72 free, high-quality past-paper-style questions on Quantum physics (HL) for IB Physics HL, each with a full mark scheme. Covers: Intensity and the photoelectric current; Compton scattering; Electrons through a double slit; and more. For when you have finished the past papers, or want to practise one topic at a time.Read the E.2 notes →My revision plan →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.328 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked
1E-1A-28Intensity and the photoelectric currentPaper 1AEasy1 mark
Monochromatic light falls on the cathode of the photocell shown. The pd of the supply is adjusted and the current is measured. The intensity of the light is then doubled, with the frequency unchanged.
Which row gives the effect on the maximum (saturation) current and on the stopping potential?
The supply can make the collector positive or negative with respect to the cathode.
Saturation currentStopping potential
Show mark scheme
Marking point
Mark
Notes
Step 1Doubling the intensity at fixed frequency doubles the number of photons per second, so twice as many electrons are emitted per second: the saturation current doubles.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2Each photon still has the same energy hf, so the maximum kinetic energy hf − Φ of the electrons is unchanged.
—
Step 3The stopping potential Vs = (hf − Φ)/e therefore stays the same.
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: more photons, same photon energy.
BThe stopping potential depends only on the frequency (and the work function), not on the intensity.
CIntensity affects the number of electrons, not their energy; the current does change.
DThe current changes because the number of photons per second changes.
Photons are scattered by free electrons. Which graph shows how the shift Δλ in the wavelength of the scattered photons varies with the scattering angle θ?
Four graphs of the shift Δλ in the wavelength of the scattered photon against the scattering angle θ (λC is the Compton wavelength of the electron).Show mark scheme
Marking point
Mark
Notes
Step 1Δλ = (h/mec)(1 − cos θ) = λC(1 − cos θ).
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2θ = 0: Δλ = 0; θ = 90°: Δλ = λC; θ = 180°: Δλ = 2λC, and Δλ is never negative.
—
Step 3Only graph D has these values.
✓ 1
Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThis is λC cos θ: the "1 −" is missing, giving a shift at θ = 0 and a negative shift (shorter wavelength) for back-scattering.
BThis is λC sin θ: the trigonometric function has been swapped, putting the largest shift at 90°.
CThis is λC(1 + cos θ): a sign error that gives the largest shift for photons that are not deflected at all.
DCorrect: λC(1 − cos θ) rises from 0 to 2λC at 180°.
3IBP2H-037-01-01Compton scatteringPaper 1AEasy1 mark
In a Compton scattering experiment, X-rays of wavelength λ are scattered by free electrons. A detector at a scattering angle of 60° measures a wavelength shift Δλ. The source is replaced by one that emits X-rays of wavelength 2λ, and the detector stays at 60°. What is the new wavelength shift?
Show mark scheme
Marking point
Mark
Notes
The Compton shift is Δλ = (h/mec)(1 − cos θ). It depends only on the scattering angle and the electron mass, not on the incident wavelength. At the same angle the shift is still Δλ, even though the fractional change in wavelength is now smaller.
✓ 1
Answer B
Answer: B
Common error: Do not assume the shift scales with the incident wavelength. The absolute shift at a given angle is fixed. Only the fractional shift Δλ/λ changes.
4IBP2H-037-01-02Intensity and the photoelectric currentPaper 1AEasy1 mark
A photocell is illuminated by monochromatic light, and the stopping potential is measured as V. The intensity of the light is tripled without changing its frequency. What are the new stopping potential and the new saturation photocurrent?
Show mark scheme
Marking point
Mark
Notes
The stopping potential is set by the maximum kinetic energy, eVs = hf − Φ, which depends only on the frequency. Tripling the intensity triples the number of photons arriving per second. This triples the number of photoelectrons emitted per second, so the saturation current triples while Vs stays at V.
✓ 1
Answer B
Answer: B
Exam tip: Link each quantity to its cause. Frequency controls the energy per photon and therefore the stopping potential. Intensity controls the photon rate and therefore the current.
5IBP2H-037-00-01Intensity and the photoelectric currentPaper 1AEasy1 mark
An automatic door sensor contains a photocell. Monochromatic light with a frequency above the threshold frequency falls on the photocathode. The intensity of the light is doubled and the frequency is kept constant. What happens to the maximum kinetic energy of the photoelectrons and to the saturation photoelectric current?
Show mark scheme
Marking point
Mark
Notes
Each photoelectron absorbs the energy of one photon, hf. The frequency is unchanged, so the photon energy and the maximum kinetic energy hf − Φ are unchanged. Doubling the intensity at fixed frequency doubles the number of photons arriving per second. This doubles the number of electrons emitted per second, so the saturation current doubles. Options A and C use the classical wave idea that a more intense wave gives each electron more energy. Option D wrongly assumes the current depends only on frequency.
✓ 1
Answer B
Answer: B
Common error: Intensity controls how many photons arrive each second, not the energy of each one. Only frequency changes the maximum kinetic energy.
6IBP2H-037-00-08Electrons through a double slitPaper 1AEasy1 mark
Which observation gives direct evidence that electrons have wave properties?
Show mark scheme
Marking point
Mark
Notes
The rings are a diffraction pattern, produced when the electron waves are diffracted by the regularly spaced atomic layers in the graphite. Diffraction is a wave behaviour, so this shows that electrons behave as waves. A and B are evidence for the particle (photon) nature of electromagnetic radiation. D (thermionic emission) is a thermal effect and says nothing about wave behaviour.
✓ 1
Answer C
Answer: C
Exam tip: Photoelectric effect and Compton scattering show that light behaves as particles. Electron diffraction shows that matter behaves as waves. Keep the two directions of wave–particle duality separate.
7IBP3H-E2-20r38-01Intensity and the photoelectric currentPaper 1AEasy1 mark
Ultraviolet light, with frequency above the threshold frequency of the metal, falls on a clean metal plate in the calibration unit of a space telescope. The intensity of the light is doubled and its frequency is unchanged. What happens to the number of photoelectrons emitted per second and to their maximum kinetic energy?
Show mark scheme
Marking point
Mark
Notes
Doubling the intensity at constant frequency doubles the number of photons arriving per second, so twice as many photoelectrons are released each second. Each photon still has energy hf, so the maximum kinetic energy hf − φ is unchanged. Saying the kinetic energy doubles confuses intensity with photon energy; saying the number is unchanged ignores that each photon can release at most one electron, so more photons mean more electrons.
✓ 1
Answer C
Answer: C
Common error: Intensity changes the number of photons per second, not the energy of each photon; maximum kinetic energy depends only on frequency and work function.
8IBP3H-E2-20r38-02The photoelectric effectPaper 1AEasy1 mark
The cathode of a UV sensor at an athletics stadium is made of a metal with work function 2.5 eV. Use hc = 1240 eV nm. What is the longest wavelength of light that can cause photoelectric emission from this metal?
Show mark scheme
Marking point
Mark
Notes
Emission just occurs when the photon energy equals the work function, so λ = hc/φ = 1240 / 2.5 = 496 nm. The value 3100 nm comes from multiplying 1240 by 2.5 instead of dividing. The value 248 nm comes from dividing by 5.0 (twice the work function), and 620 nm comes from halving 1240 without using the work function at all.
✓ 1
Answer A
Answer: A
Exam tip: Longer wavelength means lower photon energy, so the threshold wavelength is the maximum wavelength that still works.
9IBP3H-E2-20r38-08Electron diffractionPaper 1AEasy1 mark
A beam of electrons passes through a thin graphite film and produces a pattern of concentric rings on a screen. What does this observation show?
Show mark scheme
Marking point
Mark
Notes
Concentric rings are a diffraction pattern, and diffraction by the regular spacing of atoms in graphite is a property of waves, so the pattern shows that electrons behave as waves. Deflection by a magnetic field shows only that electrons are charged and moving. A fixed charge and a small mass are properties of electrons as particles and do not produce a diffraction pattern.
✓ 1
Answer C
Answer: C
Exam tip: Diffraction and interference are wave behaviours; when matter produces them, it supports de Broglie's wave description.
14E-1A-35The photoelectric effectPaper 1AMedium1 mark
In the classical wave model of light, the energy of the light is spread evenly over each wavefront and an electron in a metal surface absorbs energy continuously from the wave.
Two experiments are carried out on the same clean metal surface: Experiment 1: very dim light of a frequency well above the threshold frequency Experiment 2: very intense light of a frequency below the threshold frequency
Which row gives the predictions of the classical wave model?
Experiment 1Experiment 2
Show mark scheme
Marking point
Mark
Notes
Step 1Experiment 1: in the wave model a very dim beam delivers energy to each electron slowly, so an electron must absorb for some time before it has gained the work function: a delay is predicted.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2Experiment 2: in the wave model the energy delivered depends on the intensity, not the frequency, so a very intense beam soon gives electrons enough energy: emission is predicted at any frequency.
—
Step 3Both predictions disagree with what is observed (immediate emission; no emission below the threshold frequency), which the photon model explains: one photon of energy hf gives all its energy to one electron.
✓ 1
Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThese are the observations, which the photon model explains; the wave model predicts the opposite in both experiments.
BThis credits the wave model with a threshold frequency. In the wave model only the energy delivered per second matters, so intense light of any frequency should release electrons.
CThis credits the wave model with immediate emission. A dim wave spreads its energy over the whole surface, so each electron needs time to collect enough energy.
DCorrect: a dim wave needs time to deliver the work function, and an intense wave of any frequency eventually delivers it.
15E-1A-36The photoelectric effectPaper 1AMedium1 mark
Ultraviolet radiation of wavelength 250 nm falls on a metal surface and the maximum kinetic energy of the emitted electrons is 1.20 eV. The radiation is replaced by radiation of wavelength 350 nm. (h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1, e = 1.60 × 10−19 C)
What is the new maximum kinetic energy of the emitted electrons?
Show mark scheme
Marking point
Mark
Notes
Step 1hc/λ for 250 nm = 4.97 eV, so Φ = 4.97 − 1.20 = 3.77 eV.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2hc/λ for 350 nm = 3.55 eV.
—
Step 33.55 eV < Φ: the frequency is below the threshold frequency, so no photoelectrons are released.
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: the 350 nm photons have less energy than the work function.
BThe maximum kinetic energy has been scaled in proportion to 1/λ (1.20 × 250/350), ignoring the work function.
CThe misconception that the maximum kinetic energy depends only on the metal (or on intensity), not on the frequency of the radiation.
DThe maximum kinetic energy has been scaled in proportion to λ (1.20 × 350/250) — inverted ratio.
17E-1A-39The photoelectric effectPaper 1AMedium1 mark
Monochromatic light of wavelength λ and power P falls on a photocathode. On average, only a fraction η of the incident photons each release one photoelectron, and every photoelectron released reaches the collector.
What is the photoelectric current?
Show mark scheme
Marking point
Mark
Notes
Step 1Photon energy = hc/λ, so the number of photons arriving per second is Pλ/hc.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2Electrons released per second = ηPλ/hc.
—
Step 3Current = charge per second = ηPλe/hc.
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: I = e × (electrons per second) = ηPλe/hc.
BThis divides by the fraction η instead of multiplying, giving more electrons than photons.
CThis assumes every photon releases an electron: the fraction η has been ignored.
DThis writes the photon energy as hcλ instead of hc/λ.
18E-1A-64The photoelectric effectPaper 1AMedium1 mark
Monochromatic light falls on a metal surface. The wavelength λ of the light is varied while the intensity is kept constant.
