Edexcel International A Level Physics · Unit 2 · Waves and Electricity
Photons, the photoelectric effect and line spectra
40 free, high-quality past-paper-style questions on Photons, the photoelectric effect and line spectra for Edexcel IAL Physics, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Physics notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.40 new this month · 149 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked
1IALP-U2-5-A-01Unit 2 · ASEasy1 mark
Which observation is evidence for the wave model of light rather than the photon model?
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Interference fringes form where waves from the two slits arrive in phase (bright) and in antiphase (dark). Particles cannot cancel each other out, so two-slit interference supports the wave model. The other three are explained by photons.
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Answer A
Answer: A
Exam tip: Wave evidence: diffraction, interference, polarisation. Photon evidence: the photoelectric effect and line spectra (emission or absorption by electrons).
Monochromatic light above the threshold frequency falls on a metal surface and photoelectrons are emitted. The intensity of the light is doubled without changing its frequency. Which row describes the effect on the photoelectrons?
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Doubling the intensity doubles the number of photons arriving per second, so twice as many electrons are emitted each second. Each photon still has energy hf, so the maximum kinetic energy hf − φ is unchanged, with the number per second doubled.
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Answer B
Answer: B
Exam tip: Frequency sets the energy of each photon; intensity sets how many photons arrive each second.
A metal used for the light-sensitive surface of a photocell has a work function of 2.9 eV.
(a)
Define work function.
(1)
(b)
Define threshold frequency.
(1)
(c)
Calculate the threshold frequency for this metal.
(2)
(d)
Red light of wavelength 650 nm falls on the metal. State and explain whether electrons are emitted.
(1)
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Part (a)
the minimum energy needed to remove an electron from the surface of the metal
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Part (b)
the minimum frequency of electromagnetic radiation that will release electrons from the surface of the metal
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Part (c)
f0 = φ/h = 2.9 × 1.60 × 10−19 ÷ 6.63 × 10−34
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= 7.0 × 1014 Hz
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Part (d)
no: each photon has energy hc/λ = 3.06 × 10−19 J = 1.9 eV, less than the 2.9 eV work function / the frequency 4.6 × 1014 Hz is below the threshold frequency
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Exam tip: “Minimum” is the key word in both definitions. Without it, neither definition is complete.
Figure 1 shows four experiments, P, Q, R and S, carried out with light or ultraviolet radiation. Which experiment gives evidence for the particle (photon) nature of electromagnetic radiation?
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In R, photons of ultraviolet each give enough energy to one electron to free it from the zinc, so the plate loses negative charge at once. Emission that is immediate and depends on frequency is explained by photons, so R. Diffraction (P), polarisation (Q) and dispersion by a prism (S) are all explained by the wave model.
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Answer C
Answer: C
Exam tip: Sort the evidence: diffraction, interference and polarisation need waves; the photoelectric effect and line spectra need photons.
The graph shows how the energy E of a photon depends on its frequency f. Point P lies on the line. What value of the Planck constant does the graph give?
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E = hf, so h is the gradient. Using P: h = 4.97 × 10−19 ÷ 7.5 × 1014 = 6.6 × 10−34 J s.
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Answer A
Answer: A
Common error: Reading the numbers off the axes and forgetting the powers of ten in the axis labels: E / 10−19 J means each number must be multiplied by 10−19.
A metal surface is illuminated by radiation of one frequency, greater than the threshold frequency. The intensity is gradually increased from a low value. Which graph, A to D, shows how the maximum kinetic energy of the photoelectrons varies with the intensity?
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Maximum kinetic energy = hf − φ, which depends only on the frequency. A higher intensity means more photons per second, each with the same energy, so the maximum kinetic energy stays constant: graph A.
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Answer A
Answer: A
Exam tip: Intensity controls how many photons arrive each second; frequency controls the energy of each photon.
The graph shows the spectrum of the light emitted by a white light-emitting diode (LED) in a desk lamp. The blue light comes from the LED chip itself; the broad peak comes from a phosphor coating on the chip.
(a)
Use the graph to state the wavelength at which the blue emission is greatest.
