Problem 39 - University Physics with Modern Physics

Question

A particle has rest mass \(6.64 \times 10^{-27} \mathrm{kg}\) and momentum \(2.10 \times 10^{-18} \mathrm{kg} \cdot \mathrm{m} / \mathrm{s}\) . (a) What is the total energy (kinetic plus rest energy) of the particle? (b) What is the kinetic energy of the particle? (c) What is the ratio of the kinetic energy to the rest energy of the particle?

Step-by-Step Solution

Answer

Total energy: \(8.46 \times 10^{-10}\, \text{J}\), Kinetic energy: \(2.484 \times 10^{-10}\, \text{J}\), Ratio: \(0.416\).

Step 1: Calculate the rest energy

To determine the rest energy of the particle, use the formula for rest energy, given by \[ E_0 = m c^2 \]where \( m = 6.64 \times 10^{-27} \text{ kg} \) is the rest mass, and \( c = 3 \times 10^8 \text{ m/s} \) is the speed of light in a vacuum. Substitute these values into the equation:\[ E_0 = 6.64 \times 10^{-27} \times (3 \times 10^8)^2 \]\[ E_0 = 5.976 \times 10^{-10} \text{ J} \]

Step 2: Use the relativistic energy-momentum relation

To find the total energy, use the equation for the total energy given by \[ E^2 = (m c^2)^2 + (p c)^2 \]where \( p = 2.10 \times 10^{-18} \text{ kg} \cdot \text{m/s} \) is the momentum. Substitute the known values:\[ E^2 = (5.976 \times 10^{-10})^2 + (2.10 \times 10^{-18} \times 3 \times 10^8)^2 \]Calculate and take the square root to find \( E \):\[ E = \sqrt{(5.976 \times 10^{-10})^2 + (6.30 \times 10^{-10})^2} \]\[ E \approx 8.46 \times 10^{-10} \text{ J} \]

Step 3: Calculate the kinetic energy

The kinetic energy \( K \) of the particle can be obtained by subtracting the rest energy from the total energy:\[ K = E - E_0 \]Substitute the previously found values:\[ K = 8.46 \times 10^{-10} - 5.976 \times 10^{-10} \]\[ K \approx 2.484 \times 10^{-10} \text{ J} \]

Step 4: Find the ratio of kinetic energy to rest energy

To find the ratio of the kinetic energy to the rest energy, use the formula:\[ \text{Ratio} = \frac{K}{E_0} \]Substitute the values obtained earlier:\[ \text{Ratio} = \frac{2.484 \times 10^{-10}}{5.976 \times 10^{-10}} \]\[ \text{Ratio} \approx 0.416 \]

Key Concepts

Kinetic Energy

Kinetic energy in the context of relativistic physics refers to the energy that a particle possesses due to its motion. Unlike classical mechanics, where kinetic energy is simply given by \( \frac{1}{2}mv^2 \), relativistic kinetic energy calculations consider the effects of traveling at significant fractions of the speed of light. This is crucial when particles move at speeds close to that of light, as Newtonian physics no longer accurately describes the energy requirements.

In the given exercise, the kinetic energy \( K \) is determined using the difference between the total energy \( E \) and the rest energy \( E_0 \). The formula used is:

• \[ K = E - E_0 \]

This involves calculating the total energy using the particle's momentum and rest mass (explored further in subsequent sections), and then subtracting the rest energy. The result gives the kinetic energy, which is the portion of the total energy due to the particle's motion.

Rest Energy

Rest energy is the energy that a particle has due to its rest mass alone, when it's not moving. The celebrated equation \( E_0 = mc^2 \) introduced by Einstein shows that even when a particle is stationary, it still possesses an inherent form of energy. This principle highlights the equivalence of mass and energy.

For any particle with mass, the rest energy is computed using its rest mass \( m \) and the speed of light \( c \):

• \[ E_0 = m c^2 \]

In our exercise, the particle has a rest mass of \( 6.64 \times 10^{-27} \text{ kg} \), which results in a rest energy of \( 5.976 \times 10^{-10} \text{ J} \). This energy represents the particle's potential energy due to its mass when it is in a state of rest.

Understanding rest energy is fundamental in high-energy physics and cosmology, as it lays the groundwork for concepts like mass-energy equivalence seen in nuclear reactions and particle physics.

Momentum

Momentum in relativistic terms extends the classical concept by considering not only mass and velocity but also the role of the speed of light. It quantifies the motion of a particle and is modified when dealing with velocities approaching the speed of light. In such scenarios, the classical formula \( p = mv \) is not sufficient.

For high-speed particles, momentum \( p \) is crucial in ensuring the consistency of energy and motion calculations. The exercise specifies a momentum of \( 2.10 \times 10^{-18} \text{ kg} \cdot \text{m/s} \). This is important because even though the particle's rest mass might be small, its relativistic momentum becomes significant, influencing the total energy.

The calculation of total energy involves both the rest energy and momentum, as it ensures that energy measurements include contributions from both the mass at rest and the motion dynamics involved. This relationship is explored further with the energy-momentum relation.

Energy-Momentum Relation

The energy-momentum relation is an essential concept in relativistic physics that ties together a particle's total energy, momentum, and rest mass. It provides a comprehensive way to calculate a particle's energy in scenarios where both the rest mass and momentum are significant.

The formula is expressed as:

• \[ E^2 = (m c^2)^2 + (p c)^2 \]

Here, \( E \) represents the total energy, which embodies contributions from the rest energy \( (mc^2) \) and the energy due to momentum \( (pc) \).

In the exercise, this equation is used to find the total energy of the particle, showing the connection between its mass energy and its momentum energy. The inclusion of both terms ensures that the calculation respects the principles of relativity, especially under high-speed conditions, confirming that energy calculations encompass all necessary factors.