• Tags: comparison, photon-electron, wavelength
• Difficulty: Advanced

Same energy E: Photon λ_p = hc/E. Electron λ_e = h/√(2mE). Ratio: λ_p/λ_e = c√(2m/E)/1 = c√(2m)/√E. For E = 1 eV: λ_p = 1240 nm, λ_e = 1.227 nm. The photon wavelength is ~1000× larger. This ratio increases with decreasing energy.

Same momentum p: Both have the same de Broglie wavelength λ = h/p. However, their energies differ: E_photon = pc >> E_electron = $p^{2}$/(2m) for non-relativistic electrons. The photon has much more energy than the electron at the same momentum.

Same wavelength λ: Both have the same momentum p = h/λ. The photon energy E_p = hc/λ; electron energy E_e = $h^{2}$/(2mλ^{2}). Ratio E_p/E_e = 2$mc^{2}$λ/h = 2mcλ/h. For λ = 1 Å: E_p ≈ 12.4 keV, E_e ≈ 150 eV. The photon has ~80× more energy.