Which graph shows how the maximum kinetic energy Emax of the emitted electrons varies with λ?
Four graphs of the maximum kinetic energy of the photoelectrons against the wavelength of the incident light (λ₀ is the threshold wavelength).Show mark scheme
Marking point
Mark
Notes
Step 1Emax = hc/λ − Φ: this is not linear in λ; it falls steeply at short wavelengths and more gently near the threshold.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2Emax = 0 at the threshold wavelength λ₀ = hc/Φ.
—
Step 3For λ > λ₀ the photons have too little energy and no electrons are emitted, so there is no value of Emax to plot (it is never negative). Only B shows this.
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis treats Emax as a linear function of λ, as it is of frequency; Emax depends on 1/λ.
BCorrect: Emax = hc/λ − Φ, falling to zero at λ₀, with no emission at longer wavelengths.
CThis uses the right equation but extends it beyond λ₀, giving a negative kinetic energy; above λ₀ no electrons are emitted at all.
DThis copies the shape of the Emax–frequency graph onto a wavelength axis: longer wavelength means lower photon energy, not higher.
19IBP2H-037-01-03Stopping potentialPaper 1AMedium1 mark
A graph of stopping potential Vs against light frequency f is a straight line for photocathode metal X. Metal X is replaced by metal Y, which has a larger work function. How do the gradient and the intercept on the frequency axis change?
Show mark scheme
Marking point
Mark
Notes
From eVs = hf − Φ, we get Vs = (h/e)f − Φ/e. The gradient h/e is a combination of universal constants, so it is the same for every metal. The frequency intercept is the threshold frequency f0 = Φ/h, which increases when Φ is larger.
✓ 1
Answer A
Answer: A
Common error: The gradient is h/e for every metal. Only the intercepts depend on the work function.
A flat solar sail faces the Sun and is perpendicular to the sunlight. At first its surface absorbs all incident light. The spacecraft then deploys a coating that reflects all the light straight back. At the same time, it moves to a position where the intensity of sunlight is half its original value. What is the ratio (new radiation force) / (original radiation force)?
Show mark scheme
Marking point
Mark
Notes
An absorbing sail receives force IA/c. A perfectly reflecting sail receives 2IA/c, because each photon's momentum is reversed rather than just removed. The new force is 2(I/2)A/c = IA/c, which equals the original force. The ratio is therefore 1.
✓ 1
Answer C
Answer: C
Common error: Reflection doubles the momentum change per photon, so it doubles the force. Apply both changes before comparing the forces.
21IBP2H-037-01-05Electron diffractionPaper 1AMedium1 mark
In an electron diffraction tube, electrons pass through a thin graphite film and form rings on a screen. When the accelerating potential difference is 2.0 kV, a particular ring has radius r. The potential difference is increased to 8.0 kV. Assuming small diffraction angles and non-relativistic electrons, what is the new radius of this ring?
Diagram not to scale
Show mark scheme
Marking point
Mark
Notes
The kinetic energy is eV = p²/2m, so p ∝ √V and λ = h/p ∝ 1/√V. Quadrupling V halves λ. For small angles the ring radius is proportional to λ, so the radius becomes r/2.
✓ 1
Answer B
Answer: B
Common error: The wavelength scales as 1/√V, not 1/V. Choosing r/4 comes from treating λ as inversely proportional to V.
22IBP2H-037-01-06Compton scatteringPaper 1AMedium1 mark
High-energy photons are Compton-scattered by free electrons in a radiation detector. At which scattering angle of the photon is the kinetic energy transferred to the recoiling electron greatest?
Show mark scheme
Marking point
Mark
Notes
The wavelength shift (h/mec)(1 − cos θ) is largest when cos θ = −1, that is θ = 180° (back-scattering). The scattered photon then has its longest wavelength and lowest energy. By energy conservation, the electron receives the most kinetic energy. At 90° the shift equals h/mec, which is only half the maximum.
✓ 1
Answer D
Answer: D
Exam tip: The electron's energy gain equals the photon's energy loss. Maximise the wavelength shift and you maximise the electron's kinetic energy.
23IBP2H-037-00-02The photoelectric effect and a charged conductorPaper 1AMedium1 mark
A small, isolated, initially uncharged metal sphere on a space probe is in a vacuum. It is illuminated with ultraviolet radiation. The photoelectrons are emitted with a maximum kinetic energy of 2.0 eV. The illumination continues for a long time. Which statement describes the final state of the sphere?
Show mark scheme
Marking point
Mark
Notes
Each emitted electron removes negative charge, so the sphere becomes positive. A positive sphere attracts the escaping electrons. An electron must do work eV to escape to infinity. Once eV equals the maximum kinetic energy of 2.0 eV, no electron can escape, so net emission stops at V ≈ +2.0 V. Option A ignores that electrons leave the sphere. Option B ignores the retarding effect of the sphere's own potential. Option D has the wrong sign: losing electrons makes the sphere positive.
✓ 1
Answer C
Answer: C
Exam tip: For an isolated conductor, the limiting potential in volts is numerically equal to the maximum kinetic energy of the photoelectrons in electronvolts.
In a research laboratory, a beam of electrons and a beam of protons each have a kinetic energy of 5 keV. Relativistic effects can be ignored. The proton mass is about 1840 times the electron mass. What is the ratio \(\frac{\text{de Broglie wavelength of an electron}}{\text{de Broglie wavelength of a proton}}\)?
Show mark scheme
Marking point
Mark
Notes
For a non-relativistic particle, p = √(2mEK) and λ = h/p = h/√(2mEK). At equal kinetic energy, λ ∝ 1/√m. So λe/λp = √(mp/me) = √1840 ≈ 43. The lighter electron has the longer wavelength. Option D comes from forgetting the square root, which would only apply at equal speeds. Options A and B invert the ratio.
✓ 1
Answer C
Answer: C
Common error: At equal speed λ ∝ 1/m, but at equal kinetic energy λ ∝ 1/√m. Check which quantity is held constant.
25IBP2H-037-00-04Electrons through a double slitPaper 1AMedium1 mark
An electron source is made so weak that only one electron at a time is in the apparatus. Each electron passes a double slit and is recorded by a position-sensitive detector. Which statement describes what is observed?
Show mark scheme
Marking point
Mark
Notes
Each electron is detected as a localised particle at a single, unpredictable point. The probability of arriving at each point is given by the wave (the square of the amplitude), and this wave passes through both slits. After many electrons have arrived, the distribution of impacts shows the double-slit fringes. The interference cannot be between different electrons, because only one is present at a time, so D is wrong. A wrongly spreads a single electron over the whole screen. C applies classical particle paths.
✓ 1
Answer B
Answer: B
Exam tip: Describe both behaviours: detection as particles at points, and distribution as a wave pattern that needs many electrons to appear.
26IBP2H-037-00-06Stopping potentialPaper 1AMedium1 mark
A camera designer compares two photocathode materials, P and Q. The work function of Q is greater than that of P. For each material, the maximum kinetic energy of the photoelectrons is plotted against the frequency of the incident light. Which statement about the two straight lines is correct?
Show mark scheme
Marking point
Mark
Notes
EK,max = hf − Φ. The gradient is the Planck constant h for every metal, so the lines are parallel. The frequency-axis intercept is the threshold frequency Φ/h. This is larger for Q, which has the larger work function. The other options wrongly make the gradient depend on the material, or fail to link the work function to the threshold frequency.
✓ 1
Answer A
Answer: A
Exam tip: On an EK,max–f graph: gradient = h, frequency intercept = f₀, and the extrapolated energy-axis intercept = −Φ.
27IBP3H-E2-20r38-03Stopping potentialPaper 1AMedium1 mark
In a sprint-timing gate a photocell is illuminated by light of frequency f, and the maximum kinetic energy of the photoelectrons is 1.0 eV. When the light is replaced by light of frequency 2f, the maximum kinetic energy is 3.4 eV. What is the work function of the metal?
Show mark scheme
Marking point
Mark
Notes
The photoelectric equation gives hf = φ + 1.0 eV and 2hf = φ + 3.4 eV. Doubling the first equation gives 2hf = 2φ + 2.0 eV, so 2φ + 2.0 = φ + 3.4 and φ = 1.4 eV. The value 2.4 eV is the photon energy hf of the first light, not the work function. The value 1.7 eV comes from halving 3.4 eV, which wrongly treats kinetic energy as proportional to frequency. The value 4.4 eV comes from adding the two kinetic energies.
✓ 1
Answer D
Answer: D
Exam tip: Write hf = φ + Emax for each frequency and eliminate hf; the kinetic energies are not proportional to the frequencies.
28IBP3H-E2-20r38-04Matter wavesPaper 1AMedium1 mark
A proton and an electron are each accelerated from rest through the same potential difference. The mass of the proton is 1840 times the mass of the electron. What is the approximate value of (de Broglie wavelength of the proton) ÷ (de Broglie wavelength of the electron)?
Show mark scheme
Marking point
Mark
Notes
Each particle gains the same kinetic energy eV, and momentum is p = √(2m·eV), so λ = h/p is proportional to 1/√m. The ratio is therefore 1/√1840 ≈ 1/43. The value 1/1840 wrongly takes λ proportional to 1/m, as it would be at equal speeds. A ratio of 1 wrongly assumes the same energy gives the same wavelength. A value of 43 inverts the ratio; the heavier particle has the shorter wavelength.
✓ 1
Answer B
Answer: B
Common error: At equal speeds λ ∝ 1/m, but after acceleration through the same potential difference it is the kinetic energy that is equal, so λ ∝ 1/√m.
29IBP3H-E2-20r38-05Electrons through a double slitPaper 1AMedium1 mark
Electrons accelerated through a potential difference produce interference fringes of separation 0.80 mm at a detector behind a double slit. The accelerating potential difference is increased by a factor of 4; the slit separation and the slit-to-detector distance are unchanged. What is the new fringe separation?
Show mark scheme
Marking point
Mark
Notes
The momentum is proportional to √V, so quadrupling V doubles the momentum and halves the de Broglie wavelength. Fringe separation is proportional to λ, so it halves to 0.40 mm. The value 0.20 mm assumes λ is proportional to 1/V. The value 0.80 mm assumes electrons are particles whose fringes do not depend on speed. The value 1.6 mm has the wavelength increasing, which is the wrong direction for faster electrons.
✓ 1
Answer B
Answer: B
Exam tip: Fringe separation is proportional to λ, and for electrons accelerated from rest λ is proportional to 1/√V.
X-ray photons of wavelength 70 pm and gamma-ray photons of wavelength 2.0 pm are each scattered through 90° by free electrons that are initially at rest.
Which row compares the shift in wavelength of the scattered photons and the fraction of the incident photon energy that is transferred to the electron?
Shift in wavelengthFraction of the photon energy transferred to the electron
Show mark scheme
Marking point
Mark
Notes
Step 1Δλ = (h/mec)(1 − cos θ) depends only on the angle, not on the incident wavelength: at 90° both shifts equal λC = 2.43 pm.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2The photon energy is hc/λ, so the fraction transferred is 1 − λ/(λ + Δλ) = Δλ/(λ + Δλ).
—
Step 3X-ray: 2.43/(70 + 2.43) = 0.033 (about 3 %); gamma ray: 2.43/(2.0 + 2.43) = 0.55 (about 55 %). The same shift is a much larger fraction of a short wavelength.
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: the shift is fixed by the angle alone, and the same shift is a far larger fraction of the shorter (gamma-ray) wavelength, so that photon loses a larger fraction of its energy.
