References only, not question text. Open Official paper ↗ where Cambridge publishes the paper,
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school. “Paper 42” means Paper 4, variant 2.
Practice questions
Original questions written by HomiLabs in the style of the exam, with answers and mark schemes. They are not
Cambridge past paper questions and have not been reviewed by Cambridge. Try each one before opening the answer.
Q1Photons & photoelectric effect · Core · Paper 4 style · 4 marks
(a)State what is meant by a photon.[1]
(b)A green laser pointer emits light of wavelength 532 nm. Calculate the energy of one photon in joules.[2]
(c)Express this energy in electronvolts.[1]
Show answer
(a) A discrete quantum (packet) of electromagnetic energy.
B1
(b) 3.74 × 10⁻¹⁹ J
C1 E = hc/λ; A1 3.74 × 10⁻¹⁹ J
(c) 2.34 eV
3.74 × 10⁻¹⁹/1.60 × 10⁻¹⁹ = 2.34 eV (A1)
Q2Photons & photoelectric effect · Core · Paper 4 style · 4 marks
(a)Define work function.[1]
(b)Define threshold frequency.[1]
(c)The work function of zinc is 4.30 eV. Calculate its threshold frequency.[2]
Show answer
(a) The minimum energy needed to remove an electron from the surface of the metal.
B1
(b) The minimum frequency of electromagnetic radiation that causes electrons to be emitted from the surface.
Q3Photons & photoelectric effect · Core · Paper 4 style · 4 marks
(a)Calculate the momentum of a photon of red light of wavelength 650 nm.[2]
(b)Show that the energy E of a photon is related to its momentum p by E = pc, and use this to find the photon energy.[2]
Show answer
(a) 1.02 × 10⁻²⁷ N s
C1 p = h/λ; A1 1.02 × 10⁻²⁷ N s
(b) E = hf = hc/λ = (h/λ)c = pc; E = 3.06 × 10⁻¹⁹ J
B1 derivation; B1 value
Q4Photons & photoelectric effect · Core · Paper 4 style · 5 marks
(a)Describe three observations from photoelectric emission experiments that cannot be explained by a wave theory of light.[3]
(b)Explain how the photon model accounts for the existence of a threshold frequency.[2]
Show answer
(a) Any three: no emission below a threshold frequency however intense the light; emission is instantaneous even at low intensity; the maximum kinetic energy of the electrons depends on frequency, not intensity; increasing intensity increases the rate of emission only.
B1 each, max 3
(b) Each electron absorbs the energy of one photon, E = hf; if hf is less than the work function no electron can escape, however many photons arrive.
B1 one photon to one electron; B1 hf < φ means no emission
Q5Photons & photoelectric effect · Core · Paper 4 style · 4 marks
(a)Define the electronvolt.[1]
(b)Convert 3.5 eV into joules.[1]
(c)Calculate the wavelength of a photon of energy 3.5 eV and state the region of the electromagnetic spectrum.[2]
Show answer
(a) The energy gained (transferred) when an electron moves through a potential difference of one volt.
B1
(b) 5.6 × 10⁻¹⁹ J
3.5 × 1.60 × 10⁻¹⁹ (A1)
(c) 3.55 × 10⁻⁷ m; ultraviolet
B1 λ = hc/E value; B1 ultraviolet
Q6Photons & photoelectric effect · Core · Paper 4 style · 3 marks
(a)An FM radio station in Islamabad broadcasts at 98.6 MHz. Calculate the energy of one of its photons.[1]
(b)Explain why radio waves cannot cause photoelectric emission from a metal, however intense they are.[2]
Show answer
(a) 6.54 × 10⁻²⁶ J
E = hf (A1)
(b) Each photon's energy is far below any metal's work function (a few eV ≈ 10⁻¹⁹ J), and an electron absorbs only one photon, so no electron gains enough energy.