(1)
(b)
Calculate the energy of one photon at this wavelength. Give your answer in eV.
(2)
(c)
The phosphor absorbs blue photons and emits photons at the wavelength of the broad peak. Calculate the energy, in eV, that is not re-emitted when one blue photon is replaced by one photon at the broad peak.
Figure 1 shows a clean zinc plate fixed to a gold-leaf electroscope. The plate and the leaf are negatively charged, so the leaf is raised. When the ultraviolet lamp is switched on, the leaf falls steadily.
(a)
Explain why the leaf falls.
(2)
(b)
Calculate the longest wavelength of electromagnetic radiation that can make the leaf fall.
(2)
(c)
The ultraviolet lamp is replaced by a very bright lamp that emits only visible light. Explain why the leaf does not fall.
(1)
(d)
The electroscope is now given a positive charge and the ultraviolet lamp is switched on again. Suggest why the leaf stays raised.
(1)
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Part (a)
each ultraviolet photon has enough energy (at least the work function) to release an electron from the zinc
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the plate, rod and leaf lose negative charge, so the repulsion between leaf and rod decreases and the leaf falls
visible photons have energies below about 3.3 eV, less than 4.3 eV, so no single photon can release an electron; a brighter lamp only sends more of these photons
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Part (d)
any electrons released are attracted back to the positively charged plate, so the plate does not lose its charge
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Common error: Writing that the radiation “gives the plate energy”. Say that each photon releases one electron, and that removing electrons reduces the negative charge.
Figure 1 shows two spectra of the same element. Spectrum 1 is from a lamp containing the hot gas. Spectrum 2 is seen when white light passes through a cool sample of the vapour.
(a)
Explain how the bright lines in spectrum 1 are produced.
(2)
(b)
Use the scale to determine the wavelength of line L, and calculate the energy difference, in eV, between the two levels that produce it.
(3)
(c)
Suggest why spectrum 2 has dark lines at only two of the three wavelengths in spectrum 1.
(1)
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Part (a)
electrons in excited atoms move down to lower energy levels
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each transition emits one photon with hf = E1 − E2; the levels are discrete, so only certain frequencies (wavelengths) are emitted
in the cool vapour almost all atoms are in the ground state, so only photons that lift electrons from the ground state are absorbed; line L comes from a transition between two excited levels
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Exam tip: Emission lines come from electrons falling between any two levels; absorption lines in a cool gas come only from transitions that start at the ground state.
The stopping potential for photoelectrons emitted from a metal surface is 1.4 V. Mass of electron = 9.11 × 10−31 kg. What is the maximum speed of the photoelectrons?
A water-purifying unit contains a lamp that emits ultraviolet radiation of wavelength 254 nm. The lamp's electrical input power is 4.0 W and 30% of this is emitted as radiation of this wavelength.
(a)
Calculate the energy of one photon of wavelength 254 nm.
(2)
(b)
Calculate the number of these photons emitted per second.
(2)
(c)
Radiation from this lamp can break chemical bonds in the genetic material of bacteria, but light from a much more powerful red lamp cannot. Explain why, using the photon model.
(2)
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Part (a)
E = hc/λ = 6.63 × 10−34 × 3.00 × 108 ÷ 254 × 10−9
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= 7.83 × 10−19 J
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Part (b)
power emitted at 254 nm = 0.30 × 4.0 = 1.2 W
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N = 1.2 ÷ 7.83 × 10−19 = 1.5 × 1018 s−1
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Part (c)
bonds are broken by single photons, one photon per interaction, so what matters is the energy of each photon, not the power of the beam
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ultraviolet photons have a much higher frequency and so more energy (E = hf) than red photons, enough to break a bond; red photons have too little, however many arrive
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Common error: Using the electrical input power instead of the power emitted at the stated wavelength. Only 30% of the 4.0 W becomes 254 nm photons.
An energy level diagram for an atom shows the following levels: ground state −6.40 eV; first excited state −3.10 eV; second excited state −1.85 eV; third excited state −0.90 eV; ionisation 0 eV.
Figure: energy level diagram for the atom (not to scale)
(a)
Explain why the energy levels are given negative values.
