A photon and an electron have the same de Broglie wavelength. Choose the true statement.

Question

BlackTom AI · Accepted Answer

λ = h/p (the de Broglie relation) so equal de Broglie wavelengths imply equal momenta.

1) Relevant knowledge points and concepts
- de Broglie wavelength: λ = h/p, valid for both matter waves (electrons) and photons (treating photon momentum p = h/λ).
- Photon energy-momentum relation: E_photon = pc = hc/λ.
- Electron energy (nonrelativistic): KE = p^2/(2m) = h^2/(2mλ^2). Relativistic energy: E = sqrt(p^2c^2 + m^2c^4).
- Photon rest mass = 0; electron rest mass = m_e. Photon speed = c; electron speed v < c.

2) Step-by-step reasoning
- Start from λ = h/p. If a photon and an electron have the same λ, then by rearranging p = h/λ both have the same momentum p.
- Use photon relations: for that p, E_photon = pc = (h/λ)c.
- Use electron relations: for the same p, electron kinetic energy (nonrelativistic) = p^2/(2m) = h^2/(2mλ^2). Even relativistically, electron total energy ≠ pc because of the rest mass term.

3) Intermediate calculation examples
- Momentum equality: p_photon = h/λ, p_electron = h/λ => p_photon = p_electron.
- Photon energy: E_photon = pc = (h/λ)c.
- Electron KE (NR): KE_e = p^2/(2m) = (h^2)/(2mλ^2). These two expressions are manifestly different for finite m.

4) Why the correct answer is correct
- Because λ = h/p directly ties wavelength to momentum for all particles/waves, same λ forces same p. Thus option (b) is correct.

5) Why the other options are incorrect
- (a) Same energy: false — E_photon = pc while electron energy depends on p and m; these are not equal in general.
- (c) Same mass: false — photon rest mass = 0, electron mass = m_e ≠ 0.
- (d) Same velocity: false — photon speed = c, electron speed = p/m (or relativistic v < c); they are not equal.

6) Conclusion
Equal de Broglie wavelength implies equal momentum, so the correct answer is (b), matching the provided answer.

A photon and an electron have the same de Broglie wavelength. Choose the true statement.单项选择题

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