BThe shift is right, but an equal shift in wavelength does not mean an equal fraction of the energy: the fraction is Δλ/(λ + Δλ), which depends on λ.
CThis takes the more energetic photon to be shifted further. The Compton shift does not depend on the incident wavelength or energy.
DThis takes the shift to be proportional to the incident wavelength (a fixed fractional shift), which would make the fraction of energy transferred the same for both.
Electrons, accelerated from rest through a potential difference V, pass through a narrow slit. The first minimum of the diffraction pattern is at a small angle θ to the central direction.
The potential difference is reduced to V/4. What is the new angle of the first minimum?
Show mark scheme
Marking point
Mark
Notes
Step 1The kinetic energy eV = p²/2m, so p = √(2meV) ∝ √V.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2λ = h/p ∝ 1/√V: dividing V by 4 doubles λ.
—
Step 3For a single slit θ ≈ λ/b, so the angle doubles to 2θ.
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis takes λ ∝ V — a slower electron is wrongly given a shorter wavelength, and the square root is missing.
BThis takes λ ∝ √V: the dependence is inverted.
CCorrect: λ ∝ 1/√V and θ ∝ λ.
DThis takes λ ∝ 1/V, omitting the square root in p = √(2meV).
32E-1A-63Intensity and the photoelectric currentPaper 1AHard1 mark
The same photocathode is illuminated in turn by two monochromatic light sources, P and Q. The graph shows how the photoelectric current I varies with the potential V of the anode relative to the cathode. For both sources, each incident photon has the same probability of releasing a photoelectron.
Which row compares the frequency and the intensity of light Q with those of light P?
Photoelectric current against anode potential for lights P and Q (graph drawn to scale)
Frequency of QIntensity of Q
Show mark scheme
Marking point
Mark
Notes
Step 1The current for Q falls to zero at −1.60 V and for P at −0.80 V: the stopping potential, and so Emax = eVs, is greater for Q.
—
All 4 steps must be completed — there is no mark for a part-answer.
Step 2From Emax = hf − Φ with the same Φ, the photons of Q have the higher frequency.
—
Step 3The saturation currents are equal, so equal numbers of photoelectrons, and therefore equal numbers of photons, arrive each second.
—
Step 4Each photon of Q carries more energy, so the same photon rate means a greater energy per second per unit area: Q has the greater intensity.
✓ 1
Answer C
Answer: C · 4 stages of work, one mark
Every option, and why
AThis gets the frequency right but assumes that equal saturation currents mean equal intensities; the current fixes the photon rate, and each photon of Q carries more energy.
BThis reads the curve that reaches zero at the more negative potential as having the smaller stopping potential (lower frequency), and then reasons that a lower-frequency light needs a greater intensity to give the same current.
CCorrect: larger stopping potential → higher frequency; same photon rate with more energy per photon → greater intensity.
DThis misreads the stopping potentials and also equates equal currents with equal intensities.
33IBP2H-037-01-07Matter waves: neutron diffractionPaper 1AHard1 mark
A thermal neutron and an electron have the same kinetic energy, and both are non-relativistic. The neutron mass is about 1840 times the electron mass. What is the approximate value of (de Broglie wavelength of electron) / (de Broglie wavelength of neutron)?
Show mark scheme
Marking point
Mark
Notes
For equal kinetic energy E, the momentum is p = √(2mE), so λ = h/√(2mE) ∝ 1/√m. The ratio λe/λn = √(mn/me) = √1840 ≈ 43. The lighter electron has the longer wavelength.
✓ 1
Answer C
Answer: C
Common error: Equal kinetic energy does not mean equal momentum. The momentum ratio is the square root of the mass ratio, so the answer is not 1840.
A photon and a non-relativistic electron have the same wavelength. Which statement is correct?
Show mark scheme
Marking point
Mark
Notes
Both obey p = h/λ, so equal wavelengths mean equal momenta. The photon's energy is E = pc. The electron's kinetic energy is p²/2m = pc × (v/2c). Since v < c, this is less than pc, so the photon has the greater energy.
✓ 1
Answer A
Answer: A
Exam tip: The relation p = h/λ holds for both photons and matter. The energy–momentum relation is different: E = pc for photons, but Ek = p²/2m for slow particles.
Electrons accelerated from rest through a potential difference V pass through a double slit. Fringes of spacing s form on a distant screen. The accelerating potential difference is increased to 4V, and the slit separation is halved. The slit-to-screen distance is unchanged and relativistic effects can be ignored. What is the new fringe spacing?
Show mark scheme
Marking point
Mark
Notes
λ = h/√(2meV), so increasing V by a factor of 4 halves λ. The fringe spacing is s = λD/d. Halving λ and halving d leaves λ/d unchanged, so the spacing stays s. Option B comes either from using λ ∝ 1/V or from ignoring the change in d. Option A combines λ ∝ 1/V with treating the spacing as proportional to d. Option D treats the spacing as proportional to √V.
✓ 1
Answer C
Answer: C
Common error: The de Broglie wavelength depends on 1/√V, not 1/V. Because momentum is √(2meV), a factor-of-4 change in V gives only a factor-of-2 change in λ.
36IBP2H-037-00-07Intensity and the photoelectric currentPaper 1AHard1 mark
Light falls on a metal of work function 2.0 eV. The stopping potential is 1.0 V. The wavelength of the light is then halved. What is the new stopping potential?
Show mark scheme
Marking point
Mark
Notes
Photon energy = EK,max + Φ = 1.0 eV + 2.0 eV = 3.0 eV. Halving the wavelength doubles the photon energy to 6.0 eV. The new EK,max = 6.0 − 2.0 = 4.0 eV, so the stopping potential is 4.0 V. Option B doubles the stopping potential instead of the photon energy. Option A treats the stopping potential as proportional to wavelength. Option D forgets to subtract the work function.
✓ 1
Answer C
Answer: C
Common error: The stopping potential is not proportional to frequency. Double the photon energy first, then subtract the work function.
37IBP3H-E2-20r38-06The photoelectric effect and a charged conductorPaper 1AHard1 mark
A small, isolated metal sphere in space is initially uncharged and far from all other objects. Its work function is 4.3 eV. It is illuminated by ultraviolet light from a star, each photon having energy 5.6 eV. Photoelectrons are emitted until the emission stops. What is the final potential of the sphere relative to a point far away?
Show mark scheme
Marking point
Mark
Notes
The sphere loses electrons, so it becomes positively charged and its potential rises. Emission stops when even the most energetic photoelectrons cannot escape, that is when eV equals the maximum kinetic energy 5.6 − 4.3 = 1.3 eV, so V = +1.3 V. A negative value ignores that electrons leave the sphere. The value +4.3 V uses the work function as the stopping energy, and +5.6 V uses the whole photon energy, which ignores the energy needed to leave the surface.
✓ 1
Answer D
Answer: D
Common error: The stopping potential depends on the maximum kinetic energy, hf − φ, not on the photon energy or the work function alone.
38IBP3H-E2-20r38-07Compton scatteringPaper 1AHard1 mark
A photon of X-radiation scatters from a free electron at rest and is deflected through 60°. Use h = 6.63×10−34 J s, me = 9.11×10−31 kg and c = 3.00×108 m s−1, with Δλ = (h/mec)(1 − cos θ). What is the increase in wavelength of the photon?
Show mark scheme
Marking point
Mark
Notes
The factor h/(mec) = 2.43 pm and 1 − cos 60° = 0.50, so Δλ = 1.21 pm. The value 2.43 pm omits the angle factor. The value 2.10 pm uses sin 60° in place of 1 − cos 60°. The value 3.64 pm uses 1 + cos 60° = 1.5, a sign error in the bracket.
✓ 1
Answer A
Answer: A
Common error: The shift involves 1 − cos θ, which is 0 at θ = 0, 1 at 90° and 2 at 180°; do not replace it with sin θ or 1 + cos θ.
Monochromatic X-rays of wavelength 70.9 pm are directed at a block of graphite. A crystal spectrometer on a rotating arm measures the wavelength λf of the X-rays scattered through angle θ; each value is uncertain by ±0.15 pm. The Compton formula predicts that λf = λi + (h/mec)(1 − cos θ), where λi is the incident wavelength.
The graph shows λf against (1 − cos θ) with error bars and the line of best fit. (h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1)
θ / °
1 − cos θ
λf / pm
30
0.134
71.3
60
0.500
72.1
90
1.000
73.4
120
1.500
74.5
150
1.866
75.5
180
2.000
75.7
Scattered wavelength against (1 − cos θ) (graph drawn to scale)
(a)
(i)
Determine the gradient of the line.
(2)
(b)
(i)
Determine the mass of the electron from your answer to (a), and compare it with the accepted value 9.11 × 10−31 kg.
(2)
(c)
(i)
Explain why the reading at θ = 30° contributes least to the determination of the gradient, and suggest how the student could improve the experiment using the same spectrometer.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Two well-separated points on the line, e.g. (0, 71.0) and (2.00, 75.7)
✓ 1
Gradient = 2.40 pm
✓ 1
Accept 2.30–2.50 pm (2.4 × 10−12 m).
Part (b)(i)
me = h/(gradient × c) = 6.63 × 10−34/(2.40 × 10−12 × 3.00 × 108) = 9.2 × 10−31 kg
✓ 1
Allow ECF from (a). Accept 8.8–9.6 × 10−31 kg.
About 1 % above the accepted value — close agreement, supporting the Compton model
✓ 1
The comparison must use the candidate's own value.
Part (c)(i)
Shift at 30° = gradient × 0.134 = 0.32 pm, so ±0.15 pm is about 47 % of the shift (only ≈ 3 % at 180°)
✓ 1
Allow ECF from (a).
Take more (repeated) readings at large angles, 120°–180°, where the shift is largest, and fewer at small angles
✓ 1
Do not accept "use a more precise spectrometer".
Answers: (a)(i) 2.4 pm · (b)(i) 9.2 × 10−31 kg (the remaining parts are explanations — see the table above)
In a vacuum photocell, light from a white source passes through one of six narrow-band filters and falls on a cathode coated with potassium. For each filter the student increases the reverse potential difference between anode and cathode until a sensitive ammeter in the circuit just reads zero; this is the stopping potential Vs. The frequency f passed by each filter is given by the manufacturer. Because the current falls to zero gradually, each Vs is uncertain by ±0.10 V.
The graph shows the data with error bars and the line of best fit. (e = 1.60 × 10−19 C)
f / 1014 Hz
6.0
7.0
8.0
9.0
10.0
11.0
Vs / V
0.26
0.62
1.08
1.51
1.87
2.31
Stopping potential against frequency (graph drawn to scale)
(a)
(i)
Determine, using the line of best fit, a value for the Planck constant.
(2)
(b)
(i)
Draw the lines of maximum and minimum gradient that are consistent with the error bars. Hence determine the absolute uncertainty in your value of the Planck constant.
(2)
(c)
(i)
Determine, by extrapolating the line of best fit, the work function of potassium in eV.
(1)
(d)
(i)
The accepted value of the Planck constant is 6.63 × 10−34 J s. Comment on your result.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Gradient from two well-separated points on the line = 4.12 × 10−15 V s
✓ 1
Accept 4.0–4.2 × 10−15 V s.
h = e × gradient = 1.60 × 10−19 × 4.12 × 10−15 = 6.6 × 10−34 J s
✓ 1
Allow ECF from the gradient. Accept 6.4–6.7 × 10−34 J s.
