A-Level Physics May/June 2025 Q6(c)(i): Show that the minimum wavelength of these electromagnetic waves is 21 pm.

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  The minimum wavelength of the emitted electromagnetic waves (X-ray photons) corresponds to the maximum possible photon energy.

  This occurs when an electron loses all its kinetic energy in a single interaction to create one photon. The kinetic energy of the electron is gained from the accelerating potential difference, $V$.

  Kinetic energy of electron, $E_k = eV$ Energy of photon, $E_{photon} = \frac{hc}{\lambda}$

  For the minimum wavelength, $\lambda_{min}$, the maximum photon energy is equal to the electron's kinetic energy: $eV = \frac{hc}{\lambda_{min}}$

  Rearranging for $\lambda_{min}$: $\lambda_{min} = \frac{hc}{eV}$

  Substituting the values (assuming the accelerating potential is 58 kV from the question context): $\lambda_{min} = \frac{(6.63 \times 10^{-34}) \times (3.00 \times 10^8)}{(1.60 \times 10^{-19}) \times (58 \times 10^3)}$

  $\lambda_{min} = \frac{1.989 \times 10^{-25}}{9.28 \times 10^{-15}}$

  $\lambda_{min} = 2.143... \times 10^{-11} \text{ m}$

  To convert from metres to picometres (pm), we use the fact that $1 \text{ pm} = 10^{-12} \text{ m}$: $\lambda_{min} = (2.143... \times 10^{-11}) \times \frac{1 \text{ pm}}{10^{-12} \text{ m}} = 21.43... \text{ pm}$

  Therefore, the minimum wavelength is $21 \text{ pm}$ (to 2 significant figures).

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  How the marks are awarded

  • C1 — Stating the correct relationship that equates the electron's kinetic energy ($eV$) with the photon's energy (rac{hc}{lambda}).
  • M1 — Correctly substituting all numerical values for Planck's constant, the speed of light, the elementary charge, and the potential difference (in Volts) into the equation.
  • A1 — Showing the final calculation leading to a value in metres and a clear, correct conversion to picometres to arrive at the required answer of 21 pm.

  Common mistakes

  • Forgetting to convert the accelerating potential from kilovolts (kV) to volts (V), i.e., using 58 instead of 58 × 10³.
  • Making a unit conversion error when converting the final answer from metres to picometres, such as using 10⁻⁹ or dividing by 10¹² instead of 10⁻¹².
  • Using an incorrect or incomplete energy formula, for example only stating E = hf without relating it to the electron's energy eV.
  • Mixing up the concepts of minimum wavelength and maximum frequency, leading to an incorrect initial relationship.

  Examiner tip: For problems involving particle-photon energy conversion, always start by equating the energy expressions for the particle (e.g., $E=qV$) and the photon (e.g., $E=hc/\lambda$).

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