B1 photon energy much less than work function; B1 one photon per electron / intensity only increases number of photons
Q7Photons & photoelectric effect · Standard · Paper 4 style · 6 marks
Ultraviolet radiation of wavelength 250 nm falls on a clean potassium surface. The work function of potassium is 2.30 eV.
(a)Calculate the energy of each photon in eV.[2]
(b)Calculate the maximum kinetic energy of the emitted electrons in eV.[1]
(c)Calculate the maximum speed of the emitted electrons.[2]
(d)Explain why most emitted electrons have less than this kinetic energy.[1]
Show answer
(a) 4.97 eV
C1 E = hc/λ = 7.96 × 10⁻¹⁹ J; A1 4.97 eV
(b) 2.67 eV
KE_max = 4.97 − 2.30 = 2.67 eV (A1)
(c) 9.69 × 10⁵ m s⁻¹
C1 ½mv² = 4.28 × 10⁻¹⁹ J; A1 v = 9.69 × 10⁵ m s⁻¹
(d) Electrons below the surface lose energy (in collisions) before escaping, or need more than the work function to escape.
B1
Q8Photons & photoelectric effect · Standard · Paper 5 style · 7 marks
In a photoelectric experiment, light of different frequencies f falls on a metal surface, and the maximum kinetic energy KE_max of the emitted electrons is measured in electronvolts.f / 10¹⁴ Hz | KE_max / eV6.0 | 0.257.0 | 0.648.0 | 1.069.0 | 1.5010.0 | 1.88
(a)Show that a graph of KE_max (in eV) against f should be a straight line, and state expressions for its gradient and its intercept on the KE_max axis.[2]
(b)Using the first and last rows (or a best-fit line), determine a value for the Planck constant.[2]
(c)Determine the work function of the metal in eV.[2]
(d)Determine the threshold frequency.[1]
Show answer
(a) hf = φ + KE_max, so KE_max = (h/e)f − φ/e with KE in eV; gradient = h/e, intercept = −φ (in eV).
B1 rearranged equation; B1 gradient h/e and intercept −φ/e
(b) h ≈ 6.52 × 10⁻³⁴ J s (accept 6.4–6.9 × 10⁻³⁴)
M1 gradient (in V s) × e; A1 h in range with unit
(c) φ ≈ 2.19 eV (accept 2.1–2.4 eV)
M1 φ = hf/e − KE_max for a point on the line (or −intercept); A1 φ
(d) ≈ 5.39 × 10¹⁴ Hz (accept 5.3–5.5 × 10¹⁴ Hz)
f₀ = φ/h, or the intercept on the f axis: 6.0 × 10¹⁴ − 0.25/(gradient 4.075 × 10⁻¹⁵ V s) (A1, consistent with (b) and (c))
Q9Photons & photoelectric effect · Standard · Paper 4 style · 5 marks
A red laser pointer emits light of wavelength 650 nm with an output power of 5.0 mW.
(a)Calculate the number of photons emitted per second.[3]
(b)A second laser of the same power emits blue light of wavelength 450 nm. State and explain whether it emits more or fewer photons per second.[2]
Show answer
(a) 1.63 × 10¹⁶ s⁻¹
C1 E = hc/λ = 3.06 × 10⁻¹⁹ J; M1 N = P/E; A1 1.63 × 10¹⁶ s⁻¹
(b) Fewer: each blue photon has more energy (E = hc/λ, smaller λ), so fewer photons are needed for the same power.
B1 fewer; B1 reason
Q10Photons & photoelectric effect · Standard · Paper 4 style · 5 marks
A metal with work function 2.20 eV is illuminated with light containing only two wavelengths, 420 nm and 610 nm.
(a)Calculate the threshold wavelength for this metal.[2]
(b)State which of the two wavelengths causes photoelectric emission.[1]
(c)Calculate the maximum speed of the emitted electrons.[2]
Show answer
(a) 5.65 × 10⁻⁷ m
C1 λ₀ = hc/φ; A1 5.65 × 10⁻⁷ m
(b) 420 nm only (it is shorter than the threshold wavelength; 610 nm is longer).