Part (b)(i)
Both lines pass through every error bar; gradients ≈ 4.5 × 10−15 V s and 3.7 × 10−15 V s
✓ 1
Accept 4.4–4.6 × 10−15 and 3.6–3.9 × 10−15 V s.
Δh = e(gmax − gmin)/2 = ±0.6 × 10−34 J s
✓ 1
Allow ECF from (a) if the fractional route Δh/h = Δg/g is used. Accept ±0.5 to ±0.7 × 10−34 J s; must be given to 1 s.f.
Part (c)(i)
Intercept on the Vs axis = −2.2 V, so Φ = 2.2 eV
✓ 1
Accept 2.1–2.4 eV. ALT: f0 = 5.4 × 1014 Hz from the f axis and Φ = hf0/e using the value from (a); Allow ECF from (a).
Part (d)(i)
6.63 × 10−34 J s lies inside 6.6 ± 0.6 (× 10−34 J s), so the result is consistent with the accepted value
✓ 1
Allow ECF from (a) and (b). The conclusion must agree with the candidate's own value and uncertainty.
Answers: (a)(i) 6.6 × 10−34 J s · (b)(i) ±0.6 × 10−34 J s · (c)(i) 2.2 eV (the remaining parts are explanations — see the table above)
41E-1B-06Electron diffractionPaper 1BMedium6 marks
In an electron diffraction tube, electrons are accelerated from rest through a potential difference V and then pass through a very thin layer of polycrystalline graphite. Rings appear on a screen a distance L = 0.135 m beyond it. For each V the student measures the radius r of the innermost ring. For small angles r = Lλ/d, where λ is the de Broglie wavelength of the electrons and d is the spacing of the planes of atoms that produce the ring.
The graph shows r against V−1/2 with a line of best fit. (h = 6.63 × 10−34 J s, me = 9.11 × 10−31 kg, e = 1.60 × 10−19 C)
V / V
2000
3000
4000
5000
6000
r / cm
1.75
1.41
1.24
1.09
1.01
V−1/2 / 10−3 V−1/2
22.36
18.26
15.81
14.14
12.91
Ring radius against V−1/2 (graph drawn to scale)
(a)
(i)
Determine the gradient of the graph, giving its unit.
(2)
(b)
(i)
Determine d.
(3)
(c)
(i)
Calculate the de Broglie wavelength of the electrons when V = 4000 V and use it to comment on whether the small-angle approximation is justified.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Gradient = 0.78 (from e.g. 1.95 cm at 25 × 10−3 V−1/2)
✓ 1
Accept 0.76–0.80 m V1/2 (76–80 cm V1/2).
Unit: m V1/2 (or cm V1/2 with a consistent value)
✓ 1
Part (b)(i)
eV = p²/2me, so p = √(2meeV) and λ = h/√(2meeV)
✓ 1
Work done by the field equals the kinetic energy gained.
Hence r = [Lh/(d√(2mee))] V−1/2: gradient = Lh/(d√(2mee))
A beam of fullerene molecules, known to be either C60 (mass 720 u) or C70 (mass 840 u), passes through a velocity selector and then through a single slit of width b = 0.20 μm. A detector scanned across the beam L = 1.50 m beyond the slit records a diffraction pattern; the distance y from its centre to the first minimum is measured for different molecular speeds v. The first minimum is at angle θ ≈ λ/b, so that y = (hL/mb)(1/v) for molecules of mass m.
The graph shows y against 1/v with the line of best fit. (h = 6.63 × 10−34 J s, 1 u = 1.66 × 10−27 kg)
v / m s−1
1/v / 10−3 s m−1
y / μm
100
10.00
43.5
125
8.00
34.5
150
6.67
29.5
175
5.71
25.1
200
5.00
22.8
225
4.44
19.9
250
4.00
18.5
y against 1/v (graph drawn to scale)
(a)
(i)
Determine the gradient of the line in SI units.
(2)
(b)
(i)
Determine the mass of one molecule in u and deduce which fullerene was used.
(2)
(c)
(i)
Another student uses only the reading at 250 m s−1 and assumes that y is proportional to 1/v. Determine the mass he obtains, and explain why the method in (b) is more reliable.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Two well-separated points on the line, e.g. (0, 1.7 μm) and (10.0 × 10−3, 43.2 μm)
Gradient = 18.5/4.00 = 4.62 μm per 10−3 s m−1, so m = 648 u — matching neither fullerene
✓ 1
Accept 640–660 u.
The line has an intercept of ≈ 1.7 μm (a zero offset, e.g. the centre of the pattern located wrongly); a single point through the origin includes this offset, whereas the gradient of the full line in (b) is unaffected by it
✓ 1
Allow ECF from (b). Accept intercept 1.2–2.0 μm.
Answers: (a)(i) 4.2 × 10−3 m² s−1 · (b)(i) 721 u, C60 · (c)(i) 648 u (the remaining parts are explanations — see the table above)
43IBP2H-037-01-15The photoelectric effectPaper 2Easy9 marks
A UV-index sensor for testing sunscreen uses a zinc photocathode, which has a work function of 4.30 eV. When ultraviolet radiation of a single wavelength falls on the cathode, the stopping potential is 1.20 V. (h = 6.63×10−34 J s, c = 3.00×108 m s−1, me = 9.11×10−31 kg, e = 1.60×10−19 C)
(a)
Explain why the stopping potential does not change when the intensity of the UV radiation is increased.
(2)
(b)
Determine the wavelength of the UV radiation.
(3)
(c)
Calculate the maximum speed of the emitted photoelectrons.
(2)
(d)
Calculate the de Broglie wavelength of the fastest photoelectrons.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A higher intensity means more photons per second, but the energy of each photon (hf) is unchanged .
✓ 1
The maximum kinetic energy hf − Φ depends only on frequency, so eVs and hence Vs are unchanged .
✓ 1
Part (b)
Photon energy = Φ + eVs = 4.30 + 1.20 = 5.50 eV .
✓ 1
= 5.50 × 1.60×10−19 = 8.80×10−19 J .
✓ 1
λ = hc/E = 1.989×10−25/8.80×10−19 = 2.26×10−7 m .
✓ 1
Part (c)
Ek,max = 1.20 × 1.60×10−19 = 1.92×10−19 J .
✓ 1
v = √(2Ek/me) = 6.49×105 m s−1 .
✓ 1
Part (d)
λ = h/(mev) .
✓ 1
λ = 6.63×10−34/(9.11×10−31 × 6.49×105) = 1.12×10−9 m . ECF from (c).
✓ 1
Exam tip: Working in eV keeps the energy bookkeeping simple: photon energy = work function + eVs. Convert to joules only when you use h, c or me.
44IBP2H-037-00-09The photoelectric effectPaper 2Easy7 marks
The photocathode of a night-vision image tube has a work function of 1.50 eV. Red light of wavelength 650 nm falls on the photocathode. h = 6.63×10−34 J s, c = 3.00×108 m s−1, e = 1.60×10−19 C, me = 9.11×10−31 kg.
(a)
State what is meant by the work function of a metal.
(1)
(b)
Calculate, in eV, the energy of one photon of this light.
(2)
(c)
Calculate, in J, the maximum kinetic energy of the emitted photoelectrons.
(2)
(d)
Calculate the maximum speed of the photoelectrons.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
States that it is the minimum energy needed to remove an electron from the surface of the metal
45IBP3H-E2-20r38-09The photoelectric effectPaper 2Easy6 marks
A detector in a sports-science laboratory uses a metal cathode of work function 2.10 eV. Light of wavelength 450 nm is incident on the cathode. Use h = 6.63×10−34 J s, c = 3.00×108 m s−1, e = 1.60×10−19 C and 1 eV = 1.60×10−19 J. Use unrounded values from earlier parts.
(a)
Calculate the energy of one photon of this light, in eV.
(2)
(b)
Calculate the maximum kinetic energy of the emitted photoelectrons, in joules.
(2)
(c)
The light is changed to wavelength 700 nm and its intensity is increased. Determine whether photoelectrons are emitted.
47E-2-02The photoelectric effect and a charged conductorPaper 2Medium12 marks
A zinc sphere of radius 2.0 cm is mounted on an insulating stand in a vacuum. The photoelectric properties of zinc were first measured in a photocell: the graph shows the stopping potential Vs against the frequency f of the radiation incident on a zinc cathode.
(c = 3.00 × 108 m s−1, e = 1.60 × 10−19 C, k = 8.99 × 109 N m² C−2)
Stopping potential against frequency for zinc (graph drawn to scale)
(a)
(i)
Determine, using the graph, a value for the Planck constant.
(2)
(ii)
Determine the work function of zinc in eV.
(2)
(iii)
On the graph, draw the line that would be obtained for a metal with a larger work function.
(1)
(b)
The sphere, initially uncharged, is now illuminated with ultraviolet radiation of wavelength 200 nm.
(i)
Determine the maximum kinetic energy, in eV, of the electrons emitted while the sphere is uncharged.
(2)
(ii)
Explain why, after some time, no more electrons leave the sphere, and state the potential of the sphere when this happens.
(2)
(iii)
Calculate the charge on the sphere when emission stops.
(2)
(iv)
Hence determine the number of electrons that have left the sphere.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Gradient = ΔVs/Δf ≈ 4.1 × 10−15 V s
✓ 1
Accept 4.0–4.3 × 10−15 V s from two well-separated points on the line.
h = e × gradient = 6.6 × 10−34 J s
✓ 1
Accept 6.4–6.9 × 10−34 J s.
Part (a)(ii)
Threshold frequency from the intercept on the f-axis: f0 = 1.04 × 1015 Hz
✓ 1
Accept 1.02–1.06 × 1015 Hz.
Φ = hf0 = 6.9 × 10−19 J = 4.3 eV
✓ 1
Allow ECF from (a)(i). Accept 4.2–4.4 eV. ALT: extrapolating the line to f = 0 gives Vs = −Φ/e.
Part (a)(iii)
A straight line parallel to the given line (same gradient h/e), displaced to higher frequency
✓ 1
Allow ECF from (a)(i). Both features needed.
Part (b)(i)
Photon energy = hc/λ = 9.95 × 10−19 J = 6.22 eV
✓ 1
Emax = 6.22 − 4.3 = 1.9 eV
✓ 1
Allow ECF from (a)(ii).
Part (b)(ii)
Each emitted electron leaves the sphere more positive, so its potential rises and an escaping electron must do work eV against the attraction of the sphere
✓ 1
Emission stops when eV = Emax: V = +1.9 V
✓ 1
Allow ECF from (b)(i). MP2 needs the value.
Part (b)(iii)
V = kQ/r, so Q = Vr/k = 1.92 × 0.020/8.99 × 109
✓ 1
Q = 4.3 × 10−12 C
✓ 1
Allow ECF from (b)(ii). Accept 4.2–4.4 × 10−12 C.
Part (b)(iv)
N = Q/e = 2.7 × 107
✓ 1
Allow ECF from (b)(iii).
Answers: (a)(i) 6.6 × 10−34 J s · (a)(ii) 4.3 eV · (b)(i) 1.9 eV · (b)(ii) +1.9 V · (b)(iii) 4.3 × 10−12 C · (b)(iv) 2.7 × 107(the remaining parts are explanations — see the table above)
48E-2-06Matter waves: neutron diffractionPaper 2Medium9 marks
Neutrons from a research reactor pass through a block of heavy water at 320 K and emerge in thermal equilibrium with it. They are then diffracted by a crystal to reveal the arrangement of its atoms.
