Problem 35 - College Physics

Question

\(\bullet\) (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.

Step-by-Step Solution

Answer

(a) 2.05 million volts, (b) 3.29 x 10^-13 J or 2.05 million eV.

Step 1: Define the problem and known values

An electron is being accelerated from rest to a speed of \(0.980c\), where \(c\) is the speed of light in a vacuum. We need to find the potential difference required to achieve this speed and the kinetic energy of the electron.

Step 2: Use the relativistic energy-momentum relation

To find the energy, we use the relation: \[ E = \gamma mc^2 \]where \( \gamma = \dfrac{1}{\sqrt{1 - \left( \dfrac{v}{c} \right)^2}} \) is the Lorentz factor, \( m \) is the rest mass of the electron \((9.11 \times 10^{-31} \text{ kg})\), and \( v = 0.980c \).

Step 3: Calculate the Lorentz factor

Calculate \( \gamma \) for \( v = 0.980c \):\[ \gamma = \dfrac{1}{\sqrt{1 - (0.980)^2}} = \dfrac{1}{\sqrt{1 - 0.9604}} = \dfrac{1}{\sqrt{0.0396}} \approx 5.025 \]

Step 4: Compute the relativistic energy

Substitute \( \gamma \) into the energy formula:\[ E = 5.025 \times (9.11 \times 10^{-31} \text{ kg}) \times (3.0 \times 10^8 \text{ m/s})^2 \]Simplifying gives: \[ E \approx 4.11 \times 10^{-13} \text{ J} \]

Step 5: Find the rest energy

Calculate the rest energy:\[ E_0 = mc^2 = (9.11 \times 10^{-31} \text{ kg}) \times (3.0 \times 10^8 \text{ m/s})^2 = 8.19 \times 10^{-14} \text{ J} \]

Step 6: Calculate kinetic energy

The kinetic energy \( K \) is the total energy minus the rest energy:\[ K = E - E_0 = (4.11 \times 10^{-13} - 8.19 \times 10^{-14}) \text{ J} = 3.29 \times 10^{-13} \text{ J} \]

Step 7: Convert kinetic energy into electronvolts

Convert joules into electronvolts knowing \( 1 \text{ eV} = 1.602 \times 10^{-19} \text{ J} \):\[ K \approx \dfrac{3.29 \times 10^{-13} \text{ J}}{1.602 \times 10^{-19} \text{ J/eV}} \approx 2.05 \times 10^6 \text{ eV} \]

Step 8: Calculate the potential difference

The potential energy gained \((eV)\) equals the kinetic energy \((K)\):\[ e \times V = K \Rightarrow V = \dfrac{K}{e} = \dfrac{3.29 \times 10^{-13} \text{ J}}{1.602 \times 10^{-19} \text{ C}} \approx 2.05 \times 10^6 \text{ V} \]

Key Concepts

Lorentz factor

The Lorentz factor, denoted by \( \gamma \), is a crucial concept in relativistic physics, especially when dealing with objects moving at speeds close to that of light, \( c \). As velocities increase, classical physics becomes less accurate, and relativistic equations must be used.

• The Lorentz factor is defined as: \( \gamma = \dfrac{1}{\sqrt{1 - \left( \dfrac{v}{c} \right)^2}} \). Here, \( v \) represents the object's velocity, and \( c \) is the speed of light, approximately \( 3.0 \times 10^8 \text{ m/s} \).
• When the velocity \( v \) is much less than \( c \), the Lorentz factor is approximately equal to 1, indicating Newtonian mechanics is valid.
• At higher velocities, \( \gamma \) exceeds 1, significantly affecting mass, energy, and time calculations.

For an electron moving at \( 0.980c \), the Lorentz factor was calculated to be approximately 5.025. This means the electron exhibits about five times more relativistic mass and energy than it would at rest.

Kinetic energy

In relativistic physics, the concept of kinetic energy extends beyond the classical formula \( \dfrac{1}{2}mv^2 \) because velocities involved approach the speed of light. The total energy \( E \) of a particle in motion includes its rest energy and additional energy due to its motion.

• The total relativistic energy is expressed as: \( E = \gamma mc^2 \).
• To find the kinetic energy \( K \), subtract the rest energy \( E_0 = mc^2 \) from the total energy:
  \[ K = E - E_0 \]
• In this scenario, the electron's total energy was \( 4.11 \times 10^{-13} \text{ J} \), and its rest energy was \( 8.19 \times 10^{-14} \text{ J} \).
• Thus, the kinetic energy comes out to be \( 3.29 \times 10^{-13} \text{ J} \) or approximately \( 2.05 \times 10^6 \text{ eV} \), which are energy units ideal for particles.

Understanding these calculations helps to grasp how moving at relativistic speeds significantly increases the energy content of particles.

Potential difference

Potential difference, often referred to as voltage, is essential to understanding energy transfer in electric fields. In particle physics, it plays a key role when accelerating particles like electrons from rest.

• Potential difference \( V \) relates to the energy change of a charge \( e \) moving through said difference.
• When an electron accelerates through a potential difference, it gains kinetic energy, making the relationship \( eV = K \) — where \( e \) is the charge of the electron.
• For the electron in the problem, accelerating to \( 0.980c \), the gained kinetic energy was \( 3.29 \times 10^{-13} \text{ J} \) or \( 2.05 \times 10^6 \text{ eV} \).
• This means it was accelerated across a potential difference of approximately \( 2.05 \times 10^6 \text{ V} \).

Thus, understanding the potential difference aids in determining the required energy and conditions to bring particles up to high speeds in particle accelerators.