(1)
(b)
State the ionisation energy of the atom in joules.
(1)
(c)
Calculate the wavelength of the photon emitted when an electron falls from the second excited state to the ground state.
(2)
(d)
Photons of energy 3.00 eV pass through a gas of these atoms, all in the ground state. Explain whether the photons are absorbed.
(2)
(e)
State the number of different photon energies that could be emitted by atoms excited to the third excited state.
(1)
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Part (a)
zero is defined as the energy of an electron that has just been removed from the atom; a bound electron has less energy than this, so its energy is negative
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Part (b)
6.40 × 1.60 × 10−19 = 1.02 × 10−18 J
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Part (c)
ΔE = −1.85 − (−6.40) = 4.55 eV = 7.28 × 10−19 J
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λ = hc/ΔE = 2.73 × 10−7 m
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Part (d)
a photon is absorbed only if its energy exactly equals the difference between two levels
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the smallest gap from the ground state is 3.30 eV, so 3.00 eV photons are not absorbed (they pass through)
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Part (e)
6
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Exam tip: An energy-level question nearly always needs the gap ΔE = E1 − E2, converted from eV to J before using h.
In a teaching laboratory, white light is passed through a glass cell containing a cool metal vapour and then through a spectrometer. The continuous spectrum observed is crossed by two dark lines, at 525 nm and 390 nm.
Figure: continuous spectrum crossed by dark lines at 390 nm and 525 nm (not to scale)
(a)
Explain how the dark lines are formed.
(3)
(b)
Calculate the energy difference, in eV, that corresponds to the 525 nm line.
(2)
(c)
The vapour is then heated strongly in the dark. State and explain what is seen through the spectrometer.
(2)
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Part (a)
electrons in the vapour atoms can only occupy discrete energy levels
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a photon is absorbed only if its energy equals the difference between two levels, raising an electron to a higher level
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so light at only these wavelengths is removed from the beam; the atoms re-emit the energy in all directions, so these wavelengths are much fainter in the original direction
bright lines on a dark background (an emission spectrum)
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including lines at the same wavelengths, 525 nm and 390 nm, because electrons fall back across the same energy gaps (other lines may also appear, from transitions between higher levels)
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Model sketch
Figure: emission spectrum from the hot vapour: bright lines at the same two wavelengths, with no continuous spectrum between them (not to scale)
Exam tip: Absorption lines and emission lines of the same element share wavelengths because the same pairs of energy levels are involved.
A student estimates the Planck constant using light-emitting diodes (LEDs). For each LED, the student slowly increases the potential difference across it and records the value V at which it just starts to emit light. The student assumes that at this p.d. the energy eV transferred to each electron is just enough to produce one photon of the LED's wavelength λ. For a red LED of wavelength 630 nm, V = 1.95 V.
Figure: V against 1/λ for the five LEDs: straight line through the origin, gradient 1.24 × 10⁻⁶ V m
(a)
Show that the student's assumption leads to h = eVλ/c.
(2)
(b)
Calculate the value of h obtained from the red LED.
(2)
(c)
Using five LEDs, the student plots V against 1/λ and obtains a straight line through the origin with gradient 1.24 × 10−6 V m. Determine h from this gradient.
(2)
(d)
The voltmeter has a resolution of 0.01 V. Calculate the percentage uncertainty in the reading of 1.95 V.
(1)
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Part (a)
energy transferred to one electron = eV; photon energy = hf = hc/λ
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eV = hc/λ, so h = eVλ/c
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Part (b)
h = 1.60 × 10−19 × 1.95 × 630 × 10−9 ÷ 3.00 × 108
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= 6.55 × 10−34 J s
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Part (c)
V = (hc/e)(1/λ), so gradient = hc/e
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h = 1.24 × 10−6 × 1.60 × 10−19 ÷ 3.00 × 108 = 6.61 × 10−34 J s
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Part (d)
half the resolution = 0.005 V; 0.005 ÷ 1.95 × 100 = 0.26% (allow 0.01 ÷ 1.95 × 100 = 0.51% if the full resolution is taken as the uncertainty)
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Exam tip: Rearrange the physics into the form y = mx to see what the gradient of a graph represents before you calculate anything from it.
