Ranking de Broglie wavelengths: electron, proton, photon

Problem

Three particles have the same kinetic energy:

• A: electron
• B: proton
• C: photon

Order the de Broglie wavelengths from smallest to largest. (Hint: for a photon $p = E/c$; for non-relativistic particles $p = \sqrt{2mE_K}$.)

Solution

Principle. In every case $\lambda = h/p$: whoever has the greater momentum has the smaller $\lambda$. For equal energy $E$:

• Electron and proton (massive, non-relativistic): $p = \sqrt{2 m E}$. For equal $E$, the momentum grows with $\sqrt{m}$. Since $m_p \approx 1836\,m_e$, the proton has much greater $p$, so $\lambda_B < \lambda_A$.
• Photon (massless): $p = E/c$. Comparing with the electron: $p_\text{fotone}/p_e = (E/c)/\sqrt{2 m_e E} = \sqrt{E/(2 m_e c^2)}$. For typical energies $E \ll m_e c^2 = 511\;\text{keV}$ this ratio is $\ll 1$: the photon has the smallest momentum, and hence the largest $\lambda$.

Ranking by momentum: $p_B > p_A > p_C$, so

$\ev{\lambda_C > \lambda_A > \lambda_B}$

The photon has the longest wavelength, the proton the shortest.

Links

Topics: Quantum physics Concepts: de Broglie wavelength · Photon Skills: Ranking reasoning