/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 33 Calculate What is the kinetic en... [FREE SOLUTION] | 91Ó°ÊÓ

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Calculate What is the kinetic energy of a neutron whose de Broglie wavelength is \(3.1 \times 10^{-11} \mathrm{~m}\) ? Give your answer in electron volts (eV).

Short Answer

Expert verified
The kinetic energy of the neutron is approximately 0.085 eV.

Step by step solution

01

Understanding the Relationship

The de Broglie wavelength \( \lambda \) is related to the momentum \( p \) of the particle by the equation: \[ \lambda = \frac{h}{p} \] where \( h \) is the Planck constant (\( 6.626 \times 10^{-34} \mathrm{~m}^2 \mathrm{~kg} / \mathrm{s} \)). We need to find the momentum using this equation.
02

Calculate Momentum

Rearrange the de Broglie wavelength equation to solve for the momentum \( p \): \[ p = \frac{h}{\lambda} \] Substitute the given wavelength \( \lambda = 3.1 \times 10^{-11} \mathrm{~m} \) into the formula to find \( p \): \[ p = \frac{6.626 \times 10^{-34}}{3.1 \times 10^{-11}} = 2.137 \times 10^{-23} \mathrm{~kg} \cdot \mathrm{m/s} \]
03

Relate Momentum to Kinetic Energy

The kinetic energy \( K \) of a particle can be expressed in terms of its momentum \( p \) by the equation: \[ K = \frac{p^2}{2m} \] where \( m \) is the mass of the neutron (\( 1.675 \times 10^{-27} \mathrm{~kg} \)).
04

Calculate Kinetic Energy in Joules

Using the momentum value, calculate the kinetic energy: \[ K = \frac{(2.137 \times 10^{-23})^2}{2 \times 1.675 \times 10^{-27}} = 1.36 \times 10^{-20} \mathrm{~J} \]
05

Convert Energy to Electron Volts

To convert the energy from joules to electron volts, use the conversion factor: \[ 1 \mathrm{~eV} = 1.602 \times 10^{-19} \mathrm{~J} \]. Thus, the kinetic energy in electron volts is: \[ K = \frac{1.36 \times 10^{-20}}{1.602 \times 10^{-19}} \approx 0.085 \mathrm{~eV} \]

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Key Concepts

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

Kinetic Energy Calculation
Kinetic energy is the energy that a particle has due to its motion. Calculating kinetic energy is fundamental in understanding how a particle, such as a neutron, behaves under given conditions. In physics, the kinetic energy (\[ K \] ) of a particle is given by the formula: \[ K = \frac{1}{2} m v^2 \] , where \( m \) is the mass and \( v \) is the velocity of the particle. However, when the relationship with momentum is needed, we often use the formula: \[ K = \frac{p^2}{2m} \] . Here, \( p \) represents the momentum, and \( m \) is still the mass of the particle. Given that momentum can be derived from the de Broglie wavelength, we can substitute the momentum value into this kinetic energy equation. By finding the momentum first, we then proceed to calculate the kinetic energy, ensuring accurate results.
Momentum
Momentum is a key concept in physics that describes the amount of motion a particle has. It is the product of the mass and velocity of an object, represented by the symbol \( p \). In the context of wave-like particles such as neutrons, momentum can also be related to the de Broglie wavelength using the formula: \[ p = \frac{h}{\lambda} \] , where \( h \) is the Planck constant and \( \lambda \) is the de Broglie wavelength. Once we know the wavelength, calculating momentum becomes straightforward. For a neutron with a wavelength of \( 3.1 \times 10^{-11} \) meters, we substitute this value into the formula to find its momentum. This approach is a bridge to further determining the kinetic energy linked with the momentum of waving particles.
Neutron Mass
Neutrons are neutral subatomic particles with a specific mass crucial for calculations in nuclear and particle physics. The mass of a neutron is approximately \( 1.675 \times 10^{-27} \) kg. Neutron mass is an integral part of many quantum mechanics equations involving energy and momentum. In calculations like those of kinetic energy, knowing the precise mass of the neutron helps accurately determine the energy values. Especially in particles governed by quantum mechanics, such as neutrons, the mass factor becomes essential for high-precision calculations, linking experiments to theoretical predictions.
Planck Constant
The Planck constant \( h \) is a fundamental constant in quantum mechanics that relates the energy of a photon to the frequency of its associated electromagnetic wave. Its value is approximately \( 6.626 \times 10^{-34} \text{ m}^2 \text{ kg/s} \). This constant is not only pivotal in the realm of photons but also in particles that exhibit wave-particle duality, such as electrons and neutrons. Using the Planck constant, we derive essential quantities like momentum when the de Broglie wavelength is provided. Therefore, the Planck constant is a bridge between the micro-world of atoms and the macro-world of classical physics. Consistently knowing its precise value is crucial for conducting and validating experiments in various fields, from optics to quantum theory.

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Most popular questions from this chapter

Summarize Give a brief description of the ultraviolet catastrophe.

An electron and a proton are confined to boxes of the same size. Is the uncertainty in momentum of the electron greater than, less than, or equal to the uncertainty in momentum of the proton? Explain.

Calculate What is the frequency of (a) a photon with an energy of \(4.2 \times 10^{-19} \mathrm{~J}\) and (b) a photon with an energy of \(1.9 \mathrm{eV}\) ?

Think \& Calculate A typical lightbulb contains a tungsten filament that reaches a temperature of about \(2850 \mathrm{~K}\), roughly half the surface temperature of the Sun. (a) Treating the filament as a blackbody, determine the frequency for which its radiation is a maximum. (b) Do you expect the lightbulb to radiate more energy in the visible or in the infrared part of the spectrum? Explain.

Predict \& Explain A source of yellow light has a higher power than a source of blue light. (a) Is the number of photons emitted per second by the yellow source greater than, less than, or equal to the number of photons emitted per second by the blue source? (b) Choose the best explanation from among the following: A. The yellow source emits more photons per second, because (1) it emits more energy per second than the blue source and (2) its photons have less energy than those from the blue source. B. The yellow source has the higher power, which means its photons have higher energy than the photons from the blue source. Therefore, the yellow source emits fewer photons per second. C. The two sources emit the same number of photons per second because the higher power of the yellow source compensates for the higher energy of the blue photons.

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