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

How many electron states of the H atom have the quantum numbers \(n=3\) and \(\ell=1 ?\) Identify each state by listing its quantum numbers.
In the Davisson-Germer experiment (Section \(28.2),\) the electrons were accelerated through a \(54.0-\mathrm{V}\) potential difference before striking the target. (a) Find the de Broglie wavelength of the electrons. (b) Bragg plane spacings for nickel were known at the time; they had been determined through x-ray diffraction studies. The largest plane spacing (which gives the largest intensity diffraction maxima) in nickel is \(0.091 \mathrm{nm} .\) Using Bragg's law [Eq. ( \(25-15\) )], find the Bragg angle for the first-order maximum using the de Broglie wavelength of the electrons. (c) Does this agree with the observed maximum at a scattering angle of \(130^{\circ} ?\) [Hint: The scattering angle and the Bragg angle are not the same. Make a sketch to show the relationship between the two angles.]
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