If light, long thought to be a wave, also behaves as particles (photons), could matter — long thought to be particles — also behave as waves? In 1924 Louis de Broglie made exactly this bold proposal. Nature, he argued, is symmetric: every moving particle has a wave associated with it, called a matter wave or de Broglie wave. This idea of wave-particle duality says that both light and matter show wave-like and particle-like character, with whichever aspect dominating depending on the experiment.
Starting from the photon relation $p = \dfrac{h}{\lambda}$, de Broglie proposed the same relation for any particle of momentum $p$. The de Broglie wavelength is $\lambda = \dfrac{h}{p} = \dfrac{h}{mv}$, where $m$ is the mass and $v$ the speed of the particle. The wavelength is inversely proportional to momentum: heavy, fast objects have astronomically tiny wavelengths (a moving cricket ball has $\lambda \sim 10^{-34}\,\text{m}$, far too small to detect), whereas light, slow particles such as electrons have measurable wavelengths.
For a particle of kinetic energy $K = \dfrac{1}{2}mv^2 = \dfrac{p^2}{2m}$, the momentum is $p = \sqrt{2mK}$, so the wavelength can be written as $\lambda = \dfrac{h}{\sqrt{2mK}}$. Notice that the de Broglie wavelength does not depend on the charge of the particle, only on its mass and kinetic energy.
A very common case is an electron accelerated through a potential difference $V$. The electron gains kinetic energy $K = eV$, so its momentum is $p = \sqrt{2meV}$ and $\lambda = \dfrac{h}{\sqrt{2meV}}$. Putting in the numbers for an electron gives the handy formula $\lambda = \dfrac{12.27}{\sqrt{V}}\,\text{\u00c5}$, with $V$ in volts. For example, electrons accelerated through $100\,\text{V}$ have $\lambda \approx 1.227\,\text{\u00c5}$ — comparable to atomic spacings, which is why electron beams diffract through crystals.
The reality of matter waves was confirmed by the Davisson-Germer experiment (1927). Electrons from a heated filament were accelerated through a known voltage and directed at a single crystal of nickel. Instead of scattering uniformly, the electrons showed a strong peak in intensity at a particular scattering angle, exactly as waves do when they diffract off the regular planes of atoms in the crystal. The wavelength worked out from the diffraction angle agreed closely with the de Broglie prediction $\lambda = \dfrac{12.27}{\sqrt{V}}\,\text{\u00c5}$, proving that electrons really do behave as waves. This duality is one of the foundations of modern quantum mechanics.
Find the de Broglie wavelength of an electron moving with a speed of $2\times10^{6}\,\text{m/s}$. (Take $m_e = 9.1\times10^{-31}\,\text{kg}$, $h = 6.63\times10^{-34}\,\text{J s}$.)
Solution- de Broglie wavelength: $\lambda = \dfrac{h}{mv}$.
- $mv = (9.1\times10^{-31})(2\times10^{6}) = 1.82\times10^{-24}\,\text{kg m/s}$.
- $\lambda = \dfrac{6.63\times10^{-34}}{1.82\times10^{-24}}$.
- $\lambda = 3.64\times10^{-10}\,\text{m} = 3.64\,\text{\u00c5}$.
Answer: The de Broglie wavelength of the electron is about $3.64\,\text{\u00c5}$.
An electron is accelerated through a potential difference of $100\,\text{V}$. Find its de Broglie wavelength.
Solution- For an accelerated electron: $\lambda = \dfrac{12.27}{\sqrt{V}}\,\text{\u00c5}$.
- Here $V = 100\,\text{V}$, so $\sqrt{V} = 10$.
- $\lambda = \dfrac{12.27}{10}$.
- $\lambda = 1.227\,\text{\u00c5}$.
Answer: The de Broglie wavelength is about $1.227\,\text{\u00c5}$, comparable to atomic spacings.
Calculate the de Broglie wavelength of a particle of kinetic energy $K$ in terms of $h$, $m$ and $K$, and use it to find the wavelength of a proton of kinetic energy $1.0\,\text{keV}$. (Take $m_p = 1.67\times10^{-27}\,\text{kg}$.)
Solution- $K = \dfrac{p^2}{2m}$, so $p = \sqrt{2mK}$ and $\lambda = \dfrac{h}{\sqrt{2mK}}$.
- $K = 1.0\,\text{keV} = 1.0\times10^{3}\times1.6\times10^{-19} = 1.6\times10^{-16}\,\text{J}$.
- $2mK = 2(1.67\times10^{-27})(1.6\times10^{-16}) = 5.34\times10^{-43}$; $\sqrt{2mK} = 7.31\times10^{-22}$.
- $\lambda = \dfrac{6.63\times10^{-34}}{7.31\times10^{-22}} = 9.1\times10^{-13}\,\text{m}$.
Answer: $\lambda = \dfrac{h}{\sqrt{2mK}}$, giving about $9.1\times10^{-13}\,\text{m}$ for the proton.
Why is the wave nature of a moving cricket ball ($m = 0.15\,\text{kg}$, $v = 30\,\text{m/s}$) never observed? Support with a calculation.
Solution- $\lambda = \dfrac{h}{mv} = \dfrac{6.63\times10^{-34}}{(0.15)(30)}$.
- $mv = 4.5\,\text{kg m/s}$.
- $\lambda = \dfrac{6.63\times10^{-34}}{4.5} = 1.47\times10^{-34}\,\text{m}$.
- This is far smaller than any atomic dimension, so no diffraction or interference can ever be seen.
Answer: The wavelength ($\sim 1.5\times10^{-34}\,\text{m}$) is unimaginably small, so the ball's wave nature is completely unobservable.
An electron and a proton have the same de Broglie wavelength. Which one has the greater speed, and why?
Solution- Same wavelength means same momentum: $\lambda = \dfrac{h}{mv}$ gives $m_e v_e = m_p v_p$.
- So $\dfrac{v_e}{v_p} = \dfrac{m_p}{m_e}$.
- Since $m_p \gg m_e$ (about $1836$ times heavier), $v_e \gg v_p$.
Answer: The electron has the greater speed, because for equal momentum the lighter particle must move faster.
In the Davisson-Germer experiment, electrons accelerated through $54\,\text{V}$ showed a diffraction peak. Find the de Broglie wavelength predicted by theory.
Solution- For an accelerated electron: $\lambda = \dfrac{12.27}{\sqrt{V}}\,\text{\u00c5}$.
- $V = 54\,\text{V}$, so $\sqrt{54} = 7.35$.
- $\lambda = \dfrac{12.27}{7.35}$.
- $\lambda = 1.67\,\text{\u00c5}$.
Answer: The predicted wavelength is about $1.67\,\text{\u00c5}$, which matched the experimental diffraction result, confirming matter waves.