A beam of neutrons has the same de Broglie wavelength as a...

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 $\left.E^{2}=E_{0}^{2}+(p c)^{2} .\right]$

Key Concepts

Kinetic Energy of Massive Particles

The kinetic energy of a massive particle is defined as the total energy minus the rest mass energy. For a neutron, this becomes K = ?((pc)² + m²c?) ? mc². When comparing this kinetic energy to the energy of a photon (which is given solely by its momentum), one must take into account the extra rest mass energy term. This distinction is pivotal in understanding why, even if a photon and a neutron share the same de Broglie wavelength (and hence momentum magnitude), their energies differ fundamentally.

Relativistic Energy-Momentum Relation

For particles with nonzero rest mass, such as neutrons, the complete energy is given by the relativistic relation E² = E?² + (pc)², where E? is the rest mass energy and p is the momentum. This relation is crucial when considering high-speed particles or when comparing energy contributions from rest mass and momentum, highlighting that massive particles have an inherent energy term that photons do not have.

de Broglie Wavelength

The de Broglie wavelength is a fundamental concept that relates a particle’s momentum to its wavelength through the expression ? = h/p, where h is Planck’s constant. This idea, originally proposed for matter waves, shows that all particles, even massive ones like neutrons, can exhibit wave-like behavior, and it forms a cornerstone of quantum mechanics.

Photon Energy Relation

For photons, which are massless, the energy is directly related to the wavelength by the formula E = hc/?, where c is the speed of light. This linear relationship contrasts with the energy expressions for massive particles and is central to understanding how electromagnetic radiation carries energy.

Transcript

00:01 In the sector size.

00:01 We have electrons that are accelerated through a potential difference off 54 votes.

00:06 And in question eight, we have to find what is the liberally wavelength linda of the electorate.

00:14 Well, we know that the kinetic energy of electrons is equal to p squared, divided by two m.

00:24 So from here you get the p is equal to the square root of 2 a.m. times a kinetic energy, and we also have that the momentum off an election according to the dev really formula.

00:37 It's aged, divided by lunda.

00:41 So from here we gather, lunda is equal to the to age, divided by the square root of two times aimed m times.

00:50 A kinetic energy ke remember that the kinetic energy of electrons that are accelerated through a potential difference.

01:01 Delta v is the chart of the election e times out of e.

01:07 So london is equal to age, divided by the square root of two and e.

01:14 Delta v.

01:18 So this is 6.63 times 10 to the minus 34 year old second.

01:24 That's the punks constant, divided by the square root of two times the mass of the electorate which is 9.1 times 10 to the minus 31 kilograms times the charge, which is 1.6 times sent to the minors.

01:41 19 coombs times the potential difference.

01:46 Delta v, which is 54 votes.

01:51 So linda is equal to 1.67 times 10 to the minus 10 meters, which is the same a 0.167 millimeters.

02:09 Okay, so this is the answer to oppression...

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