Flashcards · 9702 · A Level Topic 22

Quantum physics (A Level) flashcards

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Photon: A quantum (discrete packet) of electromagnetic energy. Electromagnetic radiation has a particulate nature.
Energy of a photon: E = hf = hc ÷ λ, where h = 6.63 × 10⁻³⁴ J s.
Momentum of a photon: p = E ÷ c = h ÷ λ.
Find the momentum of a photon of wavelength 500 nm.: p = h ÷ λ = 6.63 × 10⁻³⁴ ÷ 500 × 10⁻⁹ = 1.33 × 10⁻²⁷ kg m s⁻¹.
Find the wavelength of a photon of energy 2.0 eV.: E = 2.0 × 1.60 × 10⁻¹⁹ = 3.2 × 10⁻¹⁹ J. λ = hc ÷ E = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) ÷ 3.2 × 10⁻¹⁹ = 6.2 × 10⁻⁷ m (622 nm).

Work function and threshold frequency: Work function Φ: the minimum energy needed to remove an electron from the metal surface. Threshold frequency f₀: the minimum frequency for emission, Φ = hf₀.
Photoelectric equation: hf = Φ + Ek(max): the photon energy provides the work function and the maximum kinetic energy of the emitted electron.
Why does intense red light eject no electrons from a metal when faint blue light does?: One photon gives all its energy to one electron. Each red photon has less energy than Φ, and the energy is not accumulated, so more photons do not help.
Why is Ek(max) independent of intensity, while the photoelectric current is proportional to it?: Ek(max) depends on the energy of each photon, which depends only on frequency. Higher intensity means more photons per second, so more electrons per second (a larger current).
Φ = 2.0 eV; light of 8.0 × 10¹⁴ Hz. Find Ek(max) in eV.: hf = 6.63 × 10⁻³⁴ × 8.0 × 10¹⁴ = 5.30 × 10⁻¹⁹ J = 3.31 eV. Ek(max) = 3.31 − 2.0 = 1.31 eV.
Threshold wavelength: The maximum wavelength that causes emission: λ₀ = hc ÷ Φ. Longer wavelengths (lower frequencies) cause no emission, whatever the intensity.

Evidence that electromagnetic radiation has both wave and particle natures: Particle: the photoelectric effect. Wave: interference and diffraction. So light has a dual wave–particle nature.
de Broglie wavelength: λ = h ÷ p = h ÷ mv: a moving particle has a wavelength set by its momentum.
Evidence that electrons have wave properties: Electron diffraction: electrons passed through a thin graphite (crystal) film produce rings on a screen. Diffraction is a wave property.
Find the de Broglie wavelength of an electron accelerated from rest through 100 V.: p = √(2meV) = √(2 × 9.11 × 10⁻³¹ × 1.60 × 10⁻¹⁹ × 100) = 5.4 × 10⁻²⁴ kg m s⁻¹. λ = h ÷ p = 1.2 × 10⁻¹⁰ m.
Why do electrons diffract from crystals but everyday objects never show diffraction?: Electrons of a few hundred eV have wavelengths comparable to atomic spacing. Everyday objects have so much momentum that their wavelength is far smaller than any gap.

Electron energy levels in an isolated atom: Discrete (only certain values allowed) and negative, because zero energy is defined for an electron at infinity (free of the atom).
Photon emitted or absorbed in a transition between levels: hf = E₁ − E₂: the photon energy equals exactly the difference between the two levels.
Emission line: An electron falls from a higher to a lower energy level, emitting a photon of energy exactly ΔE.
Absorption line: A photon of energy exactly ΔE is absorbed, lifting an electron to a higher energy level.
Why do line spectra show that energy levels are discrete?: Only photons whose energy equals a difference between two levels are emitted or absorbed, so only certain wavelengths appear. Each element’s pattern is unique.
An electron falls from −3.4 eV to −13.6 eV. Find the photon wavelength.: ΔE = 10.2 eV = 1.63 × 10⁻¹⁸ J. λ = hc ÷ ΔE = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) ÷ 1.63 × 10⁻¹⁸ = 1.22 × 10⁻⁷ m (ultraviolet).