^2 \]Where:
- \( \gamma \) is the Lorentz factor (depends on speed).
- \( m_0 \) represents the rest mass of the electron (\( 9.11 \times 10^{-31} \text{ kg}\)).
- \( c \) is the speed of light (\( 3 \times 10^8 \text{ m/s} \)).
Relativistic kinetic energy increases rapidly as speed approaches the speed of light, a significant deviation from linear predictions of classical physics. This is because, as speed increases, so does the relativistic mass, amplifying the exact kinetic energy achieved compared to lower speed predictions.
Speed of Light
The speed of light, denoted as \( c \), is a fundamental constant of nature with a value of \( 3 \times 10^8 \text{ m/s} \). It serves as a crucial reference point in Einstein's theory of relativity.
Light speed defines the maximum speed at which information and matter can travel through space.
  • Objects moving near this speed exhibit relativistic effects like time dilation and mass increase.
Cosmic travel and high-speed phenomena often reference \( c \), emphasizing its role as a universal speed limit in physics. In calculations, \( c \) guides transformations where dimensional and relativistic adjustments occur, such as determining particles' behaviors like electrons in accelerators or cosmic rays.
Electronvolt Conversion
Understanding energy units is essential for interpreting results in modern physics. The unit "electronvolt" (eV) is commonly used, especially in fields dealing with atomic and subatomic particles.
  • It's defined as the amount of energy gained by an electron when accelerated through a potential difference of 1 volt.
The conversion between joules and electronvolts helps bridge the gap between microscopic and macroscopic energy scales, allowing easier understanding and communication of energy values in different contexts.
  • 1 eV equals \( 1.602 \times 10^{-19} \text{ J} \), a crucial factor for converting energy values calculated in joules into electronvolts and vice versa.
When specific calculations in joules are made (like kinetic energy in relativistic physics), conversion to eV allows the results to be more relatable and easily interpretable in terms of everyday electronics and particle physics applications.

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