(kB = 1.38 × 10−23 J K−1, mass of neutron = 1.675 × 10−27 kg, me = 9.11 × 10−31 kg, h = 6.63 × 10−34 J s, e = 1.60 × 10−19 C)
(a)
(i)
Show that the average kinetic energy of the neutrons is about 6.6 × 10−21 J.
(1)
(ii)
Determine the de Broglie wavelength of a neutron with this kinetic energy.
(2)
(iii)
Explain why these neutrons are suitable for studying the arrangement of atoms in a crystal.
(1)
(b)
(i)
Electrons with the same de Broglie wavelength could also be used. Deduce, without calculating a wavelength, the ratio (kinetic energy of the electron)/(kinetic energy of the neutron).
(2)
(ii)
Hence calculate the potential difference through which the electrons must be accelerated from rest.
(2)
(iii)
Suggest why electrons of this energy are diffracted only by the outermost layers of atoms of a crystal, whereas the neutrons pass through the whole crystal.
49E-2-19The photoelectric effectPaper 2Medium12 marks
An ultraviolet flame detector contains a photocell whose metal cathode has a work function of 4.70 eV. The detector is used in a factory to confirm that a gas burner is alight. (h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1, e = 1.60 × 10−19 C) (me = 9.11 × 10−31 kg)
(a)
Threshold.
(i)
Calculate the threshold frequency for the cathode.
(1)
(ii)
Sunlight at the Earth's surface contains no wavelengths shorter than about 295 nm. Explain why the detector does not respond to sunlight, however bright it is.
(2)
(b)
Photoelectrons.
(i)
The flame emits ultraviolet radiation of wavelength 220 nm. Determine the maximum kinetic energy, in eV, of the photoelectrons.
(2)
(ii)
Calculate the maximum speed of the photoelectrons.
(1)
(c)
The wave model.
(i)
The intensity of the 220 nm radiation at the cathode is 2.0 × 10−6 W m−2. In the classical wave model, the energy is spread uniformly over the wavefront and an electron can absorb only the energy falling on an area of about 1.0 × 10−19 m² around it. Estimate the time for an electron to absorb enough energy to escape.
(2)
(ii)
The detector actually responds within a few milliseconds of the burner igniting. Explain how the photon model accounts for this.
(2)
(d)
The current.
(i)
The cathode has an area of 1.5 × 10−4 m² and one photon in ten releases an electron. Determine the photoelectric current.
v = √(2Emax/me) = √(2 × 0.95 × 1.60 × 10−19/9.11 × 10−31) = 5.8 × 105 m s−1
✓ 1
ECF from (b)(i).
Part (c)(i)
Power absorbed by one electron = 2.0 × 10−6 × 1.0 × 10−19 = 2.0 × 10−25 W
✓ 1
t = Φ/power = 4.70 × 1.60 × 10−19/2.0 × 10−25 = 3.8 × 106 s (about 44 days)
✓ 1
Work function must be in joules.
Part (c)(ii)
The energy of the radiation is concentrated in photons, each of energy hf (5.65 eV)
✓ 1
A single photon is absorbed by a single electron in one event, so an electron can be emitted as soon as the first photon arrives — no accumulation of energy is needed
✓ 1
Part (d)(i)
Photons per second = (2.0 × 10−6 × 1.5 × 10−4)/9.04 × 10−19 = 3.3 × 108 s−1
✓ 1
I = 0.10 × 3.3 × 108 × 1.60 × 10−19 = 5.3 × 10−12 A
✓ 1
ECF.
Answers: (a)(i) 1.1 × 1015 Hz · (b)(i) 0.95 eV · (b)(ii) 5.8 × 105 m s−1 · (c)(i) 3.8 × 106 s · (d)(i) 5.3 × 10−12 A (the remaining parts are explanations — see the table above)
50IBP2H-037-01-09Stopping potentialPaper 2Medium8 marks
An engineer tests a new photocathode for a night-vision image tube. The cathode is illuminated with monochromatic light of several frequencies, and the stopping potential Vs is measured each time.
f / 1014 Hz: 6.00, 7.00, 8.00, 9.00 Vs / V: 0.19, 0.60, 1.02, 1.43
(e = 1.60×10−19 C)
Data points from the table
(a)
State what is meant by the stopping potential.
(1)
(b)
Determine a value for the Planck constant from the data.
(3)
(c)
Determine the threshold frequency of the photocathode.
(2)
(d)
Determine the work function of the photocathode in eV.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The minimum potential difference needed to stop even the most energetic photoelectrons from reaching the collector, so that the photocurrent falls to zero .
✓ 1
Part (b)
eVs = hf − Φ, so the gradient of a Vs against f line is h/e .
✓ 1
Gradient = (1.43 − 0.19)/(3.00×1014) = 4.13×10−15 V Hz−1 .
✓ 1
h = 4.13×10−15 × 1.60×10−19 ≈ 6.6×10−34 J s .
✓ 1
Part (c)
Extrapolate to Vs = 0: f0 = 6.00×1014 − 0.19/(4.13×10−15) .
✓ 1
f0 ≈ 5.5×1014 Hz . ECF from (b).
✓ 1
Part (d)
Φ/e = (h/e)f − Vs = 4.13×10−15 × 6.00×1014 − 0.19, or equivalently Φ = hf0 .
✓ 1
Φ ≈ 2.3 eV (2.29 eV) . ECF from (b) and (c).
✓ 1
Exam tip: Use widely separated data points, or a best-fit line, for the gradient. Keep the gradient in V Hz⁻¹: multiplying by e gives h, and multiplying by a frequency gives a potential in volts.
51IBP2H-037-01-11Electron diffractionPaper 2Medium9 marks
A materials scientist examines a thin polycrystalline film of a new battery-cathode material. Electrons are accelerated from rest through a potential difference of 4.00 kV and pass through the film. Rings appear on a fluorescent screen. The first ring is at an angle θ = 8.9° to the undeviated beam. Assume that this ring satisfies d sin θ = λ, where d is the spacing of the atomic planes. (h = 6.63×10−34 J s, me = 9.11×10−31 kg, e = 1.60×10−19 C)
Diagram not to scale
(a)
Outline why the appearance of the rings is evidence that electrons have wave properties.
(2)
(b)
Show that the de Broglie wavelength of the electrons is about 1.9×10−11 m.
(3)
(c)
Calculate the spacing d of the atomic planes.
(2)
(d)
The accelerating potential difference is increased. State and explain the effect on the angle of the first ring.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Diffraction and interference are wave phenomena, which particles alone would not produce .
✓ 1
The electron wavelength is comparable to the atomic spacing, so waves scattered from the regular atomic planes interfere constructively at particular angles .
✓ 1
Part (b)
Ek = eV = 1.60×10−19 × 4.00×103 = 6.40×10−16 J .
✓ 1
p = √(2meEk) = 3.41×10−23 kg m s−1 .
✓ 1
λ = h/p = 1.94×10−11 m .
✓ 1
Part (c)
d = λ/sin θ = 1.94×10−11/sin 8.9° .
✓ 1
d ≈ 1.25×10−10 m . ECF from (b).
✓ 1
Part (d)
The electron momentum increases, so λ = h/p decreases .
✓ 1
Since sin θ = λ/d with d fixed, θ decreases and the ring becomes smaller .
✓ 1
Common error: Convert kV to V before calculating eV. Also find the momentum before using λ = h/p: do not substitute the kinetic energy directly.
A small demonstration spacecraft of total mass 5.00 kg uses a perfectly reflecting solar sail of area 80.0 m². Near Earth's orbit, the sail is held perpendicular to sunlight of intensity 1360 W m−2. Ignore gravitational forces. (h = 6.63×10−34 J s, c = 3.00×108 m s−1)
(a)
Explain, in terms of photons, how sunlight exerts a force on the sail and why reflection gives a larger force than absorption.
(2)
(b)
Show that the force on the sail is about 7.3×10−4 N.
(2)
(c)
Taking a mean wavelength of 550 nm, calculate the number of photons striking the sail per second.
(2)
(d)
Assuming the force stays constant, determine the increase in the speed of the spacecraft after 30.0 days.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Each photon carries momentum p = h/λ = E/c, and the force on the sail equals the rate of change of the photons' momentum (Newton's second and third laws) .
✓ 1
On reflection each photon's momentum reverses, a change of 2p, whereas absorption gives a change of only p, so the force is doubled .
✓ 1
Part (b)
Power incident = IA = 1360 × 80.0 = 1.09×105 W .
✓ 1
F = 2P/c = 2 × 1.088×105/3.00×108 = 7.25×10−4 N .
✓ 1
Part (c)
E per photon = hc/λ = 6.63×10−34 × 3.00×108/550×10−9 = 3.62×10−19 J .
✓ 1
N = 1.088×105/3.62×10−19 ≈ 3.0×1023 s−1 .
✓ 1
Part (d)
a = F/m = 7.25×10−4/5.00 = 1.45×10−4 m s−2 .
✓ 1
Δv = at = 1.45×10−4 × 30.0 × 86400 ≈ 3.8×102 m s−1 . ECF from (b).
✓ 1
Common error: Using F = P/c for a reflecting sail halves the answer. Also convert days to seconds: 1 day = 86400 s.
53IBP2H-037-01-16Matter waves: neutron diffractionPaper 2Medium7 marks
A semiconductor-wafer inspection facility can probe crystal structure with either electrons or neutrons. Both beams are prepared with a de Broglie wavelength of 0.150 nm. (h = 6.63×10−34 J s, me = 9.11×10−31 kg, mn = 1.67×10−27 kg, e = 1.60×10−19 C)
(a)
State the de Broglie hypothesis.
(1)
(b)
Calculate the momentum of a particle in either beam.
(1)
(c)
Determine the accelerating potential difference needed to give the electrons this wavelength, starting from rest.
(2)
(d)
Calculate the kinetic energy of the neutrons in eV, and suggest one advantage of using neutrons rather than electrons to study the wafer.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
All moving particles have an associated wave whose wavelength is λ = h/p, where p is the particle's momentum .
✓ 1
Part (b)
p = h/λ = 6.63×10−34/1.50×10−10 = 4.42×10−24 kg m s−1 .
Ek = 5.85×10−21/1.60×10−19 = 0.0366 eV . ECF from (b).
✓ 1
Any one valid advantage, for example: the neutrons' very low energy means they do not damage or displace atoms in the wafer; OR the neutrons are uncharged, so they penetrate deep into the bulk of the wafer rather than probing only the surface .
✓ 1
Exam tip: At equal wavelength the momenta are equal, so the kinetic energies are in the inverse ratio of the masses (about 1:1830). Use this to check your answers to (c) and (d).
54IBP2H-037-00-10Intensity and the photoelectric currentPaper 2Medium9 marks
A laser power meter uses a vacuum photocell. The photocathode has a work function of 2.1 eV. A violet laser of wavelength 405 nm delivers 2.0 mW to the photocathode. Only 12% of the incident photons release a photoelectron, and all released electrons are collected. h = 6.63×10−34 J s, c = 3.00×108 m s−1, e = 1.60×10−19 C.
(a)
Calculate the number of photons incident on the photocathode per second.
(2)
(b)
Calculate the photoelectric current.
(2)
(c)
The violet laser is replaced by a green laser of wavelength 520 nm, also delivering 2.0 mW. The fraction of photons releasing an electron is unchanged. Determine the new photoelectric current.
