Revise Quantum physics for IB Physics (E.2)

Shine ultraviolet light on a clean metal plate and electrons are ejected from its surface. These ejected electrons are called photoelectrons, and the phenomenon is the photoelectric effect. What made it revolutionary was not that electrons come off, but the precise WAY they come off — a way the wave model of light gets completely wrong.

Einstein resolved every one of these puzzles with a single bold idea: light itself is quantised into packets — photons — each carrying an energy fixed by the frequency. A photon is absorbed by one electron in an all-or-nothing event. This is why a below-threshold photon can never free an electron even in a bright beam: each electron only ever meets one photon at a time, and if that single photon is too weak, nothing happens.

The energy of a single photon depends only on the frequency $f$ of the light (or equivalently its wavelength $\lambda$, since $c = f\lambda$).

$E = hf = \frac{hc}{\lambda}$

To escape the metal, an electron must be supplied with at least a minimum amount of energy called the work function, $\phi$ — a property of the metal alone. The lowest-frequency photon that can just supply this energy defines the threshold frequency, through $\phi = hf_0$. When a photon of higher frequency arrives, it spends $\phi$ freeing the electron and hands the leftover energy to the electron as kinetic energy. The most weakly bound electrons keep the most, so the maximum kinetic energy of the photoelectrons is Einstein's photoelectric equation:

$E_{max} = hf - \phi$

At the quantum scale, the joule is an awkwardly large unit, so we use the electronvolt (eV): the energy an electron gains when accelerated through a potential difference of 1 volt. Since energy transferred is charge times voltage, $1\text{ eV} = e \times 1\text{ V} = 1.60\times10^{-19}$ J. To convert an energy from joules to electronvolts, divide by $1.60\times10^{-19}$; to go the other way, multiply.

How do we actually measure $E_{max}$? By making the electrons pay it back. In the experiment, a reverse potential difference is applied that pushes emitted electrons back toward the metal. As this voltage is increased, fewer electrons have enough energy to cross, until at the stopping potential $V_s$ even the fastest electron is just turned back and the current falls to zero. The work done against the field, $eV_s$, then equals the maximum kinetic energy:

E_{max} = eV_s

The photoelectric effect shows light behaving as particles, yet interference and diffraction show it behaving as a wave. Light is neither purely one nor the other — it has a dual nature, and which face it shows depends on the experiment. In 1924 Louis de Broglie made the daring proposal that this duality is symmetric: if waves can act like particles, then particles should act like waves. He assigned every particle of momentum $p = mv$ a wavelength:

$\lambda = \frac{h}{p} = \frac{h}{mv}$

De Broglie's idea would have stayed a curiosity without proof — and the proof is electron diffraction. When a beam of electrons is fired through a thin layer of graphite or a crystal, it does not simply make a single bright spot as particles would. Instead it produces concentric rings on a screen, exactly the diffraction pattern a wave makes when it passes through a regularly spaced lattice. Diffraction is a purely wave phenomenon, so the pattern is direct evidence that the electrons travel as waves with a definite wavelength. Measuring the ring spacing gives a wavelength that agrees precisely with $\lambda = h/p$, confirming de Broglie's relation.

If a particle is also a wave, it cannot be pinned to a single exact point — a wave is spread out. This intuition is captured by Werner Heisenberg's uncertainty principle: it is impossible to know both the position and the momentum of a particle with perfect precision at the same instant. The more precisely you fix one, the less precisely the other can be known. Written qualitatively, the product of the uncertainty in position $\Delta x$ and the uncertainty in momentum $\Delta p$ has a lower bound:

$\Delta x \, \Delta p \gtrsim \frac{h}{4\pi}$

In Paper 2 the marks are analytic: each is tied to a specific line of working — a method mark (M) or an answer mark (A) — and error-carried-forward (ECF) means a wrong number early on does not have to cost you the marks that follow. But that protection only exists if your method is written down. The photoelectric worked example above shows this in full: $E=hf$ earns M1, the photon energy converted to eV earns A1, applying $E_{max}=hf-\phi$ earns a second M1, and the final $1.31$ eV earns the last A1 — with ECF shielding the answer marks if an earlier number slips. Study how each mark is earned by a specific line, and write your own solutions the same way.

The photon and the matter wave are the doorway to the rest of quantum physics. Photon energy $E=hf$ reappears in atomic spectra and the emission and absorption of light by energy levels; the de Broglie wavelength underpins why atoms have quantised orbits and why electrons behave as standing waves; and the uncertainty principle sets the ground rules for the full quantum-mechanical description of confined particles. Master the habit here — identify whether the wave or the particle picture applies, keep your energy units consistent, show every line — and the quantum topics that follow become variations on a method you already own.