Question

(a) A particle with mass \(m\) has kinetic energy equal to three times its rest energy. What is the de Broglie wavelength of this particle? (\(Hint\): You must use the relativistic expressions for momentum and kinetic energy: \(E^2 = (pc^2) + (mc^2)^2\) and \(K = E - mc^2\).) (b) Determine the numerical value of the kinetic energy (in MeV) and the wavelength (in meters) if the particle in part (a) is (i) an electron and (ii) a proton.

Short answer

For an electron, the kinetic energy is 2.73 MeV and the wavelength is \(1.28 \times 10^{-12}\) m. For a proton, the kinetic energy is 2.80 GeV and the wavelength is \(1.51 \times 10^{-16}\) m.

Step-by-step solution

Identify the Rest Energy

The rest energy of a particle with mass \( m \) is given by \( E_0 = mc^2 \).

Relate Kinetic Energy to Rest Energy

According to the problem, the kinetic energy \( K \) is three times the rest energy, so \( K = 3mc^2 \).

Express Total Energy in Terms of Rest and Kinetic Energy

The total energy \( E \) is the sum of the rest energy and the kinetic energy: \( E = mc^2 + K = 4mc^2 \).

Use Relativistic Energy-Momentum Relation

According to the energy-momentum relation: \( E^2 = (pc)^2 + (mc^2)^2 \). Substitute \( E = 4mc^2 \) into the equation: \( (4mc^2)^2 = (pc)^2 + (mc^2)^2 \).

Solve for Relativistic Momentum

Simplify the equation: \( 16m^2c^4 = p^2c^2 + m^2c^4 \). Rearrange to find \( p^2c^2 = 15m^2c^4 \), leading to \( p = \sqrt{15}mc \).

Calculate de Broglie Wavelength

The de Broglie wavelength \( \lambda \) is given by \( \lambda = \frac{h}{p} \). Substitute \( p = \sqrt{15}mc \): \( \lambda = \frac{h}{\sqrt{15}mc} \).

Numerical Calculation for an Electron

Using known values: \( m_e = 9.11 \times 10^{-31} \text{ kg} \), \( c = 3 \times 10^8 \text{ m/s} \), and \( h = 6.626 \times 10^{-34} \text{ J s} \). Calculate \( K = 3m_ec^2 = 2.73 \text{ MeV} \) and \( \lambda = \frac{6.626 \times 10^{-34}}{\sqrt{15}(9.11 \times 10^{-31})(3 \times 10^8)} = 1.28 \times 10^{-12} \text{ m} \).

Numerical Calculation for a Proton

Using known values: \( m_p = 1.67 \times 10^{-27} \text{ kg} \). Calculate \( K = 3m_pc^2 = 2.80 \text{ GeV} \) and \( \lambda = \frac{6.626 \times 10^{-34}}{\sqrt{15}(1.67 \times 10^{-27})(3 \times 10^8)} = 1.51 \times 10^{-16} \text{ m} \).

Key concepts

Relativistic Mechanics

Relativistic mechanics is an extension of classical mechanics. It describes how objects behave when they travel at speeds close to the speed of light. When dealing with particles moving at significant fractions of the speed of light, traditional Newtonian mechanics no longer applies. Instead, Einstein's theory of relativity provides the framework for understanding the behavior of such particles.
Relativistic mechanics accounts for the effects of time dilation, length contraction, and the increase of mass with speed. These effects occur because, as an object approaches the speed of light, its mass seems to increase from the perspective of a stationary observer. This makes it impossible for the object to actually reach the speed of light.
Key equations include the relativistic energy-momentum relation, which helps us determine various properties of particles, like momentum and total energy. Using this relation is crucial for solving problems involving high-speed particles, as we can see in calculations involving the de Broglie wavelength.

Kinetic Energy

In physics, kinetic energy is the energy that an object possesses due to its motion. In classical mechanics, the kinetic energy formula is straightforward: \( K = \frac{1}{2} mv^2 \). However, in the realm of relativistic mechanics, the expression becomes more complex to incorporate relativistic effects.
For a particle traveling at relativistic speeds, its kinetic energy \( K \) can be defined as the difference between its total energy \( E \) and its rest energy \( E_0 \) (the energy it has when not moving): \( K = E - mc^2 \).
In the given exercise, we find that the particle's kinetic energy is three times its rest energy. This fact profoundly influences our calculations as it leads to the determination of the particle's total energy. Understanding this link between kinetic and rest energy is crucial for grasping how energy transforms at high velocities.

Energy-Momentum Relation

The energy-momentum relation is a fundamental equation in relativistic mechanics. It combines aspects of both energy and momentum for particles moving close to the speed of light. The relation is formulated as \( E^2 = (pc)^2 + (mc^2)^2 \), where \( E \) is the total energy, \( p \) is the momentum, and \( mc^2 \) is the rest energy of the particle.
This relation helps in understanding how energy is stored and transferred within a relativistic system. In the given problem, it supports deriving the relativistic momentum \( p \), which is critical in determining the de Broglie wavelength of the particle.
The de Broglie wavelength represents the wave-like nature of matter, especially noticeable in particles with small mass. Using the relativistic momentum derived from the energy-momentum equation, we find that the de Broglie wavelength \( \lambda \) is calculated as \( \lambda = \frac{h}{p} \), where \( h \) is Planck's constant. This demonstrates the centrality of the energy-momentum relation in solving problems involving relativistic particles.