A Level Physics 22.2 Photoelectric Effect MCQs – 30 Practice Questions with Answers (Cambridge 9702)
The photoelectric effect is the key experiment that forced physicists to treat light as a stream of photons. This page covers Cambridge 9702 A Level Section 22.2: the observations that wave theory cannot explain, Einstein’s photoelectric equation and the work function, and stopping-potential calculations.
Formula recap: photon energy E = h f = h c / λ · Einstein’s equation: h f = Φ + KEmax · work function Φ = h f0 · threshold wavelength λ0 = h c / Φ · stopping potential: e Vs = KEmax · h = 6.63 × 10−34 J s, c = 3.00 × 108 m/s, e = 1.6 × 10−19 C. Full formula sheet: A Level Physics Formula Sheet.
The Photoelectric Effect & Its Evidence
Q1. The photoelectric effect is the
- (A) bending of light as it passes through a narrow slit
- (B) emission of electrons from a metal surface when electromagnetic radiation falls on it
- (C) emission of light by a metal heated to a high temperature
- (D) deflection of an electron beam by a magnetic field
Show Answer
Answer: B – Electrons are emitted from the surface of a metal when electromagnetic radiation (such as ultraviolet light) shines on it. The emitted electrons are called photoelectrons.
Q2. Red light, whose frequency is below the threshold frequency of a metal, is shone on the metal at high intensity for several hours. No electrons are ever emitted. This result cannot be explained by the wave theory of light because wave theory predicts that
- (A) light of any colour should eject electrons instantly
- (B) only light above a certain frequency can carry energy
- (C) the energy delivered by the wave builds up in the metal over time, so electrons should eventually gain enough energy to escape, however weak the frequency
- (D) red light is always reflected by metals
Show Answer
Answer: C – In the wave picture, energy arrives continuously and spreads over the wavefront, so given enough intensity or enough time an electron should accumulate enough energy to escape. The observation that below the threshold frequency no emission occurs at ANY intensity or ANY exposure time contradicts this and is a central piece of evidence for photons.
Q3. Light of a fixed frequency above the threshold frequency falls on a metal. The intensity of the light is doubled. What happens to the emitted electrons?
- (A) The number emitted per second doubles, but the maximum kinetic energy of the electrons is unchanged
- (B) The maximum kinetic energy doubles, but the number emitted per second is unchanged
- (C) Both the number per second and the maximum kinetic energy double
- (D) No electrons are emitted because the light is too intense
Show Answer
Answer: A – Greater intensity at the same frequency means more photons arriving per second. Each photon still carries the same energy h f, so each electron gains the same energy and KEmax is unchanged; but more photons release more electrons per second, so the photoelectric current increases.
Q4. Which observation is evidence that photoelectric emission is instantaneous?
- (A) Electrons are emitted only when the frequency is high enough
- (B) Increasing the intensity increases the photoelectric current
- (C) Electrons from below the surface emerge with less kinetic energy
- (D) Even with very low-intensity light above the threshold frequency, electrons are emitted immediately, with no measurable time delay
Show Answer
Answer: D – Wave theory predicts a long delay for dim light while an electron gradually accumulates energy. In fact emission starts at once: a single photon hands over its whole energy to a single electron in one interaction.
Q5. The frequency of the radiation falling on a metal (already above threshold) is increased, with the intensity kept the same. What happens?
- (A) The number of electrons emitted per second increases and KEmax stays the same
- (B) KEmax of the emitted electrons increases
- (C) The work function of the metal increases
- (D) Emission stops because the photons are too energetic
Show Answer
Answer: B – KEmax = h f − Φ, so a higher frequency means each photon brings more energy and the fastest electrons leave with more kinetic energy. The work function Φ is a property of the metal and does not change.
Q6. In the photon model of the photoelectric effect, the energy transfer is described as
- (A) each photon shares its energy equally among many electrons
- (B) each electron collects energy from many photons until it can escape
- (C) one photon gives all of its energy to one electron in a one-to-one interaction
- (D) the whole wavefront transfers energy to the whole surface at once
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Answer: C – A single photon is absorbed by a single electron, which receives the photon’s entire energy h f. Part of this energy (at least the work function Φ) is used to escape from the metal; the rest becomes kinetic energy.
