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Quantum Physics: Question 3

Syllabus 22.3

Structured A2 8 marks

In an electron diffraction tube, electrons are accelerated from rest through a potential difference of U=2.50 kVU = 2.50\text{ kV} before striking a thin sheet of graphite.

(a) Show that the kinetic energy gained by each electron is 4.00×10−16 J4.00\times10^{-16}\text{ J}. [2]

(b) Calculate the momentum of an electron after acceleration. (mass of electron =9.11×10−31 kg=9.11\times10^{-31}\text{ kg}) [2]

(c) Calculate the de Broglie wavelength of the accelerated electrons. [2]

(d) Beyond the graphite sheet, a pattern of concentric rings is observed on a fluorescent screen. State what this observation demonstrates about electrons, and explain briefly why this cannot be accounted for using a simple particle model of the electron. [2]

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Worked solution

Part (a): Show that the kinetic energy is 4.00 × 10⁻¹⁶ J

Each electron of charge ee accelerated from rest through a potential difference UU gains kinetic energy equal to the work done on it by the electric field: Ek=eUE_k = eU

Substituting e=1.60×10−19 Ce=1.60\times10^{-19}\text{ C} and U=2.50×103 VU=2.50\times10^{3}\text{ V}: Ek=1.60×10−19×2.50×103=4.00×10−16 JE_k = 1.60\times10^{-19}\times2.50\times10^{3} = 4.00\times10^{-16}\text{ J}

Recompute as a check, multiplying in a different order: 2.50×103×1.60×10−19=2.50×1.60=4.002.50\times10^{3}\times1.60\times10^{-19}=2.50\times1.60=4.00, with powers of ten 103×10−19=10−1610^{3}\times10^{-19}=10^{-16}, giving the same 4.00×10−16 J4.00\times10^{-16}\text{ J}.

So Ek=4.00×10−16 JE_k=4.00\times10^{-16}\text{ J}, as required to show.

Part (b): Momentum of the electron

Since Ek=12mv2=p22mE_k=\tfrac12mv^2=\dfrac{p^2}{2m}, the momentum is: p=2mEkp=\sqrt{2mE_k}

Substituting m=9.11×10−31 kgm=9.11\times10^{-31}\text{ kg} and Ek=4.00×10−16 JE_k=4.00\times10^{-16}\text{ J}: p=2×9.11×10−31×4.00×10−16=7.288×10−46=2.70×10−23 kg m s−1p=\sqrt{2\times9.11\times10^{-31}\times4.00\times10^{-16}}=\sqrt{7.288\times10^{-46}}=2.70\times10^{-23}\text{ kg m s}^{-1}

Recompute as a check: (2.70×10−23)2=7.29×10−46(2.70\times10^{-23})^2=7.29\times10^{-46}, matching the value under the square root above (small rounding difference only). Consistent.

Part (c): De Broglie wavelength

The de Broglie wavelength is: λ=hp\lambda=\frac{h}{p}

Substituting h=6.63×10−34 J sh=6.63\times10^{-34}\text{ J s} and p=2.70×10−23 kg m s−1p=2.70\times10^{-23}\text{ kg m s}^{-1}: λ=6.63×10−342.70×10−23=2.46×10−11 m\lambda=\frac{6.63\times10^{-34}}{2.70\times10^{-23}}=2.46\times10^{-11}\text{ m}

Recompute as a check using the combined formula directly, λ=h/2mEk=h/2meU\lambda=h/\sqrt{2mE_k}=h/\sqrt{2meU}: substituting all values under one square root gives the same denominator 2.70×10−23 kg m s−12.70\times10^{-23}\text{ kg m s}^{-1} found in part (b), so λ=2.46×10−11 m\lambda=2.46\times10^{-11}\text{ m} again.

This wavelength (≈0.025 nm\approx0.025\text{ nm}) is comparable to the spacing between atomic planes in graphite, which is why diffraction of the electrons is observed.

Part (d): Evidence from the diffraction pattern

The pattern of concentric rings shows that the electrons are being diffracted by the graphite (an effect that only occurs for waves, since diffraction depends on constructive and destructive interference between waves of comparable wavelength to the gaps (here, the atomic spacing) they pass through. This demonstrates that electrons, normally thought of as particles, also exhibit wave-like behaviour) evidence for wave-particle duality. A simple particle model, in which each electron is treated as a small charged mass travelling in a straight line, predicts no such interference pattern and cannot account for the rings.

Final answers

  • (a) Ek=4.00×10−16 JE_k=\boxed{4.00\times10^{-16}}\text{ J}
  • (b) p=2.70×10−23 kg m s−1p=\boxed{2.70\times10^{-23}}\text{ kg m s}^{-1}
  • (c) λ=2.46×10−11 m\lambda=\boxed{2.46\times10^{-11}}\text{ m}
  • (d) The rings show electron diffraction, evidence that electrons exhibit wave-like behaviour (wave-particle duality), which a particle-only model cannot explain.