Problem 58 - Fundamentals of Physics

Question

The mass of an electron is \(9.10938188 \times 10^{-31} \mathrm{~kg}\). To eight significant figures, find the following for the given electron kinetic energy: (a) \(\gamma\) and (b) \(\beta\) for \(K=1.0000000 \mathrm{keV},(\mathrm{c}) \gamma\) and \((\) d \() \beta\) for \(K=\) \(1.0000000 \mathrm{MeV}\), and then (e) \(\gamma\) and (f) \(\beta\) for \(K=1.0000000 \mathrm{GeV}\).

Step-by-Step Solution

Answer

Compute \(\gamma\) and \(\beta\) for each energy case using the formulas provided.

Step 1: Convert Energy Units

Convert the kinetic energy from the given units to joules.For (a) and (b), K is given as 1.0000000 keV. Using the conversion factor, 1 eV = \(1.60219 \times 10^{-19} \text{ J}\), \( K = 1.0000000 \times 10^3 \times 1.60219 \times 10^{-19} \text{ J}\).For (c) and (d), K is given as 1.0000000 MeV. \( K = 1.0000000 \times 10^6 \times 1.60219 \times 10^{-19} \text{ J}\).For (e) and (f), K is given as 1.0000000 GeV. \( K = 1.0000000 \times 10^9 \times 1.60219 \times 10^{-19} \text{ J}\).

Step 2: Calculate Total Energy

Total energy \(E\) is the sum of the rest energy \(E_0\) and the kinetic energy \(K\).The rest energy of the electron is calculated using the formula, \( E_0 = mc^2 \) where \( m = 9.10938188 \times 10^{-31} \text{ kg} \) and \( c = 3 \times 10^8 \text{ m/s} \).Calculate \(E_0 = 9.10938188 \times 10^{-31} \times (3 \times 10^8)^2 \text{ J}\).Then sum \(E_0\) and \(K\) for each case to get \(E\).

Step 3: Calculate Lorentz Factor \(\gamma\)

The Lorentz factor \(\gamma\) is calculated as \(\gamma = \frac{E}{E_0}\).Substitute the values of \(E\) and rest energy \(E_0\) calculated in the previous step.Perform the division to find \(\gamma\) for each given kinetic energy.

Step 4: Calculate Velocity Ratio \(\beta\)

\(\beta\) is given by \(\beta = \frac{v}{c} = \sqrt{1 - \frac{1}{\gamma^2}}\).Use the \(\gamma\) values obtained earlier to calculate \(\beta\) by substituting them into the formula.For each kinetic energy case, solve the equation to find \(\beta\).

Key Concepts

Lorentz Factor (γ)

In the realm of special relativity, the Lorentz factor (\(\gamma\)) plays a crucial role in understanding how time, length, and relativistic mass transform for an object moving at a significant fraction of the speed of light. It is mathematically defined as:\[\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}\]where \(v\) is the velocity of the object, and \(c\) is the speed of light in a vacuum.The Lorentz factor indicates the degree to which time dilation and length contraction occur. As an object's speed approaches the speed of light, \(\gamma\) increases dramatically, leading to significant relativistic effects.

• At rest, \(\gamma = 1\), meaning no relativistic effects.
• As velocity \(v\) increases, \(\gamma\) becomes greater than 1.
• Approaching the speed of light, \(\gamma\) approaches infinity.

For practical calculations in problems, you can express \(\gamma\) in terms of total energy \(E\) and rest energy \(E_0\) using \(\gamma = \frac{E}{E_0}\). Understanding \(\gamma\) is vital for evaluating how physics behaves differently at relativistic speeds.

Velocity Ratio (β)

The velocity ratio, known as \(\beta\), is a dimensionless quantity that represents the ratio of an object's velocity to the speed of light:\[\beta = \frac{v}{c}\]This ratio provides an intuitive measure of how fast an object is moving relative to the maximum possible speed in the universe.

• \(\beta = 0\) implies the object is at rest.
• \(\beta\) close to 1 indicates significant relativistic speeds.
• It is impossible for \(\beta\) to equal or exceed 1 as this would require infinite energy.

To find \(\beta\) using the Lorentz factor, you can apply:\[\beta = \sqrt{1 - \frac{1}{\gamma^2}}\]This relationship showcases how \(\gamma\) and \(\beta\) are interconnected, contributing to our understanding of high-speed motion and relativistic effects.

Electron Kinetic Energy

Kinetic energy is a fundamental concept that defines the energy possessed by an electron due to its motion. In relativistic physics, the kinetic energy \(K\) of an electron, moving with velocity close to the speed of light, accounts for relativistic effects:Kinetic Energy Formula:In relativistic terms,\[K = E - E_0\]where:

• \(E\) is the total energy of the electron.
• \(E_0 = mc^2\) is the electron's rest energy.

The higher the kinetic energy, the larger \(\gamma\) becomes, leading to higher relativistic effects. It’s important to remember that the kinetic energy can spike as an object’s velocity nears the speed of light, due to the need for a substantial increase in energy input.

Energy Conversion

Energy conversion is a critical process in solving problems involving electron kinetic energy. Often, kinetic energy is given in electron volts (eV), and converting these units to joules (J) is a necessary step for calculations in physics.Unit Conversion
To convert energy from eV to J, use the conversion factor:

• \(1 \text{ eV} = 1.60219 \times 10^{-19} \text{ J}\)

Example: If kinetic energy \(K\) is given as 1 keV,\[K = 1 \times 10^3 \times 1.60219 \times 10^{-19} \text{ J}\]Understanding energy conversion is crucial for accurately applying formulas and evaluating physical quantities, ensuring coherence and precision in scientific calculations.

Rest Energy of Electron

The concept of rest energy reflects the energy due to an electron’s inherent mass when it is not moving. Special relativity, introduced by Einstein, gave us the fundamental equation for rest energy:\[E_0 = mc^2\]Calculation of Rest Energy- The mass \(m\) of an electron is \(9.10938188 \times 10^{-31} \text{ kg}\).- \(c\) is the speed of light, \(3 \times 10^8 \text{ m/s}\).Thus, calculating it yields:\[E_0 = 9.10938188 \times 10^{-31} \times (3 \times 10^8)^2 \text{ J}= 8.1871 \times 10^{-14}\text{ J}\]Understanding rest energy is essential. It not only underscores the mass-energy equivalence principle but also provides a baseline energy reference for calculating relativistic energies.