/*! 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 74 A beam of neutrons has the same ... [FREE SOLUTION] | 91影视

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Chapter 28: Problem 74

A beam of neutrons has the same de Broglie wavelength as a beam of photons. Is it possible that the energy of each photon is equal to the kinetic energy of each neutron? If so, at what de Broglie wavelength(s) does this occur? [Hint: For the neutron, use the relativistic energy-momentum relation \(\left.E^{2}=E_{0}^{2}+(p c)^{2} .\right]\)

Short Answer

Expert verified
Answer: The energy of each photon equals the kinetic energy of each neutron when the de Broglie wavelength is equal to the rest energy of the neutron (\(\lambda = E_0\)).

Step by step solution

01

Understand the de Broglie wavelength

The de Broglie wavelength is a characteristic of a particle that is related to its momentum, given by the formula: \(\lambda=\dfrac{h}{p}\), where \(\lambda\) is the de Broglie wavelength, \(h\) is the Planck's constant, and \(p\) is the momentum of the particle.
02

Write the energy-momentum relation for photons

The energy-momentum relation for photons is given by \(E=pc\), where \(E\) is the energy, \(p\) is the momentum, and \(c\) is the speed of light.
03

Write the energy-momentum relation for neutrons

From the given hint, the energy-momentum relation for neutrons (relativistic) is \(E^{2}=E_{0}^{2}+(pc)^{2}\), where \(E\) is the total energy, \(E_0\) is the rest energy, and \(p\) is the momentum.
04

Relate the de Broglie wavelength to the momentum

Using the de Broglie wavelength formula \(\lambda=\dfrac{h}{p}\), we can express momentum \(p\) in terms of \(\lambda\). \(p=\dfrac{h}{\lambda}\)
05

Write the energy expressions for photons and neutrons in terms of de Broglie wavelength

Substitute the expression for momentum into the energy-momentum relations: For photons, \(E=\dfrac{hc}{\lambda}\). For neutrons, \(E^2=E_0^2 + \left(\dfrac{hc}{\lambda}\right)^2\)
06

Set the energy expressions equal and solve for the de Broglie wavelength

Equate the energies of the photons and neutrons: \(\dfrac{hc}{\lambda}=\sqrt{E_0^2 + \left(\dfrac{hc}{\lambda}\right)^2}\) Simplify and solve for \(\lambda\): \(\left(\dfrac{hc}{\lambda}\right)^2 = E_0^2 + \left(\dfrac{hc}{\lambda}\right)^2 - 2E_0\dfrac{hc}{\lambda}\) Divide by \(h^2c^2\), and rearrange to form a quadratic equation: \(\lambda^2 - 2E_0\lambda + E_0^2 = 0\)
07

Solve the quadratic equation for the wavelength(s)

Use the quadratic formula to solve for \(\lambda\): \(\lambda_\pm = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a} = \dfrac{2E_0\pm\sqrt{(2E_0)^2 - 4E_0^2}}{2}\) Finally, we find the possible de Broglie wavelengths: \(\lambda_\pm = E_0\). Since \(E_0\) is a positive constant (the rest energy of the neutron), there is only one possible de Broglie wavelength, \(E_0\), where the energy of each photon is equal to the kinetic energy of each neutron.

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Most popular questions from this chapter

An image of a biological sample is to have a resolution of \(5 \mathrm{nm} .\) (a) What is the kinetic energy of a beam of electrons with a de Broglie wavelength of \(5.0 \mathrm{nm} ?\) (b) Through what potential difference should the electrons be accelerated to have this wavelength? (c) Why not just use a light microscope with a wavelength of 5 nm to image the sample?

A nickel crystal is used as a diffraction grating for x-rays. Then the same crystal is used to diffract electrons. If the two diffraction patterns are identical, and the energy of each \(\mathrm{x}\) -ray photon is \(E=20.0 \mathrm{keV},\) what is the kinetic energy of each electron?
To resolve details of an object, you must use a wavelength that is about the same size, or smaller, than the details you want to observe. Suppose you want to study a molecule that is about \(1.000 \times 10^{-10} \mathrm{m}\) in length. (a) What minimum photon energy is required to study this molecule? (b) What is the minimum kinetic energy of electrons that could study this? (c) Through what potential difference should the electrons be accelerated to reach this energy?
An electron in a one-dimensional box has ground-state energy 0.010 eV. (a) What is the length of the box? (b) Sketch the wave functions for the lowest three energy states of the electron. (c) What is the wavelength of the electron in its second excited state \((n=3) ?\) (d) The electron is in its ground state when it absorbs a photon of wavelength \(15.5 \mu \mathrm{m}\). Find the wavelengths of the photon(s) that could be emitted by the electron subsequently.
An 81 -kg student who has just studied matter waves is concerned that he may be diffracted as he walks through a doorway that is \(81 \mathrm{cm}\) across and \(12 \mathrm{cm}\) thick. (a) If the wavelength of the student must be about the same size as the doorway to exhibit diffraction, what is the fastest the student can walk through the doorway to exhibit diffraction? (b) At this speed, how long would it take the student to walk through the doorway?
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