(3)
(d)
The power of the violet laser is now doubled. Explain the effect on the stopping potential of the photocell.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Calculates the photon energy: E = hc/λ = 4.91×10−19 J
✓ 1
Calculates the photon rate: N = 2.0×10−3 / 4.91×10−19 = 4.1×1015 s−1
✓ 1
Part (b)
Calculates the electron rate: 0.12 × 4.07×1015 = 4.9×1014 s−1
✓ 1
Calculates the current: I = 4.9×1014 × 1.60×10−19 = 7.8×10−5 A
✓ 1
ECF from (a)
—
Part (c)
Shows that green photons can still cause emission: E = hc/λ = 3.83×10−19 J = 2.39 eV, which is greater than 2.1 eV
✓ 1
Uses a valid method for the new current. Either find the photon rate 2.0×10−3/3.83×10−19 = 5.23×1015 s−1, or recognise that at fixed power the current is proportional to λ, so it scales by 520/405
✓ 1
Calculates the current: I = 1.0×10−4 A
✓ 1
ECF from (b)
—
Part (d)
States that the photon energy (frequency) is unchanged, so the maximum kinetic energy hf − Φ of the photoelectrons is unchanged
✓ 1
Concludes that the stopping potential is unchanged; only the number of electrons per second (the current) doubles
✓ 1
Common error: At equal power, longer-wavelength light delivers more photons per second. If the photons are still above the threshold energy, a longer wavelength therefore gives a larger current, not a smaller one.
A transmission electron microscope in a hospital pathology laboratory is used to image viruses about 100 nm across. The electrons are accelerated from rest through a potential difference of 25 kV. Ignore relativistic effects. h = 6.63×10−34 J s, c = 3.00×108 m s−1, e = 1.60×10−19 C, me = 9.11×10−31 kg.
(a)
Show that the momentum of each electron is about 8.5×10−23 N s.
(2)
(b)
Calculate the de Broglie wavelength of the electrons.
(1)
(c)
Outline why the electron microscope can resolve detail on the viruses that a light microscope cannot.
(2)
(d)
Calculate the wavelength of an X-ray photon that has the same energy as one of these electrons.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Uses eV = p2/2me, so p = √(2meeV)
✓ 1
Substitutes: p = √(2 × 9.11×10−31 × 1.60×10−19 × 2.5×104) = 8.54×10−23 N s
States that diffraction blurs detail when the wavelength is comparable to, or larger than, the size of the detail; a smaller wavelength gives less diffraction and finer resolution
✓ 1
Compares values: electron wavelength ~10−11 m is far smaller than 100 nm, whereas visible light ~5×10−7 m is larger than the virus
✓ 1
Part (d)
Calculates the energy: E = 2.5×104 × 1.60×10−19 = 4.0×10−15 J
✓ 1
Calculates λ = hc/E = 5.0×10−11 m
✓ 1
Common error: For a photon use λ = hc/E, but for an electron use λ = h/p with p = √(2mEK). Using hc/E for an electron gives a wavelength that is far too large.
56IBP2H-037-00-14Stopping potentialPaper 2Medium8 marks
A light-gate timing system for sprint races uses a photocell. To characterise its photocathode, the stopping potential Vs is measured for light of different frequencies f: f / 1014 Hz: 6.0, 7.0, 8.0, 9.0, 10.0 Vs / V: 0.18, 0.60, 1.01, 1.43, 1.84 e = 1.60×10−19 C.
Data points from the table
(a)
Determine a value for the Planck constant using the data.
(3)
(b)
Determine the threshold frequency of the photocathode.
(2)
(c)
Determine, in eV, the work function of the photocathode.
(1)
(d)
At a frequency of 8.0×1014 Hz, the intensity of the light is doubled. State and explain the effect on the stopping potential.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Uses eVs = hf − Φ, so the gradient of Vs against f equals h/e
✓ 1
Calculates the gradient = (1.84 − 0.18)/(4.0×1014) = 4.15×10−15 V s
✓ 1
Calculates h = 4.15×10−15 × 1.60×10−19 = 6.6×10−34 J s
✓ 1
Part (b)
Sets Vs = 0, or extrapolates the line to the frequency axis
States that the stopping potential is unchanged, at 1.01 V
✓ 1
Explains that the photon energy depends only on frequency, so EK,max is unchanged; the extra intensity only increases the number of photoelectrons per second
✓ 1
Exam tip: Working in Vs against f gives a gradient of h/e. Multiplying by e gives h, and the magnitude of the intercept on the Vs axis gives Φ directly in eV.
57IBP2H-037-00-16The photoelectric effectPaper 2Medium7 marks
A smartphone ambient-light sensor relies on electrons absorbing energy from incident light. In a simple model, the sensing surface has a work function of 2.3 eV. It is illuminated by dim light of intensity 1.0 W m−2. e = 1.60×10−19 C.
(a)
Outline, using the photon model, why no electrons are emitted from a metal surface if the frequency of the light is below a threshold value, however intense the light.
(2)
(b)
In a classical wave model, a surface electron would collect energy continuously from the wave over an area of 3.0×10−20 m2. Estimate the time this model predicts before the first electron is emitted.
(3)
(c)
The light used has a frequency above the threshold frequency of the surface. Emission is observed to begin with no measurable delay. Explain how the photon model accounts for this.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
States that each electron absorbs the energy hf of a single photon
✓ 1
Explains that if hf < Φ, no single photon supplies enough energy to release an electron; greater intensity only supplies more photons, each of which is still too weak
✓ 1
Part (b)
Calculates the energy needed: Φ = 2.3 × 1.60×10−19 = 3.68×10−19 J
✓ 1
Calculates the power absorbed: P = IA = 1.0 × 3.0×10−20 = 3.0×10−20 W
✓ 1
Calculates t = Φ/P ≈ 12 s (accept 12.3 s)
✓ 1
Part (c)
States that energy arrives in discrete photons and each photon's energy hf is absorbed fully by a single electron in one interaction
✓ 1
Explains that since hf > Φ, an electron can be emitted as soon as the first photon is absorbed, with no build-up of energy needed; intensity affects only the rate of emission, not when it begins
✓ 1
Exam tip: A full explanation contrasts the two models directly: the wave model spreads energy continuously, so electrons need time to collect enough; the photon model delivers it all at once.
58IBP3H-E2-20r38-11Electrons through a double slitPaper 2Medium7 marks
Electrons are accelerated from rest through a potential difference of 2.50 kV and then pass through a double slit with slit separation 5.0×10−7 m. A detector is placed 1.2 m beyond the slits. Use h = 6.63×10−34 J s, me = 9.11×10−31 kg and e = 1.60×10−19 C. Ignore relativistic effects. The fringe separation is w = λD/s.
Diagram not to scale
(a)
Calculate the de Broglie wavelength of the electrons.
(3)
(b)
Calculate the fringe separation on the detector. Use the unrounded wavelength from (a).
(2)
(c)
The electron source is adjusted so that electrons reach the slits one at a time. Each electron is detected at a single point on the detector, yet after many electrons have arrived the fringes appear. Explain what this shows about electrons.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 kinetic energy: eV = p2/2me, so p = √(2meeV) = 2.70×10−23 kg m s−1
✓ 1
M1 λ = h/p
✓ 1
A1 2.46×10−11 m
✓ 1
Part (b)
M1 w = λD/s = 2.456×10−11 × 1.2 / 5.0×10−7 (ECF from (a))
✓ 1
A1 5.89×10−5 m (accept 5.90×10−5 m if the rounded wavelength is used)
✓ 1
Part (c)
A1 each electron is detected at one place, so it is detected as a localised particle
✓ 1
A1 the places where electrons arrive build up an interference pattern, so the probability of arrival at each point is determined by the wave properties of the electron
✓ 1
Exam tip: Electrons are detected as particles but travel and interfere as waves; mention both aspects in a duality explanation.
59IBP3H-E2-20r38-12Compton scatteringPaper 2Medium8 marks
An X-ray photon of wavelength 2.50×10−11 m from the gas around a binary star scatters from a free electron that is initially at rest. The photon is scattered through 120°. Use h = 6.63×10−34 J s, me = 9.11×10−31 kg and c = 3.00×108 m s−1. The increase in photon wavelength is Δλ = (h/mec)(1 − cos θ).
Diagram not to scale
(a)
Explain, in terms of energy, why the scattered photon has a longer wavelength.
(2)
(b)
Calculate the wavelength of the scattered photon.
(3)
(c)
Calculate the kinetic energy gained by the electron, in joules. Use the unrounded scattered wavelength from (b).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 the photon transfers some of its energy to the electron, which gains kinetic energy
✓ 1
A1 the photon's energy hc/λ is reduced, so its wavelength is longer
M1 substitutes λ = 2.50×10−11 m and λ′ = 2.864×10−11 m (ECF from (b))
✓ 1
A1 1.01×10−15 J (accept 1.00×10−15 J to 1.01×10−15 J)
✓ 1
Exam tip: The electron's kinetic energy equals the energy lost by the photon, so use the difference of the two photon energies, not the difference of the wavelengths.
60IBP3H-E2-20r38-14Stopping potentialPaper 2Medium7 marks
A student tests a photocell for a telescope camera. The stopping potential Vs is measured for light of several frequencies f. Frequency f / 1014 Hz: 6.0, 7.0, 8.0, 9.0, 10.0 Stopping potential Vs / V: 0.19, 0.60, 1.02, 1.43, 1.85 Use e = 1.60×10−19 C. The graph of Vs against f is a straight line.
Data points from the table
(a)
Show that Vs = (h/e)f − φ/e, where φ is the work function.
(2)
(b)
Use the first and last data points to determine the gradient of the line, and hence a value for the Planck constant.
(3)
(c)
Use one data point and your gradient to determine the work function of the metal, in eV.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 the stopping potential just stops the fastest electrons, so eVs = Emax
✓ 1
A1 combines with hf = φ + Emax to get Vs = (h/e)f − φ/e
✓ 1
Part (b)
M1 gradient = (1.85 − 0.19) / ((10.0 − 6.0)×1014)
✓ 1
A1 4.15×10−15 V s
✓ 1
A1 h = gradient × e = 6.64×10−34 J s (accept 6.4×10−34 to 6.9×10−34 J s from a best-fit line)
✓ 1
Part (c)
M1 φ/e = (gradient × f) − Vs = 4.15×10−15 × 6.0×1014 − 0.19 (ECF from (b))
✓ 1
A1 2.30 eV (accept 2.2 eV to 2.4 eV)
✓ 1
Exam tip: Using the first and last points of a straight-line data set gives a gradient that is less affected by rounding in the middle values.
61IBP3H-E2-20r38-15Matter waves: neutron diffractionPaper 2Medium8 marks
A beam of neutrons, each with kinetic energy 0.025 eV, is incident on a crystal surface on which rows of atoms are 0.45 nm apart. Use h = 6.63×10−34 J s, mn = 1.675×10−27 kg, me = 9.11×10−31 kg and 1 eV = 1.60×10−19 J. Use unrounded values from earlier parts.
Diagram not to scale
(a)
Calculate the de Broglie wavelength of the neutrons.
(3)
(b)
The first-order diffraction maximum obeys d sin θ = λ, where d is the spacing of the atomic rows. Calculate the angle θ.