The graph shows how the maximum kinetic energy of the electrons emitted from a metal surface varies with the frequency f of the radiation falling on it. What is the work function of the metal?
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The line meets the frequency axis at the threshold frequency f0 = 5.0 × 1014 Hz. φ = hf0 = 6.63 × 10−34 × 5.0 × 1014 = 3.3 × 10−19 J = 2.1 eV. (Point P gives the same: hf − Ek = 5.97 × 10−19 − 2.65 × 10−19 = 3.3 × 10−19 J.)
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Answer B
Answer: B
Exam tip: On a graph of maximum kinetic energy against frequency, the intercept on the frequency axis is the threshold frequency, and φ = hf0.
The energy level diagram shows four levels of an atom and four transitions, P, Q, R and S. Which transition emits the photon with the longest wavelength?
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λ = hc/ΔE, so the longest wavelength comes from the smallest energy gap. Gaps: P 6.10 eV, Q 2.50 eV, R 3.70 eV, S 3.60 eV. The smallest is Q (2.50 eV, 497 nm).
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Answer B
Answer: B
Common error: Picking the longest arrow. A bigger energy gap gives a photon of higher frequency and so a shorter wavelength.
The bar chart shows the work functions of five metals. Each metal in turn is illuminated with light of wavelength 400 nm. From how many of the metals are photoelectrons emitted?
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Photon energy = hc/λ = 6.63 × 10−34 × 3.00 × 108 ÷ 400 × 10−9 = 4.97 × 10−19 J = 3.11 eV. This is greater than the work functions of caesium, potassium and calcium only, so electrons are emitted from 3 metals.
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Answer C
Answer: C
Exam tip: Convert the photon energy into eV, then compare it with each work function. Emission needs hf ≥ φ, however bright the light.
The graph shows how the photocurrent from a photocell depends on the intensity of the light falling on its cathode. The frequency of the light is kept constant and every emitted electron reaches the anode. How many electrons leave the cathode each second when the intensity is 35 W m−2?
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From the graph, I = 7.0 µA at 35 W m−2. Number per second = I/e = 7.0 × 10−6 ÷ 1.60 × 10−19 = 4.4 × 1013 s−1.
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Answer B
Answer: B
Common error: Dropping the µ: 1 µA = 10−6 A. Convert the current to amperes before dividing by e.
The chart shows the output power and the wavelength of four lasers. Which laser emits the greatest number of photons per second?
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Photons per second N = P/E = Pλ/hc, so N ∝ Pλ. In mW nm: P 810, Q 2128, R 3196, S 2925. The greatest is laser R (1.6 × 1016 photons per second): each infrared photon carries less energy, so more photons are needed for the same power.
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Answer C
Answer: C
Exam tip: For a given power, longer-wavelength photons each carry less energy, so more of them are emitted per second.
The energy level diagram shows the four lowest energy levels of an atom, drawn to scale, with one transition, T, marked. Visible light has wavelengths from 380 nm to 700 nm.
(a)
Calculate the wavelength of the photon emitted in transition T.
(2)
(b)
Determine which of the possible downward transitions between the levels shown emit visible light.
(3)
(c)
An atom in the ground state absorbs a photon of energy 6.50 eV. State what happens to the atom.
(1)
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Part (a)
ΔE = −3.30 − (−5.50) = 2.20 eV = 3.52 × 10−19 J
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λ = hc/ΔE = 5.65 × 10−7 m
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Part (b)
visible photon energies run from hc/700 nm = 1.78 eV to hc/380 nm = 3.27 eV
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the six gaps are 2.90, 5.10, 6.50, 2.20, 3.60 and 1.40 eV
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only T (2.20 eV) and the drop from −5.50 eV to the ground state (2.90 eV) are visible
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Part (c)
an electron is raised from the ground state to the −1.90 eV level (the photon energy exactly equals this gap)
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Exam tip: Turn a wavelength range into an energy range first; then you only have to compare each energy gap with two numbers.
The timeline in Figure 1 shows some of the steps in the development of models of light.
