A Level Physics 22.4 Energy Levels in Atoms and Line Spectra MCQs – 30 Practice Questions with Answers (Cambridge 9702)
This page covers Cambridge AS & A Level Physics (9702) Section 22.4, an A Level (A2) topic on the discrete energy levels of atoms and the line spectra they produce. Work through 30 original practice questions – using a model atom with levels at −8.0 eV, −5.0 eV, −2.0 eV and −0.8 eV – covering transitions, emission and absorption spectra, and full photon calculations.
Formula recap: photon emitted or absorbed in a transition: h f = ΔE = Ehigh − Elow; λ = c / f = h c / ΔE; convert electronvolts to joules with 1 eV = 1.6 × 10−19 J; ionisation energy from a level = energy needed to reach zero (the free electron). Constants: h = 6.63 × 10−34 J s, c = 3.00 × 108 m/s. Full notes: A Level Physics Formula Sheet.
Energy Levels & Transitions
Q1. The energy levels of an atom are quoted as negative values, with the highest level (a free electron) set at zero. Why are the bound levels negative?
- (A) Because electrons carry a negative charge, all their energies are negative
- (B) Because the levels are measured downwards from the nucleus
- (C) Because energy must be supplied to a bound electron to raise it to zero (ionisation), so a bound electron has less energy than a free one
- (D) Because emitted photons carry negative energy away from the atom
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Answer: C – Zero energy is defined for the electron at rest at infinity, completely removed from the atom. A bound electron sits below this, and work must be done on it to free it, so its total energy is negative. The sign is about energy, not charge.
Q2. What is meant by the ground state of an atom?
- (A) The lowest energy state of the atom, with its electron in the lowest available level
- (B) The state in which the atom has zero energy
- (C) The most highly excited state of the atom
- (D) The state of the atom just after it has been ionised
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Answer: A – The ground state is the lowest possible energy state. For the model atom used on this page it is the level at −8.0 eV, and an unexcited atom is found in this state. Any state above it is an excited state.
Q3. The model atom has its ground level at −8.0 eV and its first excited level at −5.0 eV. A photon is absorbed and raises the electron from the ground level to the first excited level. What must the photon energy be?
- (A) Any value greater than 3.0 eV
- (B) Exactly 3.0 eV
- (C) Exactly 8.0 eV
- (D) Exactly 13.0 eV
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Answer: B – Absorption of a photon is an all-or-nothing process: the photon energy must exactly equal the gap between the two levels, here (−5.0) − (−8.0) = 3.0 eV. A photon of any other energy cannot be absorbed in this transition.
Q4. An electron in the model atom falls from the level at −2.0 eV to the level at −5.0 eV. What is the energy of the photon emitted?
- (A) 2.0 eV
- (B) 5.0 eV
- (C) 7.0 eV
- (D) 3.0 eV
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Answer: D – The photon carries away the energy lost by the electron: ΔE = (−2.0) − (−5.0) = 3.0 eV. The levels themselves are not the photon energy; only the difference is.
Q5. Which of the following transitions in the model atom results in the emission of a photon?
- (A) −8.0 eV → −5.0 eV
- (B) −0.8 eV → −2.0 eV
- (C) −5.0 eV → −2.0 eV
- (D) −8.0 eV → −0.8 eV
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Answer: B – A photon is emitted only when the electron moves DOWN to a level of lower (more negative) energy, losing energy as light. All the other transitions move the electron up, which requires absorption of a photon instead.
Q6. What is the ionisation energy of the model atom when its electron is in the ground state (−8.0 eV)?
- (A) 0.8 eV
- (B) 5.0 eV
- (C) 8.0 eV
- (D) 15.8 eV
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Answer: C – Ionisation means removing the electron completely, taking it to zero energy. From the ground level this needs 0 − (−8.0) = 8.0 eV. The ionisation energy is the magnitude of the ground-level energy.
Q7. An electron in the model atom is already in the excited level at −2.0 eV. How much additional energy is needed to ionise the atom from this level?
- (A) 2.0 eV
- (B) 6.0 eV
- (C) 8.0 eV
- (D) 10.0 eV
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Answer: A – From −2.0 eV the electron only has to be raised to zero, so 2.0 eV is enough. An excited atom is easier to ionise than one in its ground state – which is why the level the electron starts from must always be stated.