(2)
(c)
Calculate the kinetic energy, in eV, of an electron that has the same de Broglie wavelength as these neutrons.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 kinetic energy = 0.025 × 1.60×10−19 = 4.0×10−21 J, so p = √(2mnE) = 3.66×10−24 kg m s−1
✓ 1
M1 λ = h/p
✓ 1
A1 1.81×10−10 m
✓ 1
Part (b)
M1 sin θ = λ/d = 1.81×10−10 / 0.45×10−9 (ECF from (a))
✓ 1
A1 θ = 23.7°
✓ 1
Part (c)
M1 the same wavelength means the same momentum, p = 3.66×10−24 kg m s−1
✓ 1
M1 Ek = p2/2me = 7.35×10−18 J
✓ 1
A1 46.0 eV
✓ 1
Common error: Equal wavelengths mean equal momenta, not equal energies: the lighter electron needs far more kinetic energy than the neutron.
62IBP3H-E2-20r38-16The photoelectric effectPaper 2Medium8 marks
A photodetector on a space telescope receives light of wavelength 500 nm at a total power of 3.0×10−12 W. Of the photons arriving, 2.0% each release one photoelectron. Use h = 6.63×10−34 J s, c = 3.00×108 m s−1, e = 1.60×10−19 C. Use unrounded values from earlier parts.
(a)
Calculate the energy of one photon of the light.
(2)
(b)
Calculate the number of photons arriving each second.
(2)
(c)
Calculate the photoelectric current.
(2)
(d)
The power of the light is doubled at the same wavelength. State and explain the effect on the current and on the maximum kinetic energy of the photoelectrons.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 E = hc/λ = 6.63×10−34 × 3.00×108 / 500×10−9
✓ 1
A1 3.98×10−19 J
✓ 1
Part (b)
M1 rate = power / photon energy = 3.0×10−12 / 3.98×10−19 (ECF from (a))
✓ 1
A1 7.54×106 s−1
✓ 1
Part (c)
M1 electrons per second = 0.020 × 7.54×106 = 1.51×105 s−1, and current = rate × e (ECF from (b))
✓ 1
A1 2.41×10−14 A (accept 2.41 to 2.42×10−14 A)
✓ 1
Part (d)
A1 the current doubles because twice as many photons, and so twice as many photoelectrons, arrive each second
✓ 1
A1 the maximum kinetic energy is unchanged because each photon has the same energy hc/λ
✓ 1
Exam tip: Photocurrent is proportional to the number of photons per second, so find the photon rate first and then multiply by the charge of one electron.
In a hospital gamma camera, a patient is given technetium-99m, which emits gamma photons of energy 140 keV. Some photons are Compton scattered by electrons in the patient's body before reaching the camera. Treat these electrons as free and at rest. (h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1, e = 1.60 × 10−19 C) (me = 9.11 × 10−31 kg)
Diagram not to scale
(a)
A single scattering.
(i)
Calculate the wavelength of a 140 keV photon.
(1)
(ii)
A 140 keV photon is scattered through 90°. Calculate the wavelength and the energy, in keV, of the scattered photon.
(2)
(iii)
State the kinetic energy given to the electron.
(1)
(b)
The energy window.
(i)
The camera records only photons with energies above 126 keV. Determine the largest angle through which a photon can be scattered and still be recorded.
(3)
(ii)
Explain why the camera is designed to reject scattered photons.
(2)
(c)
The recoiling electron.
(i)
For the photon scattered through 90° in (a)(ii), determine the angle between the direction of motion of the recoiling electron and the direction of the incident photon.
(2)
(d)
Back-scattering.
(i)
A 140 keV photon is scattered back through 180°. Determine the magnitude of the momentum of the recoiling electron.
Δλ = (h/mec)(1 − cos 90°) = 2.43 × 10−12 m, so λf = 8.88 + 2.43 = 11.3 × 10−12 m
✓ 1
E = hc/λf = 1.76 × 10−14 J = 110 keV
✓ 1
ECF; accept 109–111 keV.
Part (a)(iii)
140 − 110 = 30 keV (energy conservation)
✓ 1
ECF.
Part (b)(i)
λf = hc/(126 keV) = 9.87 × 10−12 m, so Δλ = 9.87 − 8.88 = 0.99 × 10−12 m
✓ 1
1 − cos θ = 0.99/2.43 = 0.41, so cos θ = 0.59
✓ 1
θ = 54°
✓ 1
Accept 53–55°. Scattering through a larger angle gives a photon below 126 keV.
Part (b)(ii)
A scattered photon has changed direction, so it appears to come from a point in the body where the source is not
✓ 1
Recording it would blur the image / add background; since scattering lowers the energy, an energy threshold removes most scattered photons
✓ 1
Part (c)(i)
Momentum is conserved in two perpendicular directions: along the incident direction the electron's momentum component is h/λi; perpendicular to it, it is h/λf, opposite to the momentum of the scattered photon
✓ 1
Allow ECF from (a)(i) and (a)(ii).
tan φ = (h/λf)/(h/λi) = λi/λf = 8.88/11.3, so φ = 38° (on the opposite side of the incident direction from the scattered photon)
✓ 1
Accept 37–39°. The inverted ratio (52°) scores [1].
Part (d)(i)
λf = 8.88 + 2 × 2.43 = 13.7 × 10−12 m
✓ 1
pi = h/λi = 7.47 × 10−23 N s; pf = h/λf = 4.83 × 10−23 N s in the opposite direction
✓ 1
Momentum conservation: pe = pi + pf = 1.2 × 10−22 N s
✓ 1
Adding (not subtracting) the photon momenta is the key step; accept 1.2–1.3 × 10−22 N s.
Answers: (a)(i) 8.88 × 10−12 m · (a)(ii) 1.13 × 10−11 m, 110 keV · (a)(iii) 30 keV · (b)(i) 54° · (c)(i) 38° · (d)(i) 1.2 × 10−22 N s (the remaining parts are explanations — see the table above)
64E-2-36Electrons through a double slitPaper 2Hard12 marks
Electrons, accelerated from rest through a potential difference of 600 V, pass through two very narrow parallel slits whose centres are 2.5 × 10−7 m apart, cut in a thin membrane. A position-sensitive detector, 0.60 m beyond the slits, records the point of arrival of each electron as a single dot. The beam current is so small that the dots appear one at a time; after a long time they build up the distribution shown in the graph.
(h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1, e = 1.60 × 10−19 C, me = 9.11 × 10−31 kg)
Distribution of electron arrivals across the detector after a long time (graph drawn to scale)
(a)
(i)
Show that the de Broglie wavelength of the electrons is about 5.0 × 10−11 m.
(2)
(ii)
Determine, using the graph, the separation of adjacent maxima on the detector.
(2)
(iii)
Hence determine the wavelength of the electrons, and comment on your answer.
(2)
(b)
(i)
The beam current at the slits is 1.6 × 10−15 A. Determine the average distance between one electron and the next in the beam.
(3)
(ii)
Explain what your answer to (b)(i) shows about the formation of the pattern.
(2)
(c)
(i)
State the feature of the experiment that shows the particle nature of the electrons.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
p = √(2meV) = √(2 × 9.11 × 10−31 × 1.60 × 10−19 × 600) = 1.32 × 10−23 kg m s−1
✓ 1
λ = h/p = 5.01 × 10−11 m
✓ 1
An answer to at least 3 s.f. must be seen.
Part (a)(ii)
Measure across several spacings, e.g. the maxima at about −0.48 mm and +0.48 mm (0.95 mm apart on the graph) are 8 spacings apart
✓ 1
Measuring one spacing only: [1 max].
s = 0.95/8 = 0.119 mm
✓ 1
Accept 0.115–0.125 mm (0.96/8 = 0.120 mm accepted).
The electrons are about 1.5 km apart, so there is almost never more than one electron between the slits and the detector (0.60 m) at a time
✓ 1
Allow ECF from (b)(i).
The fringes therefore cannot come from different electrons interfering with each other: each single electron behaves as a wave that passes through both slits and interferes with itself
✓ 1
OWTTE. Dependent on MP1.
Part (c)(i)
Each electron arrives at a single point on the detector (one dot, all of its charge at one place), not spread over the pattern
✓ 1
Do not accept "electrons have mass".
Answers: (a)(ii) 0.12 mm · (a)(iii) 5.0 × 10−11 m · (b)(i) 1.5 × 103 m (the remaining parts are explanations — see the table above)
A solar sail is a large, very thin sheet that is accelerated by the momentum of sunlight. A sail of area 1600 m² has a total mass (sail and payload) of 12 kg. Sunlight strikes the sail at normal incidence and is completely reflected.
(L☉ = 3.85 × 1026 W, M☉ = 1.99 × 1030 kg, 1 AU = 1.50 × 1011 m, G = 6.67 × 10−11 N m² kg−2, h = 6.63 × 10−34 J s, c = 3.00 × 108 m s−1)
Acceleration of the sail due to sunlight against 1/d² (graph drawn to scale)
(a)
(i)
Show that the momentum of a photon of wavelength 550 nm is about 1.2 × 10−27 kg m s−1.
(1)
(ii)
Estimate the number of photons striking the sail each second when it is 1.00 AU from the Sun.
(2)
(iii)
State one assumption made in your estimate in (a)(ii).
(1)
(iv)
Show that the force exerted by the sunlight on the sail is F = 2bA/c, where b is the intensity of the sunlight and A is the area of the sail.
(2)
(v)
Determine the acceleration of the sail at 1.00 AU due to the sunlight.
(2)
(b)
(i)
Show that the ratio of the force due to sunlight to the gravitational force of the Sun on the sail does not depend on the distance d of the sail from the Sun.
(2)
(ii)
Calculate this ratio for the sail.
(1)
(iii)
Determine the greatest mass per unit area that such a sail (including its payload) could have if sunlight is to be able to balance the Sun's gravitational force on it.
(2)
(c)
(i)
During a mission the acceleration a of the sail due to sunlight alone is measured at different distances d from the Sun. The graph shows a against 1/d². Determine the gradient of the graph, giving its unit.
(2)
(ii)
Hence determine the luminosity of the Sun.
(2)
(d)
(i)
Over time the surface of the sail degrades so that it absorbs 10 % of the incident photons and reflects the rest. Determine the force on the sail at 1.00 AU.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
p = h/λ = 6.63 × 10−34/550 × 10−9 = 1.21 × 10−27 kg m s−1
✓ 1
Must see 1.21 × 10−27 or the full substitution.
Part (a)(ii)
b = L/4πd² = 1.36 × 103 W m−2; power on the sail = 2.18 × 106 W
66IBP2H-037-01-10Compton scatteringPaper 2Hard9 marks
A gamma camera in a hospital detects 140 keV photons emitted by a radioactive tracer inside a patient. Some photons are Compton-scattered by electrons in the patient's tissue before reaching the camera. Treat the electrons as free and at rest. (h = 6.63×10−34 J s, c = 3.00×108 m s−1, me = 9.11×10−31 kg, e = 1.60×10−19 C)
Diagram not to scale
(a)
Show that the wavelength of a 140 keV photon is about 8.9×10−12 m.
(2)
(b)
Calculate the wavelength of a photon scattered through 60°.
(2)
(c)
Determine, in keV, the kinetic energy of the recoiling electron in (b).
(2)
(d)
The camera electronics reject any photon with energy below 126 keV. Determine the smallest scattering angle for which a scattered photon is rejected.
Exam tip: Find h/mec = 2.43×10⁻¹² m first and reuse it. Keep at least three significant figures in the wavelengths, because (d) depends on a small difference between two of them.
