Dual Nature of Matter and Light

Electromagnetic radiation and Planck's quantum theory. Light travels as an electromagnetic wave of wavelength $\lambda$ and frequency $\nu$, related to its speed by $c = \nu\lambda$, with $c = 3\times10^{8}\ \text{m s}^{-1}$ in vacuum. The electromagnetic spectrum orders radiation from radio waves (long wavelength, low energy) through microwaves, infrared, visible light, ultraviolet, X-rays, to gamma rays (short wavelength, high energy). Max Planck's decisive break with classical physics was to propose that a body emits or absorbs energy not continuously but in discrete packets, quanta — for light, photons — each carrying energy proportional to frequency:

where: $h = 6.626\times10^{-34}\ \text{J s}$ is Planck's constant, $\nu$ is frequency, and $\lambda$ is wavelength. This single idea explained black-body radiation and, soon after, the photoelectric effect — both problems where classical wave theory failed outright.

The photoelectric effect. Shine light of frequency above a metal-specific threshold frequency $\nu_0$ onto a metal surface and electrons are ejected immediately, called photoelectrons. Three experimental facts pin down the physics: below $\nu_0$ no electrons emerge, however intense the light; above $\nu_0$ the electrons' kinetic energy rises with frequency but is independent of intensity; and above $\nu_0$, greater intensity only increases how many electrons are ejected, not their individual energy. Wave theory predicts none of this — a bright enough wave of any frequency should eventually free electrons, which experiment flatly contradicts. Einstein resolved it with Planck's quantum idea: one photon of energy $h\nu$ is absorbed by one electron; part of that energy, the work function $h\nu_0$, frees the electron from the metal, and the remainder becomes its kinetic energy:

where: $KE_{max}$ is the maximum kinetic energy of an ejected electron and $h\nu_0$ (often written $\phi_0$) is the work function, the minimum energy needed to remove an electron from that particular metal.

Dual nature of matter: the de Broglie relation. If light, ordinarily a wave, can behave as particles (photons), de Broglie reasoned by symmetry that matter, ordinarily particles, should carry an associated wave character. For a particle of mass $m$ and velocity $v$ (momentum $p = mv$), the wavelength associated with it is:

where: $\lambda$ is the de Broglie wavelength and $p$ is momentum. This wavelength is only large enough to matter for very light particles such as electrons; for everyday objects it is far too small to ever observe, which is why footballs and cars show no wave behaviour.

Question. Calculate the de Broglie wavelength of an electron ($m = 9.1\times10^{-31}\ \text{kg}$) moving with velocity $2.0\times10^{6}\ \text{m s}^{-1}$, and compare it with the threshold-frequency photoelectric problem where a metal of threshold frequency $5.0\times10^{14}\ \text{Hz}$ is struck by light of frequency $8.0\times10^{14}\ \text{Hz}$.

Step 1. Find the electron's momentum: $mv = (9.1\times10^{-31})(2.0\times10^{6}) = 1.82\times10^{-24}\ \text{kg m s}^{-1}$.

Step 2. Apply $\lambda = h/mv$: $\lambda = \dfrac{6.626\times10^{-34}}{1.82\times10^{-24}} \approx 3.64\times10^{-10}\ \text{m} = 0.364\ \text{nm}$. This is comparable to the atomic spacing in a crystal, which is exactly why fast-electron beams show measurable diffraction — the experimental confirmation of de Broglie's hypothesis.

Step 3. For the companion photoelectric case, apply $KE_{max} = h(\nu - \nu_0)$: $KE_{max} = (6.626\times10^{-34})(8.0\times10^{14} - 5.0\times10^{14}) = 6.626\times10^{-34} \times 3.0\times10^{14} = 1.99\times10^{-19}\ \text{J}$, about $1.24\ \text{eV}$.

Result: the electron's de Broglie wavelength is $\mathbf{0.364\ nm}$, and the photoelectron's maximum kinetic energy is $\mathbf{1.99\times10^{-19}\ J\ (1.24\ eV)}$ — both independent of the light's intensity, since intensity only changes the number of photons or electrons involved, not the energy of a single one.

Competitive stretch

If the same electron's speed were doubled to $4.0\times10^{6}\ \text{m s}^{-1}$, momentum doubles to $3.64\times10^{-24}\ \text{kg m s}^{-1}$, so $\lambda$ halves to $0.182\ \text{nm}$ — de Broglie wavelength is inversely proportional to speed, a quick check worth remembering for ranking questions.

• Expecting a brighter light below the threshold frequency to eventually eject electrons — intensity is irrelevant below $\nu_0$; only raising the frequency past $\nu_0$ starts ejection.
• Swapping the terms in the photoelectric equation as $KE_{max} = h(\nu_0 - \nu)$, which gives a negative, nonsensical answer whenever $\nu > \nu_0$.
• Expecting a "visible" de Broglie wavelength for everyday objects — for macroscopic masses $\lambda$ works out far below $10^{-30}\ \text{m}$, utterly unobservable; the relation is significant only for sub-atomic particles like electrons.
• Using velocity directly as momentum in $\lambda = h/p$ — remember $p = mv$, so both mass and velocity must be multiplied together first.

Ask the tutor: want a walkthrough of why intensity and frequency play such different roles in the photoelectric effect, or help ranking de Broglie wavelengths for different particles?