OCR A Physics · Module 4: Electrons, waves and photons
Quantum physics: practice questions with answers
6 exam-style questions on quantum physics. Try each one on paper first, then open the answer to see the full working and the common slips that cost marks.
- eV and joules2 marks
Calculate the energy, in eV, of a photon of wavelength 450 nm.
E = eV
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Answer E = 2.76 eV
E = hc ÷ λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) ÷ (450 × 10⁻⁹) = 4.42 × 10⁻¹⁹ J = 2.76 eV.
- Got 4.42 × 10⁻¹⁹ eV? That is the energy in joules. Divide by 1.60 × 10⁻¹⁹ to convert to eV: 4.42 × 10⁻¹⁹ ÷ 1.60 × 10⁻¹⁹ = 2.76 eV.
- Got 0.00276 eV? You have divided by a thousand too many. E = hc ÷ λ = 4.42 × 10⁻¹⁹ J = 2.76 eV.
- eV and joules3 marks
Ultraviolet radiation of frequency 1.2 × 10¹⁵ Hz is incident on a metal of work function 2.9 eV. Calculate the maximum kinetic energy of the emitted photoelectrons, in eV.
KEmax = eV
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Answer KEmax = 2.07 eV
hf = 6.63 × 10⁻³⁴ × 1.2 × 10¹⁵ = 7.96 × 10⁻¹⁹ J = 4.97 eV. KEmax = hf - φ = 2.07 eV.
- Got 4.97 eV? You forgot to subtract the work function. KEmax = hf - φ = 4.97 - 2.9 = 2.07 eV.
- Got 7.87 eV? You added the work function. The photon energy is shared between escaping (φ) and kinetic energy, so KEmax = hf - φ = 2.07 eV.
- Using the right equation3 marks
An electron is accelerated from rest through a potential difference of 150 V. Calculate its de Broglie wavelength, in nm.
λ = nm
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Answer λ = 0.1 nm
½mv² = eV gives v = √(2 × 1.6 × 10⁻¹⁹ × 150 ÷ 9.11 × 10⁻³¹) = 7.26 × 10⁶ m s⁻¹. λ = h ÷ mv = 0.1 nm.
- Got 0.142 nm? You left out the factor of 2: ½mv² = eV gives v = √(2eV/m) = 7.26 × 10⁶ m s⁻¹, so λ = h ÷ mv = 0.1 nm.
- Got 8.29 nm? You used the photon relationship λ = hc ÷ E. For an electron use λ = h ÷ p with p = mv: λ = 0.1 nm.
- eV and joules3 marks
An electron in a hydrogen atom falls from an energy level of -0.85 eV to one of -3.40 eV. Calculate the wavelength of the photon emitted, in nm.
λ = nm
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Answer λ = 488 nm
ΔE = 2.55 eV = 4.08 × 10⁻¹⁹ J. λ = hc ÷ ΔE = 488 nm.
- Got 293 nm? You added the energies. The photon energy is the difference: 3.40 - 0.85 = 2.55 eV, so λ = hc ÷ ΔE = 488 nm.
- Got 1460 nm? You used only the upper level. The photon energy is the difference, 2.55 eV = 4.08 × 10⁻¹⁹ J, so λ = 488 nm.
- Using the right equation1 mark
The intensity of monochromatic light above the threshold frequency is increased. What happens in the photoelectric effect?
- A More photoelectrons are emitted per second.
- B The maximum kinetic energy of the electrons increases.
- C The threshold frequency decreases.
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Answer A: More photoelectrons are emitted per second.
Higher intensity means more photons per second, so more electrons per second, each with the same maximum energy.
- Why not B? Maximum kinetic energy depends on the photon frequency, not the intensity.
- Why not C? The threshold frequency is a property of the metal and does not change with intensity.
- Using the right equation1 mark
Electron diffraction through a thin crystal produces a ring pattern. What does this provide evidence for?
- A The particle nature of electrons.
- B The wave nature of electrons.
- C The quantisation of energy levels in atoms.
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Answer B: The wave nature of electrons.
Diffraction is a wave behaviour, so electrons show wave-particle duality.
- Why not A? Diffraction is a wave effect; particle behaviour is shown by evidence such as the photoelectric effect for light.
- Why not C? Energy levels are shown by line spectra, not by diffraction.