Estimate of anode voltage needed to produce wavelengths of the order of…

The idea

In an electron microscope, electrons are accelerated through an anode voltage V, gaining kinetic energy eV. Their de Broglie wavelength is λ = h/p, where p is momentum. To estimate the anode voltage needed for a wavelength of atomic size, take λ ≈ 10⁻¹⁰ m. Combine eV = p²/(2m) with p = h/λ to obtain eV = h²/(2mλ²), so V = h²/(2meλ²). Substituting h = 6.63 × 10⁻³⁴ J s, m = 9.11 × 10⁻³¹ kg, e = 1.60 × 10⁻¹⁹ C and λ = 1 × 10⁻¹⁰ m gives V ≈ 150 V. This order-of-magnitude estimate shows that around 10² V is needed to reach atomic-scale wavelengths, though practical electron microscopes use much higher voltages to resolve finer details.

What earns the marks

• States that the kinetic energy gained by an electron accelerated through a potential difference V is eV.
• Uses the de Broglie relation λ = h/p to relate wavelength to electron momentum.
• Combines the kinetic energy and momentum relations to derive V = h²/(2meλ²) or an equivalent expression.
• Substitutes appropriate values for h, m, e and λ ≈ 10⁻¹⁰ m to obtain an estimate of the order of 10² V (approx 150 V).
• Recognises that the estimate is an order-of-magnitude calculation and that the exact voltage depends on the precise atomic size chosen.
• Links the result to the resolution of electron microscopes: shorter de Broglie wavelengths allow finer detail to be resolved.

Common mistakes and how to put them right

• Using the electron mass in grams instead of kilograms.

  Put it right: Use m = 9.11 × 10⁻³¹ kg in SI calculations.

• Forgetting to square the wavelength in V = h²/(2meλ²).

  Put it right: The wavelength appears squared because p = h/λ and kinetic energy depends on p².

• Confusing the electron charge magnitude with the electronvolt conversion.

  Put it right: Use e = 1.60 × 10⁻¹⁹ C for the charge, and remember that 1 eV = 1.60 × 10⁻¹⁹ J.

• Assuming the anode voltage equals the electron energy in electronvolts without converting to joules.

  Put it right: eV gives energy in joules when e is in coulombs and V in volts.

Exam tips

• Show the derivation step by step: start from eV = p²/(2m) and p = h/λ, then rearrange to make V the subject.
• Use λ ≈ 1 × 10⁻¹⁰ m as a representative atomic size and state that this is an estimate, so the answer is an order of magnitude.
• Keep full precision during substitution and round only at the end; an answer in the range of 100 V to 200 V is reasonable for atomic-scale wavelengths.
• Check that the final unit is volts and that the calculation uses SI base units throughout.