Q7. In a photocell, radiation of fixed frequency above threshold has its intensity doubled. The photoelectric current doubles because
- (A) twice as many photons arrive per second, so twice as many electrons are emitted per second
- (B) each electron now carries twice the charge
- (C) each electron now moves twice as fast to the collecting electrode
- (D) the work function of the cathode is halved
Show Answer
Answer: A – Current is the rate of flow of charge. Doubling the intensity doubles the photon arrival rate, doubling the rate of electron emission, while each electron’s maximum kinetic energy stays the same.
Q8. Which statement about the threshold frequency f0 of a metal is correct?
- (A) It increases as the intensity of the radiation increases
- (B) It decreases if the radiation is left shining for longer
- (C) It is the same for all metals
- (D) It depends only on the metal: below f0 no electrons are emitted, however intense the radiation
Show Answer
Answer: D – The threshold frequency is fixed by the work function of the metal, f0 = Φ / h. Below f0 each photon has energy less than Φ, so no single photon can release an electron, regardless of intensity or exposure time.
Q9. Wave theory of light predicts that increasing the intensity of the light should increase the maximum kinetic energy of the emitted electrons. Experiment shows instead that KEmax depends on
- (A) the intensity only
- (B) the exposure time only
- (C) the frequency of the radiation only
- (D) the area of the metal surface illuminated
Show Answer
Answer: C – KEmax = h f − Φ: it rises linearly with frequency and is independent of intensity. Brighter light releases MORE electrons, not FASTER electrons.
Q10. A negatively charged zinc plate is mounted on a gold-leaf electroscope. Ultraviolet light causes the leaf to fall (the plate discharges), but bright visible light does not. This is because
- (A) visible light cannot reach the plate through air
- (B) the frequency of visible light is below the threshold frequency of zinc, so each visible photon has energy less than the work function
- (C) visible light carries no energy
- (D) ultraviolet light heats the plate more strongly
Show Answer
Answer: B – Zinc has a work function of about 4.3 eV, corresponding to a threshold in the ultraviolet. Visible photons (about 1.8−3.1 eV) individually carry less than Φ, so no electrons are emitted however bright the visible light.
Einstein’s Equation & the Work Function
Q11. A metal has a threshold frequency of 5.0 × 1014 Hz. Its work function is
- (A) 3.3 × 10−19 J
- (B) 3.3 × 10−20 J
- (C) 1.3 × 10−19 J
- (D) 6.6 × 10−19 J
Show Answer
Answer: A – Φ = h f0 = 6.63 × 10−34 × 5.0 × 1014 = 3.3 × 10−19 J (about 2.1 eV).
Q12. Radiation of frequency 8.0 × 1014 Hz falls on a metal of work function 3.3 × 10−19 J. The maximum kinetic energy of the emitted electrons is
- (A) 8.6 × 10−19 J
- (B) 5.3 × 10−19 J
- (C) 3.3 × 10−19 J
- (D) 2.0 × 10−19 J
Show Answer
Answer: D – Photon energy h f = 6.63 × 10−34 × 8.0 × 1014 = 5.30 × 10−19 J. KEmax = h f − Φ = 5.30 × 10−19 − 3.3 × 10−19 = 2.0 × 10−19 J (about 1.24 eV).
Q13. A metal has a work function of 2.0 eV. Its threshold wavelength is
- (A) 310 nm
- (B) 620 nm
- (C) 1240 nm
- (D) 250 nm
Show Answer
Answer: B – Φ = 2.0 × 1.6 × 10−19 = 3.2 × 10−19 J. λ0 = h c / Φ = (6.63 × 10−34 × 3.00 × 108) / 3.2 × 10−19 = 6.2 × 10−7 m = 620 nm. Radiation of longer wavelength cannot cause emission.
Q14. Violet light of wavelength 400 nm falls on a metal of work function 2.0 eV. The maximum kinetic energy of the emitted electrons is
- (A) 5.1 eV
- (B) 3.1 eV
- (C) 1.1 eV
- (D) 0.55 eV
Show Answer
Answer: C – Photon energy E = h c / λ = 1.99 × 10−25 / 4.00 × 10−7 = 4.97 × 10−19 J = 3.1 eV. KEmax = 3.1 − 2.0 = 1.1 eV.
