E.2.6 (HL)—de Broglie wavelength

Use the de Broglie relation

Every moving particle has a wavelength inversely proportional to its momentum $p$. For non-relativistic motion, momentum may be written $mv$; choose the momentum relationship supported by the given data.

Worked example — moving electron

For $m_e=9.11\times10^{-31}\,\mathrm{kg}$ and $v=5.0\times10^6\,\mathrm{m\,s^{-1}}$, $p=mv=4.56\times10^{-24}\,\mathrm{kg\,m\,s^{-1}}$. Hence $\lambda=h/p=(6.63\times10^{-34})/(4.56\times10^{-24})=1.5\times10^{-10}\,\mathrm{m}$, comparable with atomic spacing.

Choose the momentum form

For non-relativistic motion, use $p=mv$, so $\lambda=h/(mv)$. If kinetic energy is given, use $E_k=p^2/(2m)$ and $p=\sqrt{2mE_k}$.

Scale with accelerating voltage

For an electron accelerated from rest through potential $V$, $eV=E_k$, so $p\propto\sqrt V$ and $\lambda\propto1/\sqrt V$. Quadrupling $V$ halves the wavelength.

Common trap

Do not use $h/E_k$ as the wavelength. The denominator is momentum, not kinetic energy.