Q11Photons & photoelectric effect · Standard · Paper 4 style · 4 marks
(a)Electrons in a dental X-ray tube are accelerated through a p.d. of 25 kV. State the maximum energy of an X-ray photon produced, in keV.[1]
(b)Calculate the minimum wavelength of the X-rays.[2]
(c)Explain why the X-rays produced have a continuous range of wavelengths above this minimum.[1]
Show answer
(a) 25 keV
An electron can lose all its kinetic energy (25 keV) to one photon (B1)
(b) 4.97 × 10⁻¹¹ m
C1 eV = hc/λ_min; A1 4.97 × 10⁻¹¹ m
(c) Electrons lose different amounts of energy in each deceleration (some in several steps), so photons of all energies up to the maximum are produced.
B1
Q12Photons & photoelectric effect · Standard · Paper 4 style · 4 marks
Graphs of the maximum kinetic energy KE_max of photoelectrons against frequency f are plotted for two metals, P and Q. Both are straight lines. The line for P meets the frequency axis at 5.5 × 10¹⁴ Hz and the line for Q meets it at 8.0 × 10¹⁴ Hz.
(a)Explain why the two lines are parallel.[1]
(b)Calculate the work function of each metal in eV.[2]
(c)State the intercept that each line would make if extended to the KE_max axis.[1]
Show answer
(a) The gradient of each line equals the Planck constant h (or h/e if KE_max is in eV), which is the same for all metals.
B1
(b) P: 2.28 eV; Q: 3.32 eV
C1 φ = hf₀; A1 both
(c) −2.28 eV for P and −3.32 eV for Q (minus the work function)
B1
Q13Photons & photoelectric effect · Standard · Paper 4 style · 3 marks
(a)A red LED emits photons of wavelength 630 nm. Each photon is produced when one electron loses energy crossing the junction. Estimate the minimum p.d. needed across the LED.[2]
(b)Explain why a blue LED needs a larger p.d. than a red LED.[1]
Show answer
(a) 1.97 V
M1 eV = hc/λ; A1 1.97 V
(b) Blue photons have shorter wavelength and so more energy, so each electron must be given more energy (eV).
B1
Q14Photons & photoelectric effect · Challenge · Paper 4 style · 6 marks
Light of wavelength 400 nm and power 2.0 µW falls on the cathode of a photocell.
(a)Calculate the number of photons arriving at the cathode per second.[2]
(b)On average, one photon in 500 releases an electron, and all released electrons reach the anode. Calculate the photoelectric current.[2]
(c)The intensity of the light is doubled at the same wavelength. State the effect on (i) the current and (ii) the maximum kinetic energy of the electrons, and explain.[2]
Show answer
(a) 4.02 × 10¹² s⁻¹
C1 E = hc/λ; A1 N = P/E = 4.02 × 10¹² s⁻¹
(b) 1.29 × 10⁻⁹ A
M1 electrons per second = 8.04 × 10⁹; A1 I = Ne = 1.29 × 10⁻⁹ A
(c) (i) current doubles, since twice as many photons arrive per second; (ii) maximum KE unchanged, since each photon's energy hf is unchanged.
B1 each
Q15Photons & photoelectric effect · Challenge · Paper 4 style · 5 marks
Photons carry momentum as well as energy. Solar sails use this to propel spacecraft.
(a)Show that the force exerted on a surface that absorbs all the light falling on it is P/c, where P is the power of the light.[2]
(b)Calculate the force on a 1.0 m² black panel facing the Sun just outside the Earth's atmosphere, where the intensity is 1.36 kW m⁻².[1]
(c)Calculate the force if the panel is replaced by a perfect mirror, and explain the difference.[2]
Show answer
(a) Each photon has momentum E/c; the photons arriving per second carry momentum P/c per second; when absorbed, rate of change of momentum = force = P/c.