Q8. Photons of energy 4.0 eV are incident on model atoms in their ground state (levels: −8.0, −5.0, −2.0, −0.8 eV). What happens?
- (A) They are absorbed and the electron moves to a level at −4.0 eV
- (B) Part of each photon is absorbed and the rest is re-emitted
- (C) They ionise the atoms
- (D) They are not absorbed, because no level lies exactly 4.0 eV above the ground level
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Answer: D – The gaps from the ground level are 3.0 eV, 6.0 eV and 7.2 eV, and there is no level at −4.0 eV. A photon can only be absorbed if its energy matches a gap exactly (or is at least the ionisation energy), so a 4.0 eV photon passes through. Note the contrast with excitation by collision, where a passing electron can transfer only part of its kinetic energy.
Q9. An electron makes a transition from a higher level Ehigh to a lower level Elow, emitting a photon of frequency f. Which equation correctly gives the photon frequency?
- (A) h f = Ehigh + Elow
- (B) h f = Elow − Ehigh
- (C) h f = Ehigh − Elow
- (D) f = h (Ehigh − Elow)
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Answer: C – The photon takes exactly the energy lost by the electron, Ehigh − Elow, which is positive for a downward transition (for example (−2.0) − (−5.0) = +3.0 eV). Option B would give a negative energy, and option D has h on the wrong side of the equation.
Q10. An electron in the model atom is excited to the top level at −0.8 eV and then returns to the ground state. Which statement about the emitted light is correct?
- (A) Only a single photon of 7.2 eV can ever be emitted
- (B) Either one photon of 7.2 eV, or a cascade of photons whose energies add up to 7.2 eV
- (C) Exactly four photons must be emitted, one per level
- (D) The cascade photons have a total energy greater than 7.2 eV
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Answer: B – The electron may fall straight to the ground level, emitting one 7.2 eV photon, or step down through intermediate levels (for example −0.8 → −2.0 → −8.0, emitting 1.2 eV and 6.0 eV). Energy conservation means the cascade energies always sum to the direct drop, 7.2 eV.
Line Spectra
Q11. A hot, low-pressure gas in a discharge tube is viewed through a diffraction grating. What is seen?
- (A) A series of bright lines at specific frequencies on a dark background
- (B) A continuous rainbow band of colour
- (C) Dark lines on a bright continuous background
- (D) A single broad band of white light
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Answer: A – This is an emission line spectrum. The excited atoms emit photons only at the frequencies corresponding to their level differences, so the light appears as separate bright lines with darkness between them.
Q12. Why does a gaseous element emit light only at certain discrete frequencies?
- (A) Because the gas is at low pressure
- (B) Because photons behave as waves
- (C) Because the atom has discrete energy levels, so only certain energy differences – and therefore only certain photon frequencies – are possible
- (D) Because the glass of the discharge tube absorbs the other frequencies
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Answer: C – Each line corresponds to one allowed transition. Since h f = ΔE and the level differences ΔE take only specific values, the emitted frequencies are discrete too. The existence of line spectra is direct evidence for quantised energy levels.
Q13. How is an absorption line spectrum of a gas produced?
- (A) By heating the gas until it glows
- (B) By passing the light from the gas through a prism twice
- (C) By exciting the gas with a high voltage
- (D) By passing white light from a continuous source through the cool gas, which absorbs photons at its characteristic frequencies
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Answer: D – The cool gas absorbs exactly those photon energies that match its level gaps, removing them from the continuous spectrum. The result is a bright continuous background crossed by dark lines at the absorbed frequencies.
Q14. For a given element, the dark lines in its absorption spectrum are found at frequencies that are…
- (A) Completely different from the frequencies of its emission lines
- (B) Exactly the same as the frequencies of its bright emission lines
- (C) Always lower than any of its emission frequencies
- (D) Random, changing from one experiment to the next
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Answer: B – Absorption and emission involve the same pairs of levels and therefore the same energy gaps. A gas absorbs precisely the frequencies it can emit – the dark lines sit exactly where the bright lines would be.
Q15. Astronomers can identify the elements present in a distant star from its spectrum because…
- (A) Each element has a unique set of energy levels, giving a unique pattern of lines that acts like a fingerprint
- (B) Every element emits lines of a single characteristic colour only
- (C) Heavier elements always produce more lines than lighter ones
- (D) The lines from a star are shifted by the same amount for every element
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Answer: A – No two elements share the same level structure, so no two produce the same line pattern. Matching the pattern of lines in starlight against laboratory spectra reveals which elements the star contains.
