A-Level Physics

Quantum physics

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Electromagnetic radiation is quantised into photons with energy E = hf = \(\dfrac{hc}{\lambda }\). Photon energies are often expressed in electronvolts, with 1 eV = 1.60×10⁻¹⁹ J. The photoelectric effect provides evidence for quantisation: a single electron absorbs energy from a single photon. Emission occurs only when photon energy is at least the work function φ, giving a threshold frequency f₀ = \(\dfrac{\phi }{h}\). The maximum photoelectron kinetic energy is \(K_{\text{max}}\) = hf − φ.

Matter also shows wave behaviour. A particle with momentum p has de Broglie wavelength λ = \(\dfrac{h}{p}\). Atomic energy levels are discrete, so transitions between them emit or absorb photons with hf equal to the energy difference. This produces line spectra rather than a continuous set of wavelengths.

Photons and quantised energy

Electromagnetic radiation exchanges energy in discrete packets called photons. Photon energy is E = hf = \(\dfrac{hc}{\lambda }\). Higher frequency means greater photon energy. Intensity and photon energy are different ideas: increasing intensity at fixed frequency means more photons arriving per second, not more energy per photon.

The photoelectric effect

Photoelectrons are emitted only when photon frequency exceeds a threshold frequency. One photon transfers its energy to one electron in the simple model. The maximum kinetic energy is given by hf = φ + \(KE_{\text{max}}\), where φ is the work function.

Increasing intensity above threshold increases the rate of photoelectron emission, but does not increase \(KE_{\text{max}}\) if frequency is unchanged. Increasing frequency increases maximum kinetic energy because each photon carries more energy. These observations cannot be explained by a purely classical continuous-wave transfer model.

Stopping potential

A reverse potential can prevent even the fastest photoelectrons reaching the collector. The stopping potential satisfies \(eV_{s}\) = \(KE_{\text{max}}\). A graph of stopping potential against frequency can therefore be used to determine Planck's constant and the work function through its gradient and intercept.

Wave-particle duality

Particles such as electrons show wave behaviour. The de Broglie wavelength is λ = \(\dfrac{h}{p}\). Increasing momentum decreases wavelength. Electron diffraction provides evidence for matter waves because the diffraction pattern depends on electron wavelength.

Wave and particle descriptions are complementary models used to predict different observations. Do not imagine an electron literally switching between being a small classical ball and a classical water wave.

Atomic energy levels and spectra

Electrons in atoms occupy discrete energy levels. Absorption occurs when a photon supplies exactly the energy difference between levels. Emission occurs when an electron drops to a lower level and releases a photon with hf = ΔE. Line spectra therefore provide evidence for quantised atomic energies.

Probability and interpretation

At school level, quantum physics predicts probabilities for measurement outcomes rather than definite classical trajectories. The success of the model comes from quantitative agreement with experiments such as photoelectric emission, electron diffraction and atomic spectra.

Exam technique

Separate photon energy from intensity. In photoelectric questions, check threshold frequency before calculating kinetic energy. In de Broglie problems, calculate momentum carefully and use SI units. For spectra, use energy differences, not the absolute value of one level alone.

Excitation, ionisation and the fluorescent tube

Excitation raises an electron to a higher energy level while it stays bound to the atom. Ionisation removes the electron completely; the ionisation energy is the energy needed to take it from the ground state to the zero level at the top of the energy-level diagram. Energy levels are usually quoted as negative values in eV.

In a fluorescent tube, electrons collide with mercury atoms and excite them. The atoms de-excite by emitting ultraviolet photons, which are absorbed by a phosphor coating; the phosphor then emits visible light as its electrons fall back through several smaller energy steps.

Worked example

Worked example 1: For f = 1.5×10¹⁵ Hz, E = hf = 6.63×10⁻³⁴ × 1.5×10¹⁵ = 9.945×10⁻¹⁹ J.

Worked example 2: For λ = 400 nm, E = \(\dfrac{hc}{\lambda }\) = \(\dfrac{6.63 \times 10^{-34} \times 3.00 \times 10^{8}}{4.00 \times 10^{-7}}\) = 4.97 × 10⁻¹⁹ J, about 3.10 eV.

Misconception check

Increasing light intensity below the threshold frequency does not release photoelectrons. It increases the number of low-energy photons, not the energy of each photon.

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State the photon-energy equation.

E = hf = \(\dfrac{hc}{\lambda }\).

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