B1 p = E/c per photon; B1 rate of change of momentum = P/c
(b) 4.53 × 10⁻⁶ N
F = 1360/3.00 × 10⁸ (A1)
(c) 9.07 × 10⁻⁶ N; each photon's momentum is reversed, so its change of momentum (and the force) is doubled.
B1 value; B1 reason
Q16Photons & photoelectric effect · Challenge · Paper 4 style · 6 marks
Monochromatic radiation of wavelength 300 nm falls on a metal surface and the maximum kinetic energy of the emitted electrons is 1.85 eV. When radiation of wavelength 450 nm is used, the maximum kinetic energy is 0.47 eV.
(a)Use these results to determine a value for the Planck constant.[3]
(b)Determine the work function of the metal, in eV.[2]
(c)State the longest wavelength that will cause emission from this metal.[1]
Q17Photons & photoelectric effect · Challenge · Paper 4 style · 6 marks
Ultraviolet light of wavelength 330 nm falls on a metal plate of work function 2.10 eV in a vacuum.
(a)Calculate the maximum kinetic energy of the emitted electrons, in eV.[2]
(b)Calculate the maximum speed of the electrons.[1]
(c)The fastest electrons enter a uniform magnetic field of flux density 1.2 × 10⁻⁴ T at right angles to the field. Calculate the radius of their circular path.[2]
(d)Describe how the paths of slower electrons from the same surface differ.[1]
Show answer
(a) 1.67 eV
C1 hc/λ = 3.77 eV; A1 1.67 eV
(b) 7.65 × 10⁵ m s⁻¹
½mv² = KE_max (A1)
(c) 0.0363 m
C1 Bev = mv²/r, so r = mv/(Be); A1 0.0363 m
(d) They move in circles of smaller radius (r ∝ v), in the same sense of rotation.
B1
Q18Photons & photoelectric effect · Challenge · Paper 4 style · 5 marks
On a dark night in the Thar desert, a dark-adapted human eye can just detect green light of wavelength 510 nm when 2.0 × 10⁻¹⁷ W enters the pupil. The pupil diameter is 7.0 mm.
(a)Calculate the number of photons per second entering the eye at this threshold.[2]
(b)Calculate the intensity at the eye at this threshold.[1]
(c)A lamp radiates 5.0 W of visible light of wavelength 510 nm uniformly in all directions. Estimate the greatest distance at which it could just be seen, ignoring absorption in the air.[2]
Show answer
(a) 51.3 s⁻¹
C1 E = hc/λ = 3.90 × 10⁻¹⁹ J; A1 51.3 s⁻¹
(b) 5.20 × 10⁻¹³ W m⁻²
I = P/(πr²) with r = 3.5 mm (A1)
(c) 8.75 × 10⁵ m (about 9 × 10⁵ m)
M1 I = P/(4πd²) set equal to 5.20 × 10⁻¹³ W m⁻²; A1 d = 8.75 × 10⁵ m
Q19Wave-particle duality & de Broglie · Core · Paper 4 style · 3 marks
(a)State the de Broglie relation, explaining the symbols.[1]
(b)Calculate the de Broglie wavelength of an electron moving at 2.5 × 10⁶ m s⁻¹.[2]
Show answer
(a) λ = h/p, where λ is the de Broglie wavelength, h the Planck constant and p the momentum of the particle.
B1
(b) 2.91 × 10⁻¹⁰ m
C1 p = mv; A1 2.91 × 10⁻¹⁰ m
Q20Wave-particle duality & de Broglie · Core · Paper 4 style · 4 marks
(a)Name one phenomenon that provides evidence for the wave nature of electrons and one that provides evidence for the particle nature of electromagnetic radiation.[2]
(b)In an electron diffraction tube, the electrons form rings on a fluorescent screen. Explain why this shows wave behaviour.[2]
Show answer
(a) Electron diffraction (by a thin graphite/crystal film); the photoelectric effect.
B1 each
(b) Rings are a diffraction/interference pattern: maxima and minima form where waves from the atomic layers reinforce or cancel, which particles alone would not produce.