Q16. In an emission spectrum, the line of highest frequency (shortest wavelength) corresponds to the transition with…
- (A) The smallest energy change
- (B) A drop ending at the first excited level
- (C) The largest energy change
- (D) An electron being captured by the ion from outside the atom
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Answer: C – Since f = ΔE / h, the largest energy drop gives the highest frequency. In the model atom this is the fall from −0.8 eV straight to the ground level, ΔE = 7.2 eV.
Q17. The model atom has four energy levels. Electrons are excited to the top level and cascade back down by all possible routes. What is the maximum number of distinct emission lines?
- (A) 4
- (B) 6
- (C) 8
- (D) 12
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Answer: B – Each line corresponds to one pair of levels. With 4 levels the pairs are: 4→3, 4→2, 4→1, 3→2, 3→1, 2→1 – that is 4 × 3 / 2 = 6 possible transitions, giving at most 6 lines.
Q18. A simpler atom has only three energy levels. What is the maximum number of distinct frequencies in its emission spectrum?
- (A) 1
- (B) 2
- (C) 6
- (D) 3
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Answer: D – The transitions are top→middle, top→bottom and middle→bottom: 3 pairs of levels, so 3 lines at most. (The general rule is n(n − 1)/2 lines for n levels.)
Q19. A hot solid filament produces a continuous spectrum, but a low-pressure gas produces a line spectrum. What is the main reason for the difference?
- (A) The filament is at a lower temperature than the gas
- (B) The gas contains only one element but the filament contains many
- (C) In a solid the atoms are packed closely together and interact strongly, so the sharp individual levels merge into continuous bands
- (D) A solid emits photons of every possible energy because it is denser
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Answer: C – In a low-pressure gas the atoms are isolated, so their levels stay sharp and distinct. In a solid the neighbouring atoms disturb each other’s levels so strongly that the allowed energies spread into bands, and the transitions overlap to give a continuous spectrum.
Q20. An electron in the model atom falls from −0.8 eV to the ground level in two steps: first to −2.0 eV, then to −8.0 eV. The two photons emitted have total energy…
- (A) Equal to the single photon from the direct fall, 7.2 eV
- (B) Greater than 7.2 eV
- (C) Less than 7.2 eV
- (D) That depends on how long the electron spends in the middle level
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Answer: A – The steps emit 1.2 eV and 6.0 eV, which sum to 7.2 eV – exactly the direct drop. Only the starting and finishing levels matter for the total energy, not the route taken.
Spectrum Calculations
Q21. An electron in the model atom falls from −2.0 eV to −5.0 eV. What is the energy of the emitted photon in joules?
- (A) 3.0 × 10−19 J
- (B) 1.9 × 10−18 J
- (C) 4.8 × 10−18 J
- (D) 4.8 × 10−19 J
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Answer: D – ΔE = 3.0 eV. Converting: E = 3.0 × 1.6 × 10−19 = 4.8 × 10−19 J. (Option C is the classic slip of treating the eV value as if the conversion were × 10−18.)
Q22. What is the frequency of the photon emitted in the −2.0 eV → −5.0 eV transition (photon energy 4.8 × 10−19 J)?
- (A) 7.2 × 1013 Hz
- (B) 7.2 × 1014 Hz
- (C) 1.4 × 1015 Hz
- (D) 2.1 × 1014 Hz
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Answer: B – f = E / h = (4.8 × 10−19) / (6.63 × 10−34) = 7.2 × 1014 Hz, in the visible/near-ultraviolet part of the spectrum.
Q23. Calculate the wavelength of the photon emitted when an electron in the model atom falls from −2.0 eV directly to the ground level (−8.0 eV).
- (A) 2.07 × 10−7 m
- (B) 4.14 × 10−7 m
- (C) 1.04 × 10−6 m
- (D) 6.22 × 10−7 m
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Answer: A – ΔE = 6.0 eV = 6.0 × 1.6 × 10−19 = 9.6 × 10−19 J. Then f = E / h = 1.45 × 1015 Hz and λ = c / f = (3.00 × 108) / (1.45 × 1015) = 2.07 × 10−7 m (207 nm, ultraviolet).
Q24. Which transition in the model atom produces the longest-wavelength emission line, and what is that wavelength?