Q15. A metal has a threshold wavelength of 500 nm. Its threshold frequency is
- (A) 6.0 × 1014 Hz
- (B) 3.0 × 1014 Hz
- (C) 1.5 × 1015 Hz
- (D) 6.0 × 1013 Hz
Show Answer
Answer: A – f0 = c / λ0 = 3.00 × 108 / 5.00 × 10−7 = 6.0 × 1014 Hz.
Q16. Photons of energy 5.0 eV strike a metal of work function 2.3 eV. The maximum kinetic energy of the photoelectrons is
- (A) 7.3 eV
- (B) 5.0 eV
- (C) 2.3 eV
- (D) 2.7 eV
Show Answer
Answer: D – Working in electronvolts is simplest: KEmax = h f − Φ = 5.0 − 2.3 = 2.7 eV.
Q17. The energy of one photon of ultraviolet radiation of wavelength 250 nm is
- (A) 2.5 eV
- (B) 5.0 eV
- (C) 7.5 eV
- (D) 2.0 eV
Show Answer
Answer: B – E = h c / λ = 1.99 × 10−25 / 2.50 × 10−7 = 7.96 × 10−19 J. Dividing by 1.6 × 10−19 gives 5.0 eV.
Q18. No electrons are emitted from a metal surface when the frequency of the radiation is below 4.5 × 1014 Hz. The work function of the metal is
- (A) 1.5 × 10−19 J
- (B) 4.5 × 10−19 J
- (C) 3.0 × 10−19 J
- (D) 6.6 × 10−19 J
Show Answer
Answer: C – The frequency given is the threshold frequency, so Φ = h f0 = 6.63 × 10−34 × 4.5 × 1014 = 3.0 × 10−19 J (about 1.9 eV).
Q19. Emitted photoelectrons have kinetic energies ranging from zero up to KEmax, rather than all having KEmax, because
- (A) electrons below the surface lose energy in collisions with other atoms as they travel to the surface; only electrons at the surface escape with the full KEmax
- (B) the photons arrive with a range of different frequencies
- (C) the work function varies randomly across the surface
- (D) some photons give their energy to two electrons at once
Show Answer
Answer: A – Every absorbed photon gives its electron h f of energy, and at least Φ is needed to escape. An electron that starts below the surface loses extra energy in collisions on its way out, so it emerges with less than KEmax; electrons deep inside may not escape at all.
Q20. Light of wavelength 350 nm falls on a metal of work function 2.3 eV. The maximum kinetic energy of the emitted electrons is approximately
- (A) 3.6 eV
- (B) 5.9 eV
- (C) 2.3 eV
- (D) 1.3 eV
Show Answer
Answer: D – Photon energy E = h c / λ = 1.99 × 10−25 / 3.50 × 10−7 = 5.68 × 10−19 J = 3.55 eV. KEmax = 3.55 − 2.3 = 1.25 eV, approximately 1.3 eV.
Stopping Potential & Photoelectric Calculations
Q21. Photoelectrons are emitted with a maximum kinetic energy of 1.6 × 10−19 J. The stopping potential is
- (A) 0.50 V
- (B) 1.0 V
- (C) 1.6 V
- (D) 2.6 V
Show Answer
Answer: B – e Vs = KEmax, so Vs = 1.6 × 10−19 / 1.6 × 10−19 = 1.0 V. (A KEmax of 1 eV always corresponds to a stopping potential of 1 V.)
Q22. The stopping potential Vs in a photoelectric experiment is the
- (A) minimum retarding potential difference that reduces the photoelectric current to zero, stopping even the fastest electrons
- (B) potential difference needed to accelerate the electrons to the collecting electrode
- (C) potential difference at which the metal starts to emit electrons
- (D) work function of the metal measured in volts
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Answer: A – A retarding potential opposes the motion of the photoelectrons. When e Vs equals KEmax, even the most energetic electrons are turned back just before reaching the collector, and the current falls to zero.