B1 diffraction/interference; B1 by the regular atomic spacing comparable with λ
Q21Wave-particle duality & de Broglie · Standard · Paper 4 style · 4 marks
(a)Show that the de Broglie wavelength of an electron accelerated from rest through a p.d. V is λ = h/√(2meV).[2]
(b)Calculate the wavelength for V = 150 V.[1]
(c)The atomic spacing in a crystal is about 2 × 10⁻¹⁰ m. Explain why electrons of this energy are diffracted significantly by the crystal.[1]
Show answer
(a) eV = ½mv² = p²/2m, so p = √(2meV); λ = h/p = h/√(2meV).
B1 KE = eV = p²/2m; B1 substitution
(b) 1.00 × 10⁻¹⁰ m
A1
(c) Their wavelength is similar to (the same order as) the atomic spacing.
B1
Q22Wave-particle duality & de Broglie · Standard · Paper 4 style · 3 marks
(a)Calculate the de Broglie wavelength of a cricket ball of mass 0.16 kg bowled at 40 m s⁻¹.[2]
(b)Explain why the ball shows no observable diffraction when it passes between the stumps.[1]
Show answer
(a) 1.04 × 10⁻³⁴ m
C1 λ = h/(mv); A1 1.04 × 10⁻³⁴ m
(b) The wavelength is far smaller than the gap or any obstacle (by more than 30 orders of magnitude), so diffraction is negligible.
B1
Q23Wave-particle duality & de Broglie · Standard · Paper 4 style · 4 marks
(a)The accelerating p.d. in an electron diffraction tube is increased from 3.0 kV to 5.0 kV. Calculate the ratio new wavelength ÷ old wavelength.[2]
(b)State and explain the effect on the diameter of the rings on the screen.[2]
Show answer
(a) 0.775
M1 λ ∝ 1/√V; A1 √(3.0/5.0) = 0.775
(b) The rings become smaller: the shorter wavelength is diffracted through smaller angles.
B1 smaller; B1 shorter wavelength, less diffraction
Q24Wave-particle duality & de Broglie · Challenge · Paper 4 style · 5 marks
(a)An electron and a proton have the same kinetic energy. Show that the ratio of their de Broglie wavelengths, λ_electron/λ_proton, equals √(m_p/m_e), and calculate its value.[3]
(b)Calculate the de Broglie wavelength of a proton accelerated from rest through 2.0 kV.[2]
Show answer
(a) 42.8
M1 p = √(2mE_k); M1 λ = h/√(2mE_k), so ratio = √(m_p/m_e); A1 42.8
(b) 6.41 × 10⁻¹³ m
C1 p = √(2m_p eV); A1 6.41 × 10⁻¹³ m
Q25Wave-particle duality & de Broglie · Challenge · Paper 4 style · 5 marks
(a)Calculate the de Broglie wavelength of a neutron of speed 2200 m s⁻¹ (mass 1.67 × 10⁻²⁷ kg).[2]
(b)Calculate its kinetic energy in eV.[2]
(c)Suggest why slow neutrons like these are useful for studying the arrangement of atoms in solids.[1]
Show answer
(a) 1.80 × 10⁻¹⁰ m
C1 λ = h/(mv); A1 1.80 × 10⁻¹⁰ m
(b) 0.0253 eV
C1 ½mv² = 4.04 × 10⁻²¹ J; A1 0.0253 eV
(c) Their wavelength is about the same as the atomic spacing, so they are diffracted by the crystal lattice (and being uncharged they penetrate the material).
B1
Q26Wave-particle duality & de Broglie · Challenge · Paper 4 style · 6 marks
(a)Calculate the energy of a photon of wavelength 0.20 nm, in eV.[2]
(b)Calculate the kinetic energy of an electron with de Broglie wavelength 0.20 nm, in eV.[3]
(c)Suggest why electrons, rather than X-rays, are used to study the structure of thin surface layers.[1]
Show answer
(a) 6220 eV
C1 E = hc/λ = 9.94 × 10⁻¹⁶ J; A1 value in eV
(b) 37.7 eV
M1 p = h/λ = 3.31 × 10⁻²⁴ N s; M1 E = p²/2m = 6.03 × 10⁻¹⁸ J; A1 value in eV
(c) Electrons of the same wavelength have much less energy, so they penetrate only a short distance and interact strongly with surface atoms (and cause less damage).