(a)
Identify the observation on the timeline that led scientists to prefer a wave model to Newton’s particle model, and explain why it did so.
(2)
(b)
Explain how Maxwell’s work in 1865 strengthened the wave model.
(1)
(c)
Hertz’s observation in 1887 was an early sign of the photoelectric effect. Explain why Einstein needed a photon model to account for the photoelectric effect.
(2)
(d)
Suggest why Millikan’s results in 1916 were important for the photon model.
(1)
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Part (a)
Young’s two-slit interference fringes (1801)
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dark fringes need waves to superpose and cancel (arrive in antiphase); streams of particles cannot cancel each other
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Part (b)
his theory predicted electromagnetic waves travelling at a speed equal to the measured speed of light, so light was identified as an electromagnetic wave
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Part (c)
in the wave model energy is spread continuously over the wave, so any frequency should eventually release electrons and brighter light should give faster electrons, which is not observed
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Einstein proposed energy arrives in photons of E = hf, each absorbed by one electron, so emission needs hf ≥ φ (a threshold frequency)
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Part (d)
they tested a prediction of the model: maximum kinetic energy = hf − φ gives a straight line of gradient h, so the results supported (gave evidence for) the photon model
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Exam tip: Models of light changed when new evidence could not be explained. Today both models are used: waves for diffraction and interference, photons for emission and absorption.
energy = area under the power–time graph = ½ × 2.0 × 8.0 + 4.0 × 8.0 + ½ × 2.0 × 8.0 = 48 mW ms
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= 4.8 × 10−5 J
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Part (c)
4.8 × 10−5 ÷ 3.75 × 10−19 = 1.3 × 1014
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Part (d)
8.0 × 10−3 ÷ 3.75 × 10−19 = 2.1 × 1016 s−1
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Part (e)
more: each red photon has less energy (E = hc/λ), so more photons carry the same total energy
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Common error: Multiplying the peak power by the whole pulse time. The power is not constant, so the energy is the area under the graph; and mW × ms = µJ (10−6 J).
Figure 1 shows a photocell connected to a d.c. supply and a microammeter. Ultraviolet radiation from an LED falls on the cathode. The wavelength, the power reaching the cathode, the work function and the meter reading are shown on the figure.
(a)
Calculate the energy of one photon in eV.
(2)
(b)
Calculate the maximum kinetic energy of the emitted electrons in eV.
(1)
(c)
Calculate the number of photons reaching the cathode per second.
(1)
(d)
Calculate the number of electrons reaching the anode per second.
(1)
(e)
Calculate the percentage of incident photons that release an electron.
(1)
(f)
Explain why the microammeter reading does not change when the supply p.d. is increased to 12 V.
2.5 × 1012 ÷ 2.75 × 1015 × 100 = 0.091% (allow 0.089% from the rounded value in (c))
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Part (f)
the anode already collects every emitted electron, so the current depends only on the rate of emission, which is set by the number of photons per second
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Exam tip: Photons per second = power ÷ photon energy; electrons per second = current ÷ e. Keep the two ideas separate.
An atom has four energy levels, including the ground state. Its electrons are excited to the highest of these levels and then return to the ground state by any possible route. How many different photon energies can be emitted?
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Every pair of levels gives a different transition: with 4 levels there are 4 × 3 ÷ 2 = 6 pairs (4→3, 4→2, 4→1, 3→2, 3→1, 2→1).
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Answer D
Answer: D
Exam tip: Draw every downward arrow between pairs of levels and count them; for n levels there are n(n − 1)/2.
Two energy levels of an atom are at −5.12 eV and −2.04 eV. An electron moves from the higher of these levels to the lower. What is the wavelength of the photon emitted?
Figure: the two energy levels and the transition (not to scale)
When ultraviolet radiation falls on a clean metal plate, electrons are emitted at once. Visible light, however bright, releases no electrons from the same plate, and making the ultraviolet more intense does not increase the maximum kinetic energy of the electrons.
(a)
Explain how these observations provide evidence for the photon model of electromagnetic radiation and why the wave model cannot account for them.
