Question

(a) Find the wavelength of a proton whose kinetic energy is equal to its interval energy. (b) The proton is usually regarded as being roughly of radius \(10^{-15} \mathrm{~m}\). Would this proton behave as a wave or as a particle?

Short answer

The precise values for the wavelength of a proton and its behaviour depend on the kinetic or interval energy E, which isn't provided in the question. Generally, if the calculated wavelength is smaller than the radius of the proton (\(10^{-15} m\)), the proton will behave more like a particle. If the calculated wavelength is larger, it will behave more like a wave.

Step-by-step solution

Calculate the wavelength of the proton

Consider that the kinetic energy (K) of the proton equals its interval energy (E). Thus, we have K=E. We know that the kinetic energy of a particle is given by the relation \(K=\frac{1}{2}mv^2\), where \(m\) is the mass and \(v\) is the velocity of the particle. The wavelength \(\lambda\) of the particle can be calculated using the de Broglie hypothesis, which states \(\lambda=\frac{h}{mv}\), where \(h\) is the Planck constant. Using these relations, we can write: \(\frac{1}{2}mv^2=\frac{h}{\lambda}\). Solving for \(\lambda\) gives \(\lambda= \frac{2h}{mv}\).

Continue calculations

We are still missing the velocity of the proton. We can get that from the kinetic energy formula, knowing that E equals the kinetic energy, so we have \(E = \frac{1}{2}mv^2\). Solving for \(v\) gives: \(v = \sqrt{\frac{2E}{m}}\).

Finalize the calculation of the wavelength

Substitute the velocity expression from step 2 into the wavelength formula obtained in step 1, giving: \(\lambda = \frac{2h}{m\sqrt{\frac{2E}{m}}}\). Simplification leads to \(\lambda = \frac{2h}{\sqrt{2mE}}\).

Determine the behavior of the proton

In order to answer the second part of the question, recall the principle of wave-particle duality which tells us that everything has both a particle and wave nature. If the wavelength of a proton is smaller than its characteristic dimension, it behaves more like a particle. If its wavelength is larger, it behaves more like a wave. So, compare the proton's radius (\(10^{-15} m\)) to the wavelength previously calculated. The behaviour of the proton depends on this comparison.

Key concepts

Wave-Particle Duality

Wave-particle duality is a fascinating concept in quantum physics. It suggests that every particle or quantum entity exhibits both wave and particle properties. This principle is fundamental to understanding the behavior of particles at microscopic scales.
For instance, light is commonly known as a wave, but it also behaves like a particle, termed as a photon, under certain conditions. Similarly, particles such as electrons and protons, which we often think of as solid and mass-bearing, can also exhibit wave-like characteristics. This dual nature is crucial when studying small particles in different scenarios.
In the given exercise, you must analyze whether a proton behaves more like a wave or a particle by comparing its de Broglie wavelength to its radius. When a proton's wavelength is larger than its size, wave properties become significant. Conversely, if the wavelength is smaller, it promotes a particle-like behavior. This comparison underscores the intricate dance between waves and particles that defines quantum mechanics.

Kinetic Energy

Kinetic energy is the energy that a particle possesses due to its motion. It is crucial in determining many properties of particles, especially in the context of wave-particle duality. The formula for kinetic energy in classical mechanics is given by:

• \(K = \frac{1}{2} mv^2\)

where \(m\) is the mass of the particle, and \(v\) is its velocity. This formula shows how a particle's kinetic energy depends on both its mass and the square of its speed.
In exploring wave-like behavior, a particle's kinetic energy ties into determining its wavelength via the de Broglie relation. Because kinetic energy relates directly to velocity, it helps in calculating properties like the wavelength, which emphasizes the particle's wave nature. Understanding this energy is vital for various applications in physics, from basic mechanics to complex quantum phenomena like wave-particle duality.

Planck Constant

The Planck constant, denoted as \(h\), is a fundamental constant in physics that relates the energy of a photon to its frequency:

• \(E = hf\)

This constant is central to the quantum world because it sets the scale of quantum effects. At its core, the Planck constant is what quantifies the discrete nature of energy levels in quantum mechanics.
When considering de Broglie wavelength in particle physics, the relation becomes:

• \(\lambda = \frac{h}{mv}\)

This links the wave and particle aspects of matter, using the Planck constant as a bridge between them. The constant's value is minuscule, which is why quantum effects are typically significant at atomic or subatomic scales, rather than in everyday experiences.
In the context of the exercise, the Planck constant enables the computation of a proton's wavelength, helping to decide whether the particle displays more characteristics of a wave or a particle.

Particle Physics

Particle physics is the branch of physics that studies the nature of particles at the smallest scales. This domain explores fundamental particles that make up the universe, such as electrons, protons, and neutrons. Understanding particle physics involves both experiments and theories that delve into how these particles interact with one another.
The given exercise touches on particle physics principles by examining the behavior of a proton. Protons, like other subatomic particles, are governed by the rules of quantum mechanics and display wave-particle duality.
The de Broglie wavelength and the resulting behavior prediction—whether more wave-like or particle-like—are typical tasks in particle physics. Through these, researchers gain insights into the foundational aspects of matter and uncover mysteries about the forces and particles governing the universe. Particle physics not only provides a deeper understanding of the cosmos but also plays a pivotal role in technological advancements.