B1
Q27Energy levels & line spectra · Core · Paper 4 style · 3 marks
An electron in a hydrogen atom falls from the energy level at −1.51 eV to the level at −3.40 eV.
(a)Calculate the energy of the photon emitted, in eV.[1]
(b)Calculate the wavelength of the emitted light and state its colour.[2]
Show answer
(a) 1.89 eV
−1.51 − (−3.40) = 1.89 eV (A1)
(b) 6.58 × 10⁻⁷ m; red
C1 λ = hc/ΔE; A1 6.58 × 10⁻⁷ m (red)
Q28Energy levels & line spectra · Standard · Paper 4 style · 5 marks
(a)Explain how a hot, low-pressure gas produces an emission line spectrum.[3]
(b)White light passes through a cool gas. Explain why the spectrum of the transmitted light contains dark lines.[2]
Show answer
(a) Electrons in the atoms are excited to higher energy levels; they fall to lower levels, emitting photons; each photon's energy equals the difference between two discrete levels, hf = E₁ − E₂, so only certain frequencies appear.
(b) Atoms absorb photons whose energies exactly equal differences between their energy levels; these are re-emitted in all directions, so those wavelengths are missing (much weaker) in the forward beam.
B1 absorb photons of energy = ΔE only; B1 re-emitted in all directions
Q29Energy levels & line spectra · Standard · Paper 4 style · 4 marks
Four energy levels of the hydrogen atom are −13.6 eV (ground state), −3.40 eV, −1.51 eV and −0.85 eV.
(a)State the number of different photon energies that can be emitted as atoms in the −0.85 eV level return to the ground state by any route.[1]
(b)Calculate the shortest wavelength emitted.[2]
(c)Explain why the energy levels are given negative values.[1]
Show answer
(a) 6
B1 (4→3, 4→2, 4→1, 3→2, 3→1, 2→1)
(b) 9.75 × 10⁻⁸ m
C1 ΔE = 13.6 − 0.85 = 12.75 eV; A1 9.75 × 10⁻⁸ m
(c) Zero energy is defined for an electron that is free of the atom (at infinity); a bound electron has less energy than this.
B1
Q30Energy levels & line spectra · Challenge · Paper 4 style · 6 marks
An electron with kinetic energy 12.2 eV collides with a hydrogen atom in its ground state. Energy levels of hydrogen: −13.6 eV, −3.40 eV, −1.51 eV, −0.85 eV.
(a)State which energy levels the hydrogen atom can be excited to by the collision. Explain your answer.[2]
(b)Calculate the kinetic energy of the electron after it excites an atom to the highest possible level.[1]
(c)Calculate the wavelengths of all the photons that could then be emitted.[3]
Show answer
(a) To −3.40 eV (needs 10.2 eV) or −1.51 eV (needs 12.09 eV), not −0.85 eV (needs 12.75 eV, more than 12.2 eV). Unlike photons, an electron can give up only part of its energy.
B1 levels −3.40 and −1.51 eV; B1 12.75 eV > 12.2 eV / electron can transfer part of its energy
(b) 0.11 eV
12.2 − 12.09 = 0.11 eV (A1)
(c) 1.03 × 10⁻⁷ m, 6.58 × 10⁻⁷ m and 1.22 × 10⁻⁷ m
B1 12.09 eV → 1.03 × 10⁻⁷ m; B1 1.89 eV → 6.58 × 10⁻⁷ m; B1 10.2 eV → 1.22 × 10⁻⁷ m
Similar questions
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References only: find each question in its paper (official link where available, otherwise through your school).