(6)
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Indicative content:
—
radiation consists of photons, each with energy E = hf
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one photon is absorbed by one electron and gives it all its energy
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an electron needs at least the work function to escape, so emission needs hf ≥ φ; visible photons have too little energy, whatever the intensity
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in the wave model energy would build up over time, so bright visible light should eventually release electrons, and dim light would show a delay
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the maximum kinetic energy is hf − φ, which depends on frequency only
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greater intensity means more photons per second, so more electrons per second but not faster ones; the wave model predicts faster electrons in brighter light
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Exam tip: In a 6-mark explanation, link each observation to a step in the photon model, and say what the wave model would have predicted instead.
A student uses a vacuum photocell and a set of filters to investigate the photoelectric effect. For each frequency f of light, the student increases the reverse potential difference across the photocell until the photocurrent is zero, and records this stopping potential Vs. f / 1014 Hz: 5.5, 6.5, 7.5, 8.5, 9.5 Vs / V: 0.36, 0.81, 1.20, 1.63, 2.03
(a)
Explain why the photocurrent falls to zero when the reverse potential difference reaches Vs.
(1)
(b)
Plot a graph of Vs against f and use it to determine a value for the Planck constant.
(3)
(c)
Determine the work function of the metal in eV.
(2)
(d)
Determine the threshold frequency.
(1)
(e)
The voltmeter has a resolution of 0.01 V. Calculate the percentage uncertainty in the smallest reading of Vs.
(1)
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Part (a)
the work done against the field, eVs, equals the maximum kinetic energy of the photoelectrons, so even the fastest electrons are stopped before reaching the anode
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Part (b)
points plotted correctly and a straight best-fit line drawn
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gradient from a large triangle ≈ 4.2 × 10−15 V s (accept 4.0–4.4 × 10−15)
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h = e × gradient = 6.7 × 10−34 J s (accept 6.4–7.0 × 10−34 J s)
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Part (c)
use a point on the line in eVs = hf − φ, or extrapolate to −φ/e on the Vs axis
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φ = 1.9 eV (accept 1.8–2.0 eV)
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Part (d)
intercept on the f axis: f0 = 4.6 × 1014 Hz (accept 4.4–4.8 × 1014 Hz)
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Part (e)
uncertainty = half the resolution = 0.005 V; 0.005 ÷ 0.36 × 100 = 1.4% (allow 0.01 ÷ 0.36 × 100 = 2.8% if the full resolution is taken as the uncertainty)
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Model sketch
Figure: Vₛ against f with the best-fit straight line; the gradient is h/e and the line meets the f-axis at the threshold frequency
Common error: Taking the gradient of Vs against f as h. Because the vertical axis is a potential difference, the gradient is h/e.
Far-ultraviolet radiation from the Sun, of wavelength 120 nm, falls on the aluminium-coated surface of a spacecraft. The work function of the coating is 4.1 eV. Mass of electron = 9.11 × 10−31 kg.
(a)
Calculate the energy of one photon of this radiation 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)
As electrons leave, the surface becomes positively charged. Explain why the emission of electrons stops once the surface is about 6.3 V positive relative to its surroundings.
(2)
(e)
Explain why visible sunlight of wavelength 500 nm releases no electrons from this surface.
The diagram shows the energy levels of the atoms in a cool gas, which are almost all in the ground state. White light passes through the gas and is then viewed with a spectrometer. Which spectrum, A, B, C or D, is seen?
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Atoms in the ground state absorb photons whose energy equals a gap from the ground state: 1.90, 2.60 and 3.10 eV, so λ = hc/ΔE = 654, 478 and 401 nm. The spectrum is continuous with three dark lines at these wavelengths: D.
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Answer D
Answer: D
Exam tip: Absorption by a cool gas starts from the ground state, so work out the gaps from the ground state to each higher level, then convert each to a wavelength.
The diagram shows three energy levels of an atom. The photons emitted in two of the transitions have the wavelengths shown. What is the wavelength of the photon emitted in the transition from level 3 to level 2?
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Energies add: E3→1 = E3→2 + E2→1, and E = hc/λ, so 1/λ = 1/400 − 1/600 = 1/1200. λ = 1200 nm (infrared).
