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IB DP Physics (2025 syllabus)

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E.2 Quantum physics

Practice Worksheet — name: ______________________ date: ____________

Formulas for this topic

Photoelectric equation Emax=hf−ΦE_{\text{max}} = hf - \Phi
Photon energy E=hf=hcλE = hf = \frac{hc}{\lambda}
de Broglie wavelength λ=hp=hmv\lambda = \frac{h}{p} = \frac{h}{mv}
Compton scattering λ′−λ=hmec(1−cos⁡θ)\lambda' - \lambda = \frac{h}{m_{e}c}(1 - \cos\theta)

Section A — Multiple choice

A1. Doubling the intensity of light above the threshold frequency shining on a metal surface:

  1. A Doubles the maximum kinetic energy of the photoelectrons
  2. B Doubles the number of photoelectrons emitted per second
  3. C Halves the work function
  4. D Doubles the stopping potential

A2. Light of photon energy 3.0 eV strikes a metal of work function 2.0 eV. The stopping potential is:

  1. A 5.0 V
  2. B 3.0 V
  3. C 2.0 V
  4. D 1.0 V

A3. An electron and a proton have the same de Broglie wavelength. They must have the same:

  1. A Speed
  2. B Kinetic energy
  3. C Momentum
  4. D Mass

Section B — Short answer

B1. State two observations of the photoelectric effect that cannot be explained by the wave model of light, and explain how the photon model accounts for each. [4 marks]

B2. Calculate the de Broglie wavelength of an electron accelerated through 150 V. [3 marks]

B3. In Compton scattering, explain why the scattered photon has a longer wavelength, and state the angle at which the shift is greatest. [3 marks]

Answer key

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Section A

A1: Doubles the number of photoelectrons emitted per second — Intensity means more photons per second, hence more electrons per second — but each photon still carries hfhf, so Emax=hf−ΦE_{max} = hf - \Phi (and the stopping potential) is unchanged. This is precisely what the wave model cannot explain.

A2: 1.0 V — Emax=3.0−2.0=1.0 eVE_{max} = 3.0 - 2.0 = 1.0\ \text{eV}, and the stopping potential in volts equals the maximum kinetic energy in electronvolts: Vs=1.0 VV_s = 1.0\ \text{V}.

A3: Momentum — λ=h/p\lambda = h/p depends only on momentum. Equal wavelengths mean equal momenta — the lighter electron must be moving much faster, and (at these non-relativistic speeds) carries far more kinetic energy.

Section B

B1: (1) A threshold frequency exists: below it no electrons are emitted regardless of intensity. Photon model: one photon delivers all its energy to one electron; if hf<Φhf < \Phi, no single photon can free an electron. (2) Emission is instantaneous even in dim light: energy arrives in concentrated packets rather than accumulating gradually over the surface. (Also acceptable: EmaxE_{max} depends on frequency, not intensity.)

B2: Ek=eV=150 eV=2.4×10−17 JE_k = eV = 150\ \text{eV} = 2.4\times10^{-17}\ \text{J}. Momentum: p=2mEk=2(9.11×10−31)(2.4×10−17)≈6.6×10−24 kg m s−1p = \sqrt{2mE_k} = \sqrt{2(9.11\times10^{-31})(2.4\times10^{-17})} \approx 6.6\times10^{-24}\ \text{kg m s}^{-1}. Then λ=h/p≈1.0×10−10 m\lambda = h/p \approx 1.0\times10^{-10}\ \text{m} — atomic spacing, which is why crystals diffract electrons.

B3: The photon transfers some of its energy and momentum to the recoiling electron in an elastic collision. Less energy means, via E=hc/λE = hc/\lambda, a longer wavelength. The shift Δλ=hmec(1−cos⁡θ)\Delta\lambda = \frac{h}{m_ec}(1-\cos\theta) is maximal at θ=180°\theta = 180° (backscattering), where it equals twice the Compton wavelength.