- (A) −0.8 eV → −8.0 eV, λ = 1.73 × 10−7 m
- (B) −2.0 eV → −5.0 eV, λ = 4.14 × 10−7 m
- (C) −0.8 eV → −2.0 eV, λ = 1.04 × 10−6 m
- (D) −0.8 eV → −5.0 eV, λ = 2.96 × 10−7 m
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Answer: C – The longest wavelength means the smallest photon energy, which is the smallest gap in the atom: (−0.8) − (−2.0) = 1.2 eV. Then λ = h c / ΔE = (6.63 × 10−34 × 3.00 × 108) / (1.2 × 1.6 × 10−19) = 1.04 × 10−6 m, in the infrared.
Q25. What is the shortest wavelength in the emission spectrum of the model atom?
- (A) 1.55 × 10−7 m
- (B) 2.07 × 10−7 m
- (C) 2.96 × 10−7 m
- (D) 1.73 × 10−7 m
Show Answer
Answer: D – The shortest wavelength comes from the largest energy drop between two levels: −0.8 eV → −8.0 eV, ΔE = 7.2 eV, giving λ = h c / ΔE = 1.73 × 10−7 m. (Option A, 155 nm, would need the full ionisation energy of 8.0 eV – that is the limit for absorption from the ground state, not an emission line between two levels.)
Q26. What is the minimum frequency of a photon that can ionise a model atom in its ground state?
- (A) 1.45 × 1015 Hz
- (B) 1.93 × 1015 Hz
- (C) 7.24 × 1014 Hz
- (D) 5.18 × 1015 Hz
Show Answer
Answer: B – The photon must supply the ionisation energy: E = 8.0 eV = 8.0 × 1.6 × 10−19 = 1.28 × 10−18 J. So f = E / h = (1.28 × 10−18) / (6.63 × 10−34) = 1.93 × 1015 Hz. Photons of higher frequency can also ionise, giving the freed electron some kinetic energy.
Q27. Model atoms in the ground state are illuminated with photons of various energies. Which photon can be absorbed by a ground-state atom?
- (A) 6.0 eV
- (B) 4.2 eV
- (C) 2.0 eV
- (D) 5.0 eV
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Answer: A – From the ground level the available gaps are 3.0 eV (to −5.0), 6.0 eV (to −2.0) and 7.2 eV (to −0.8). Only the 6.0 eV photon matches one of these exactly. (4.2 eV is a real gap, but between the two upper levels −0.8 and −5.0 – it cannot be reached from the ground state.)
Q28. A line in an emission spectrum has a wavelength of 414 nm. What is the energy difference between the two levels involved in the transition?
- (A) 0.30 eV
- (B) 1.5 eV
- (C) 3.0 eV
- (D) 4.8 eV
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Answer: C – E = h c / λ = (6.63 × 10−34 × 3.00 × 108) / (414 × 10−9) = 4.80 × 10−19 J. Converting to eV: 4.80 × 10−19 / 1.6 × 10−19 = 3.0 eV. (Option D quotes the energy in units of 10−19 J, not eV – a common trap.)
Q29. In a light-emitting diode (LED), electrons fall across a band gap of 1.8 eV and emit photons of energy approximately equal to the gap. What wavelength of light does this LED emit?
- (A) 173 nm
- (B) 690 nm
- (C) 1040 nm
- (D) 249 nm
Show Answer
Answer: B – E = 1.8 × 1.6 × 10−19 = 2.88 × 10−19 J. Then λ = h c / E = (6.63 × 10−34 × 3.00 × 108) / (2.88 × 10−19) = 6.9 × 10−7 m = 690 nm – red light. This is why the band gap of the semiconductor fixes the colour of an LED.
Q30. A discharge lamp emits a strong line at a wavelength of 546 nm. What energy change in the atoms produces this line?
- (A) 0.44 eV
- (B) 3.6 eV
- (C) 5.5 eV
- (D) 2.3 eV
Show Answer
Answer: D – E = h c / λ = (6.63 × 10−34 × 3.00 × 108) / (546 × 10−9) = 3.64 × 10−19 J = 3.64 × 10−19 / 1.6 × 10−19 = 2.3 eV. The line comes from electrons dropping between two levels separated by 2.3 eV.
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These are original practice questions written for revision. They are not Cambridge past-paper questions, and this page is not affiliated with or endorsed by Cambridge Assessment International Education.