A Level Physics: Quantum Physics — Practice Questions

These are original questions written for Marlbridge, in the style and at the standard of the examination. They are not reproduced past-paper questions — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Quantum Physics revision notes

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Section A

1. Define the work function of a metal. [1]

2. State the photoelectric equation and define each term. [3]

3. State de Broglie’s relation and give one piece of experimental evidence for it. [2]

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Section B

4. A metal has a work function of 2.30 eV. (h = 6.63 × 10⁻³⁴ J s; c = 3.00 × 10⁸ m s⁻¹; 1 eV = 1.60 × 10⁻¹⁹ J)

(a) Calculate the threshold frequency. [3]

(b) Calculate the maximum wavelength of light that will cause emission. [2]

(c) Light of wavelength 400 nm is shone on the metal. Calculate the maximum kinetic energy of the emitted electrons, in joules. [3]

(d) The intensity of the 400 nm light is doubled. State the effect on (i) the number of electrons emitted per second, (ii) their maximum kinetic energy. [2]

5. A very intense beam of red light produces no photoelectrons from a metal, while a very faint beam of blue light produces them immediately.

(a) Explain this observation. [4]

(b) Explain why the wave model of light cannot account for it. [2]

6. An electron is accelerated from rest through a potential difference of 250 V. (m_e = 9.11 × 10⁻³¹ kg; e = 1.60 × 10⁻¹⁹ C)

(a) Calculate its kinetic energy in joules. [1]

(b) Calculate its speed. [2]

(c) Calculate its de Broglie wavelength. [2]

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Section C

7. An electron in a hydrogen atom transitions from the n = 3 level (E = −1.51 eV) to the n = 2 level (E = −3.40 eV). (h = 6.63 × 10⁻³⁴ J s; c = 3.00 × 10⁸ m s⁻¹; 1 eV = 1.60 × 10⁻¹⁹ J)

(a) Calculate the energy of the photon emitted, in joules. [2]

(b) Calculate the wavelength of this photon. [2]

(c) State, with a reason, whether this photon lies within the visible spectrum. [1]

8. Line spectra are cited as direct experimental evidence for discrete atomic energy levels. Explain why. [2]

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Answers

1. The minimum energy required to remove an electron from the surface of the metal [1].

2. hf = φ + KE_max [1], where hf is the photon energy [1], φ is the work function and KE_max is the maximum kinetic energy of an emitted electron [1].

3. λ = h ÷ p (= h/mv) [1]. Evidence: electron diffraction — electrons passing through a thin graphite film produce diffraction rings [1].

4. (a) φ = 2.30 × 1.60 × 10⁻¹⁹ = 3.68 × 10⁻¹⁹ J [1]. f₀ = φ ÷ h = 3.68 × 10⁻¹⁹ ÷ 6.63 × 10⁻³⁴ [1] = 5.55 × 10¹⁴ Hz [1].

(b) λ_max = c ÷ f₀ = 3.00 × 10⁸ ÷ 5.55 × 10¹⁴ [1] = 5.41 × 10⁻⁷ m (541 nm) [1].

(c) E_photon = hc ÷ λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) ÷ (400 × 10⁻⁹) [1] = 4.97 × 10⁻¹⁹ J [1]. KE_max = 4.97 × 10⁻¹⁹ − 3.68 × 10⁻¹⁹ = 1.29 × 10⁻¹⁹ J [1].

(d) (i) Doubles — twice as many photons arrive per second [1]. (ii) Unchanged — KE_max depends only on the frequency [1].

5. (a) A single photon interacts with a single electron, transferring all its energy [1]. Red photons have a lower frequency and therefore less energy than the work function, so no electron can escape [1]. Increasing intensity supplies more photons but does not increase the energy of each [1]. Blue photons have energy greater than the work function, so emission occurs immediately even at low intensity [1].

(b) Wave theory predicts energy would be absorbed continuously and accumulate [1], so intense red light should eventually supply enough energy — which is not observed [1].

6. (a) KE = eV = 1.60 × 10⁻¹⁹ × 250 = 4.00 × 10⁻¹⁷ J [1].

(b) v = √(2KE ÷ m) = √((2 × 4.00 × 10⁻¹⁷) ÷ 9.11 × 10⁻³¹) [1] = 9.37 × 10⁶ m s⁻¹ [1].

(c) λ = h ÷ mv = 6.63 × 10⁻³⁴ ÷ (9.11 × 10⁻³¹ × 9.37 × 10⁶) [1] = 7.77 × 10⁻¹¹ m [1].

7. (a) ΔE = −1.51 − (−3.40) = 1.89 eV [1] = 1.89 × 1.60 × 10⁻¹⁹ = 3.02 × 10⁻¹⁹ J [1].

(b) λ = hc ÷ E = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) ÷ 3.02 × 10⁻¹⁹ [1] = 6.59 × 10⁻⁷ m (659 nm) [1].

(c) Yes — 659 nm falls within the visible range (roughly 400–700 nm), appearing as a red spectral line [1].

8. Only certain specific energy differences exist between the discrete levels available to an atom’s electrons [1], so only photons with those specific energies (and hence specific frequencies/wavelengths) can be emitted or absorbed, producing sharp, separated lines rather than a continuous spread of colour [1].

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Where marks are usually lost

• Saying more intense light gives faster photoelectrons.
• Forgetting to convert eV to joules before substituting.
• Not stating the one-photon-one-electron interaction when explaining the threshold.
• Using λ rather than f in the photoelectric equation without converting.
• Omitting the accumulation argument when criticising the wave model.
• In electron–positron annihilation from rest, explaining why two photons (not one) are produced by saying only that “the particles are annihilated” — the actual reason is momentum conservation: since the pair starts at rest, the total initial momentum is zero, so the two photons must carry equal and opposite momentum, travelling in opposite directions. (This specific two-photon, opposite-direction result assumes negligible initial kinetic energy; it is not a universal feature of every annihilation.)
• Confusing the two key quantum equations — hf = φ + KE_max applies only to the photoelectric effect, while hf = E₁ − E₂ applies only to energy-level transitions inside an atom; they describe different physical situations and are not interchangeable.