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(a) For what kinetic energy of a neutron will the associated de Broglie wavelength be \(1.40 \times 10^{-10} \mathrm{~m}\) ? (b) Also find the de Broglie wavelength of a neutron, in thermal equilibrium with matter, having an average kinetic energy of \((3 / 2) k T\) at \(300 \mathrm{~K}\).

Short Answer

Expert verified
(a) Kinetic energy is approximately \(1.473 \times 10^{-20} \text{ J}\). (b) De Broglie wavelength is approximately \(1.45 \times 10^{-10} \text{ m}\).

Step by step solution

01

Introduction to de Broglie Wavelength

The de Broglie wavelength \( \lambda \) is given by the formula \( \lambda = \frac{h}{p} \), where \( h \) is Planck's constant and \( p \) is the momentum of the neutron. The momentum \( p \) relates to kinetic energy \( K \) via \( p = \sqrt{2mK} \), where \( m \) is the mass of the neutron.
02

Solving for Kinetic Energy (Part a)

We start with the de Broglie wavelength equation \( \lambda = \frac{h}{\sqrt{2mK}} \). Rearranging for \( K \), we find \( K = \frac{h^2}{2m\lambda^2} \). For a neutron, \( m = 1.675 \times 10^{-27} \text{ kg} \) and \( h = 6.626 \times 10^{-34} \text{ J s} \). Substituting \( \lambda = 1.40 \times 10^{-10} \text{ m} \), we calculate \( K \).
03

Calculate K with Given Values

Substitute into the kinetic energy formula: \[ K = \frac{(6.626 \times 10^{-34})^2}{2 \times 1.675 \times 10^{-27} \times (1.40 \times 10^{-10})^2} \]Calculating this gives the kinetic energy \( K \approx 1.473 \times 10^{-20} \text{ J} \).
04

Understanding Average Kinetic Energy at Thermal Equilibrium (Part b)

When a neutron is in thermal equilibrium at temperature \( T \), the average kinetic energy \( \langle K \rangle = \frac{3}{2} k T \), where \( k = 1.38 \times 10^{-23} \text{ J/K} \) is Boltzmann's constant. At \( T = 300 \text{ K} \), we calculate \( \langle K \rangle \).
05

Calculate Average Kinetic Energy

Using \( T = 300 \text{ K} \), compute:\[ \langle K \rangle = \frac{3}{2} \times 1.38 \times 10^{-23} \times 300 \]This gives \( \langle K \rangle \approx 6.21 \times 10^{-21} \text{ J} \).
06

Finding the De Broglie Wavelength (Part b)

Having found \( \langle K \rangle \), we find \( \lambda \) using \( \lambda = \frac{h}{p} = \frac{h}{\sqrt{2m\langle K \rangle}} \). Substituting for \( \langle K \rangle \), calculate:\[ \lambda = \frac{6.626 \times 10^{-34}}{\sqrt{2 \times 1.675 \times 10^{-27} \times 6.21 \times 10^{-21}}} \]Computing gives \( \lambda \approx 1.45 \times 10^{-10} \text{ m} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinetic Energy of Neutron
The kinetic energy of a neutron is pivotal when assessing its behavior and properties. Kinetic energy, denoted as \( K \), is defined as the energy that a neutron possesses due to its motion. For a neutron, whose mass is \( 1.675 \times 10^{-27} \space \text{kg} \), the kinetic energy is calculated using the classical formula for non-relativistic particles: \[ K = \frac{1}{2} mv^2 \]However, when dealing with subatomic particles like neutrons, we often relate kinetic energy to momentum via:\[ K = \frac{p^2}{2m} \]where \( p \) is momentum. This connection is crucial when applying the de Broglie wavelength as it merges quantum mechanics with classical mechanics, allowing us to compute the kinetic energy based on wavelength measurements. For example, if the de Broglie wavelength \( \lambda \) of a neutron is known, its kinetic energy can be derived from:\[ K = \frac{h^2}{2m\lambda^2} \]This demonstrates the direct influence of the neutron’s wavelength on its energy, integrating wave-like characteristics with particle-like motion.
Thermal Equilibrium
Thermal equilibrium is a state where a particle, such as a neutron, achieves a consistent temperature throughout its environment. In this state, the neutron exhibits average kinetic energy synonymous with the temperature \( T \) of its surrounding. This concept is essential in statistical mechanics and thermodynamics, where the distribution of energy among particles reaches a balance. In thermal equilibrium, each particle’s average kinetic energy is given by the expression:\[ \langle K \rangle = \frac{3}{2} kT \]Here, \( k \) is Boltzmann's constant, and that provides a significant link between thermodynamics and kinetic theory. For neutrons at room temperature (about 300 K), this formula helps us approximate how much energy they generally have, which in turn affects their de Broglie wavelength. The equilibrium indicates a stable state with no net exchange of energy, allowing calculations to predict a particle's behavior within a thermal environment.
Planck's Constant
Planck's constant is a cornerstone of quantum mechanics, symbolized by \( h \). It characterizes the fundamental scale of quantum effects. Planck's constant appears in the equation for the de Broglie wavelength:\[ \lambda = \frac{h}{p} \] where \( \lambda \) is wavelength and \( p \) is momentum. This relationship forms the basis of wave-particle duality, exemplifying how particles like neutrons behave both as particles and waves. Planck's constant has a value of \( 6.626 \times 10^{-34} \text{ Js} \) and significantly impacts the calculations of energy, momentum, and wavelength interactions at microscopic scales.In this context, it allows for calculating a neutron's kinetic energy if its de Broglie wavelength is known. Planck's constant connects the energy of photons (in wave-like behavior) to the frequency of their waves, further demonstrating the depth and reach of quantum mechanics.
Boltzmann's Constant
Boltzmann's constant, \( k \), is a fundamental constant that enables the link between macroscopic and microscopic physical quantities. Its value is \( 1.38 \times 10^{-23} \text{ J/K} \). In the formula for average kinetic energy in thermal equilibrium:\[ \langle K \rangle = \frac{3}{2} kT \] Boltzmann's constant helps translate the temperature of a system into the energy levels of individual particles, such as neutrons. This is crucial for understanding the energy distribution in gases or materials at a given temperature. By using Boltzmann's constant, we relate the large-scale observables, like temperature, to the energetic motions of individual particles. It is a key player in the field of statistical mechanics, showing how ensemble averages work and how temperature can pertain to kinetic motion on a detailed level. This understanding aids in computations related to thermal properties and behaviors at the quantum level.

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Most popular questions from this chapter

A 100W sodium lamp radiates energy uniformly in all directions. The lamp is located at the centre of a large sphere that absorbs all the sodium light which is incident on it. The wavelength of the sodium light is \(589 \mathrm{~nm}\). (a) What is the energy per photon associated with the sodium light? (b) At what rate are the photons delivered to the sphere?

Monochromatic light of wavelength \(632.8 \mathrm{~nm}\) is produced by a helium-neon laser. The power emitted is \(9.42 \mathrm{~mW}\). (a) Find the energy and momentum of each photon in the light beam, (b) How many photons per second, on the average, arrive at a target irradiated by this beam? (Assume the beam to have uniform cross-section which is less than the target area), and (c) How fast does a hydrogen atom have to travel in order to have the same momentum as that of the photon?

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