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Answer D
Answer: D
Exam tip: Energies of transitions add and subtract, wavelengths do not. Work with 1/λ (or convert each wavelength to an energy first).
A student shines monochromatic light of two different frequencies, in turn, on the cathode of a vacuum photocell. For each frequency the student measures the photocurrent for a range of p.d.s between the anode and the cathode. The graph shows the results: curve X is for 6.0 × 1014 Hz and curve Y is for 8.0 × 1014 Hz.
(a)
Use the graph to state the stopping potential for each frequency.
(1)
(b)
Explain why each current stays constant when the anode is more than about 1 V positive.
(1)
(c)
Determine a value for the Planck constant from these results.
(3)
(d)
Determine the work function of the cathode in eV.
(2)
(e)
Calculate the threshold frequency of the cathode.
(1)
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Part (a)
X: 0.60 V and Y: 1.40 V (both needed; allow ±0.05 V)
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Part (b)
every electron emitted is already collected by the anode, so the current is limited by the rate of emission, which depends on the number of photons per second
Figure 1 shows three sets of results for photoelectric emission from one metal surface. In every experiment, electrons are emitted with no detectable delay, even at the lowest intensities used.
(a)
Explain how Graph 1 supports the photon model of electromagnetic radiation.
(2)
(b)
Use the photon model to explain the shapes of Graph 2 and Graph 3.
(2)
(c)
Explain why a wave model cannot account for Graph 1, Graph 2 and the absence of any delay.
(2)
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Answer
Mark
Notes
Part (a)
one photon gives all its energy hf to one electron
✓ 1
below f0, hf is less than the work function so no electron escapes; above it the maximum kinetic energy is hf − φ, a straight line of gradient h
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Part (b)
at a fixed frequency every photon has the same energy, so a higher intensity (more photons per second) does not change the maximum kinetic energy
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each photon can release at most one electron, so the number of electrons per second is proportional to the number of photons per second, which is proportional to intensity
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Part (c)
in a wave model energy arrives continuously, so radiation of any frequency would release electrons if intense enough (no threshold), and more intense waves would give faster electrons
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at low intensity an electron would need time to collect enough energy from the wave, so emission would be delayed
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Exam tip: For each observation, give the photon explanation and then what a wave model would have predicted. The contrast is what earns the marks.
Figure 1 shows a tube containing a gas at low pressure. Electrons from the cathode are accelerated by the 7.0 V supply and collide with gas atoms, which are all in the ground state. Assume that every collision happens after an electron has been accelerated through the full 7.0 V. The energy levels of the gas atoms are also shown.
(a)
Calculate the kinetic energy, in J, of an electron just before a collision.
(1)
(b)
Explain why these electrons can excite atoms to level n = 2 or n = 3 but not to level n = 4.
(2)
(c)
The excited atoms then emit photons. Determine all the photon energies that can be emitted and calculate the shortest wavelength.
(3)
(d)
State the kinetic energy, in eV, that an electron keeps after exciting an atom to level n = 3.
(1)
(e)
The supply p.d. is increased to 10.0 V. State one new effect that collisions can now have on atoms in the ground state.
(1)
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Answer
Mark
Notes
Part (a)
7.0 × 1.60 × 10−19 = 1.1 × 10−18 J (7.0 eV)
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Part (b)
exciting from the ground state needs 4.30 eV (n = 2) or 6.50 eV (n = 3), both less than 7.0 eV; an electron can give up just part of its kinetic energy
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reaching n = 4 needs 7.90 eV, more than the 7.0 eV available
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Part (c)
transitions 3→2, 3→1 and 2→1 give 2.20 eV, 6.50 eV and 4.30 eV
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shortest wavelength from the largest energy, 6.50 eV = 1.04 × 10−18 J
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λ = hc/E = 1.91 × 10−7 m
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Part (d)
7.0 − 6.50 = 0.50 eV
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Part (e)
10.0 eV is more than the 9.60 eV ionisation energy, so atoms can now be ionised (an electron removed completely); excitation to n = 4 also becomes possible
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Exam tip: A photon must be absorbed whole, so its energy must match a gap exactly; a colliding electron can hand over any part of its kinetic energy.
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