67IBP2H-037-01-12Matter waves: neutron diffractionPaper 2Hard8 marks
Neutrons from a research reactor are slowed in a moderator at 300 K. They are then used to study how hydrogen is distributed inside a fuel-cell membrane. Assume the neutrons' mean kinetic energy equals that of an ideal gas at the moderator temperature. (kB = 1.38×10−23 J K−1, mn = 1.67×10−27 kg, h = 6.63×10−34 J s)
(a)
Show that the mean kinetic energy of the neutrons is about 6.2×10−21 J.
(1)
(b)
Calculate the de Broglie wavelength of a neutron with this kinetic energy.
(2)
(c)
Outline two reasons why neutrons, rather than electrons of similar wavelength, are used to probe the inside of the membrane.
(2)
(d)
A rotating chopper selects only neutrons of wavelength 0.200 nm. Determine the time these neutrons take to travel along a 12.0 m flight path to the detector.
p = √(2mnEk) = √(2 × 1.67×10−27 × 6.21×10−21) = 4.55×10−24 kg m s−1 .
✓ 1
λ = h/p = 6.63×10−34/4.55×10−24 = 1.46×10−10 m . Allow ECF from (a).
✓ 1
Part (c)
Neutrons are uncharged, so they are not deflected or stopped by electric forces from the atomic electrons and nuclei, and they penetrate deep into the sample .
✓ 1
Any one of: for the same wavelength they have the same momentum but about 1840 times less kinetic energy, so they cause little damage or ionisation; OR they scatter from nuclei, and hydrogen nuclei scatter neutrons strongly, so hydrogen shows up clearly .
✓ 1
Part (d)
v = h/(mnλ) = 6.63×10−34/(1.67×10−27 × 2.00×10−10) .
✓ 1
v = 1.985×103 m s−1 (≈ 1.99×103 m s−1) .
✓ 1
t = L/v = 12.0/1.985×103 = 6.05×10−3 s; accept answers from 6.03×10−3 s to 6.05×10−3 s, or 6.0×10−3 s, which allows for rounding of v .
✓ 1
Exam tip: In time-of-flight methods, the travel time is proportional to wavelength (t = mnλL/h). Measuring the arrival time therefore sorts the neutrons by wavelength. Keep unrounded values until the final step so that rounding does not change your last digit.
68IBP2H-037-01-14Compton scatteringPaper 2Hard9 marks
An airport security scanner uses Compton back-scattering. X-ray photons of energy 60.0 keV strike a suitcase, and the scanner detects photons scattered back through 180°. Treat the scattering electrons as free and at rest. (h = 6.63×10−34 J s, c = 3.00×108 m s−1, me = 9.11×10−31 kg, e = 1.60×10−19 C)
Diagram not to scale
(a)
Outline why Compton scattering is evidence for the particle nature of electromagnetic radiation.
(2)
(b)
Calculate the wavelength of the back-scattered photons.
(3)
(c)
Determine the energy, in keV, of a back-scattered photon.
(2)
(d)
By considering conservation of momentum, determine the magnitude of the momentum of the recoiling electron.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A classical wave model predicts scattered radiation of the same wavelength as the incident radiation, but the scattered wavelength is observed to be longer and to depend on angle .
✓ 1
The shift is explained by an elastic collision between a photon of momentum h/λ and an electron, with energy and momentum conserved .
The photon reverses direction, so pe = h/λ + h/λ′ .
✓ 1
pe = 6.63×10−34(1/2.07×10−11 + 1/2.56×10−11) ≈ 5.8×10−23 kg m s−1 . ECF from (b).
✓ 1
Common error: In back-scattering the photon's momentum changes sign. The electron's momentum is therefore the sum of the two photon momenta, not their difference.
69IBP2H-037-00-11The photoelectric effect and a charged conductorPaper 2Hard8 marks
An isolated aluminium component on a space probe can be modelled as a conducting sphere of radius 0.50 m. It is initially uncharged. It is illuminated by solar ultraviolet radiation of wavelength 122 nm. The work function of aluminium is 4.1 eV. Ignore any charge collected from the surrounding space plasma. h = 6.63×10−34 J s, c = 3.00×108 m s−1, e = 1.60×10−19 C, ε0 = 8.85×10−12 F m−1.
(a)
Calculate, in eV, the maximum kinetic energy of the photoelectrons.
(2)
(b)
Explain why the sphere becomes positively charged and why its charge reaches a maximum value.
(3)
(c)
Determine the number of electrons that have left the sphere when its charge is at its maximum value.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Calculates the photon energy: E = hc/λ = 1.63×10−18 J = 10.2 eV
✓ 1
Calculates EK,max = 10.2 − 4.1 = 6.1 eV
✓ 1
Part (b)
States that the emitted electrons carry negative charge away, so the sphere becomes positive
✓ 1
States that the positive sphere attracts the emitted electrons, which must do work eV against the electric field to escape
✓ 1
Explains that when eV equals EK,max, no electron can escape, so there is no further net loss of charge
✓ 1
Part (c)
States that the maximum potential is V = 6.1 V
✓ 1
Calculates the charge: Q = 4πε0rV = 4π × 8.85×10−12 × 0.50 × 6.1 = 3.4×10−10 C
✓ 1
Calculates the number: N = Q/e = 2.1×109
✓ 1
ECF from (a)
—
Exam tip: Link the photoelectric equation to electric potential: the limiting potential V satisfies eV = EK,max. Then use V = Q/(4πε₀r) for a sphere.
70IBP2H-037-00-13Electrons through a double slitPaper 2Hard9 marks
In a university demonstration, electrons accelerated from rest through a potential difference V pass through a nanofabricated double slit with slit separation 250 nm. The fringes are recorded on a detector 1.40 m beyond the slits. The fringe spacing s is measured for several values of V: V / V: 150, 300, 600, 1200 s / μm: 562, 395, 282, 198 e = 1.60×10−19 C, me = 9.11×10−31 kg. Ignore relativistic effects.
Diagram not to scale
(a)
The beam intensity is reduced so that electrons pass through the apparatus one at a time. Outline what is observed on the detector as time passes.
(2)
(b)
Show that the data are consistent with s ∝ 1/√V.
(2)
(c)
Determine a value for the Planck constant from these data.
(3)
(d)
A detector is placed at the slits that records which slit each electron passes through. Explain the effect on the pattern.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
States that each electron is detected at a single point, at an unpredictable position
✓ 1
States that as many electrons accumulate, the impacts build up the same fringe pattern
✓ 1
Part (b)
Tests at least three data points by one of these methods :
• each doubling of V divides s by a factor of about 1.42, 1.40, 1.42 ≈ √2
—
• a graph of s against 1/√V is a straight line through the origin
—
Concludes that s√V is constant (to within about 1%), so s ∝ V−½, as expected from λ = h/√(2meeV) and s = λD/d
✓ 1
Part (c)
Combines s = λD/d with λ = h/√(2meeV) to give h = s√V · d√(2mee)/D
✓ 1
Uses the mean s√V = 6.87×10−3 m V½, or the gradient of s against 1/√V
✓ 1
Calculates h = 6.87×10−3 × 250×10−9 × 5.40×10−25 / 1.40 = 6.6×10−34 J s
✓ 1
Allow ECF from (b).
—
Part (d)
States that recording which slit each electron passes through removes the superposition of the waves from both slits, so there is no interference. Accept: the wavefunction collapses to a single-slit state.
✓ 1
States that the fringes disappear and only a broad distribution remains, equal to the sum of two single-slit patterns
✓ 1
Exam tip: Using several data points earns more credit than using a single row of the table. You can use a mean of s√V or the gradient of s against 1/√V.
71IBP2H-037-00-15Matter waves: neutron diffractionPaper 2Hard8 marks
Engineers measure residual strain in a welded aircraft wing component by neutron diffraction. The neutrons used have a de Broglie wavelength of 0.18 nm. h = 6.63×10−34 J s, mn = 1.675×10−27 kg, me = 9.11×10−31 kg, kB = 1.38×10−23 J K−1, e = 1.60×10−19 C.
(a)
Calculate the speed of the neutrons.
(2)
(b)
Calculate, in eV, the kinetic energy of a neutron.
(2)
(c)
Estimate the temperature of a gas whose particles have a mean translational kinetic energy equal to the kinetic energy in (b).
(2)
(d)
Calculate, in eV, the kinetic energy of an electron with the same de Broglie wavelength.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Uses λ = h/(mnv), so v = h/(mnλ)
✓ 1
Calculates v = 6.63×10−34 / (1.675×10−27 × 0.18×10−9) = 2.2×103 m s−1
✓ 1
Part (b)
Calculates EK = ½mnv2 (or p2/2mn) = 4.05×10−21 J
✓ 1
Converts: EK = 4.05×10−21 / 1.60×10−19 = 0.025 eV
✓ 1
ECF from (a)
—
Part (c)
Uses (3/2)kBT = EK
✓ 1
Calculates T = 2 × 4.05×10−21 / (3 × 1.38×10−23) ≈ 2.0×102 K
✓ 1
ECF from (b)
—
Part (d)
Calculates the momentum: p = h/λ = 3.68×10−24 N s
✓ 1
Calculates EK = p2/(2me) = 7.45×10−18 J = 47 eV (accept 46–47 eV)
✓ 1
Common error: At the same wavelength, the momentum is the same, but the kinetic energy p²/2m is about 1840 times larger for the electron. Do not assume equal wavelengths mean equal energies.
72IBP3H-E2-20r38-13The photoelectric effect and a charged conductorPaper 2Hard9 marks
An isolated sphere of radius 0.040 m is made of a metal of work function 4.30 eV. The sphere is initially uncharged and is far from other objects. It is illuminated by ultraviolet light of wavelength 220 nm, and photoelectrons are emitted. The potential at the surface of a charged sphere is V = kQ/r with k = 8.99×109 N m2 C−2. Use h = 6.63×10−34 J s, c = 3.00×108 m s−1, e = 1.60×10−19 C. Use unrounded values from earlier parts.
(a)
Calculate the energy of a photon of the ultraviolet light, in eV.
(2)
(b)
Explain why the emission of photoelectrons stops after a short time, and determine the potential of the sphere when this happens.
(3)
(c)
Calculate the charge on the sphere when emission stops.
(2)
(d)
Estimate the number of electrons that have left the sphere.
A1 electrons leave, so the sphere becomes positively charged and its potential rises
✓ 1
A1 emission stops when eV equals the maximum kinetic energy of the photoelectrons, so none can escape
✓ 1
A1 V = 5.65 − 4.30 = 1.35 V (ECF from (a))
✓ 1
Part (c)
M1 Q = Vr/k = 1.3506 × 0.040 / 8.99×109
✓ 1
A1 6.01×10−12 C
✓ 1
Part (d)
M1 N = Q/e = 6.009×10−12 / 1.60×10−19 (ECF from (c))
✓ 1
A1 3.76×107 electrons
✓ 1
Common error: The potential energy to remove an electron rises as the sphere charges, so the emission stops even though the light is still shining; the stopping potential is fixed by (photon energy − work function)/e.
We use essential cookies to make this site work. With your permission we'd also use analytics and advertising cookies to understand which resources help students most and how families find us. Nothing optional is set unless you accept. Privacy policy
EssentialNeeded for the site and your cookie choice to work.
Always on
AnalyticsAnonymous visit statistics that help improve the free resources.
MarketingMeasures whether ads reach the right families. Off unless you turn it on.