Q23. Ultraviolet light of wavelength 300 nm falls on a metal of work function 2.0 eV. The stopping potential is
- (A) 4.1 V
- (B) 6.1 V
- (C) 1.4 V
- (D) 2.1 V
Show Answer
Answer: D – Photon energy E = h c / λ = 1.99 × 10−25 / 3.00 × 10−7 = 6.63 × 10−19 J = 4.14 eV. KEmax = 4.14 − 2.0 = 2.14 eV, so Vs = 2.14 V, approximately 2.1 V.
Q24. A graph of KEmax of the photoelectrons against the frequency f of the incident radiation is a straight line. The gradient of this line is
- (A) the work function Φ
- (B) the threshold frequency f0
- (C) the Planck constant h
- (D) the elementary charge e
Show Answer
Answer: C – KEmax = h f − Φ is the equation of a straight line with gradient h. Measuring this gradient is one way of determining the Planck constant; the gradient is the same for every metal.
Q25. On the graph of KEmax against frequency f, the line crosses the frequency axis at the
- (A) threshold frequency f0, and its extrapolated intercept on the KEmax axis is −Φ
- (B) frequency of visible light
- (C) point where the intensity is zero
- (D) maximum frequency of the radiation source
Show Answer
Answer: A – KEmax = 0 when h f = Φ, i.e. at f = f0. Extending the line back to f = 0 gives KEmax = −Φ. Below f0 there is no emission, so the graph has no values there.
Q26. Radiation of frequency 1.0 × 1015 Hz falls on a metal of work function 3.0 eV. The stopping potential is
- (A) 4.1 V
- (B) 3.0 V
- (C) 1.1 V
- (D) 7.1 V
Show Answer
Answer: C – Photon energy h f = 6.63 × 10−34 × 1.0 × 1015 = 6.63 × 10−19 J = 4.14 eV. KEmax = 4.14 − 3.0 = 1.14 eV, so Vs is approximately 1.1 V.
Q27. Light of wavelength 200 nm falls on a metal of work function 4.5 eV. The maximum kinetic energy of the emitted electrons is
- (A) 6.2 eV
- (B) 1.7 eV
- (C) 10.7 eV
- (D) 4.5 eV
Show Answer
Answer: B – Photon energy E = h c / λ = 1.99 × 10−25 / 2.00 × 10−7 = 9.95 × 10−19 J = 6.2 eV. KEmax = 6.2 − 4.5 = 1.7 eV.
Q28. With the frequency of the incident radiation kept constant above threshold, the intensity is increased. What happens to the stopping potential and the photoelectric current?
- (A) Both increase
- (B) The stopping potential increases and the current is unchanged
- (C) The stopping potential decreases and the current increases
- (D) The stopping potential is unchanged and the current increases
Show Answer
Answer: D – Intensity changes only the number of photons per second, not the energy of each photon. KEmax, and therefore Vs, is unchanged, but more electrons are emitted per second so the current is larger.
Q29. Radiation of frequency 7.5 × 1014 Hz falls on a metal of work function 2.5 eV. The stopping potential is approximately
- (A) 0.61 V
- (B) 3.1 V
- (C) 2.5 V
- (D) 5.6 V
Show Answer
Answer: A – Photon energy h f = 6.63 × 10−34 × 7.5 × 1014 = 4.97 × 10−19 J = 3.11 eV. KEmax = 3.11 − 2.5 = 0.61 eV, so Vs is approximately 0.61 V.
Q30. Metal A has a work function of 2.0 eV and metal B has a work function of 4.0 eV. The same ultraviolet radiation falls on both. Which statement is correct?
- (A) Metal B has the lower threshold frequency and the longer threshold wavelength
- (B) Both metals emit electrons with the same maximum kinetic energy
- (C) Metal B has the higher threshold frequency and the shorter threshold wavelength, and metal A emits electrons with the larger KEmax
- (D) Neither metal can emit electrons with ultraviolet light
Show Answer
Answer: C – f0 = Φ / h, so the larger work function of B gives it a higher threshold frequency and (since λ0 = c / f0) a shorter threshold wavelength. For the same incident photons, KEmax = h f − Φ is larger for the metal with the smaller work function, A.
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