IB Physics flashcards · SL and HL · first assessment 2025
E.2 Quantum physics flashcards: IB Physics HL
Revision flashcards for E.2 Quantum physics, written for the IB Diploma Physics course first assessed in 2025. This is a Higher Level topic, so every card is HL only. Use study mode to test yourself one card at a time, or read the full list below with every answer.
The photoelectric effect, stopping potential, matter waves, electron diffraction and Compton scattering.
- 69 cards
- Definition: 14
- Equation: 13
- Concept/explain: 14
- HL-only cards marked
All 69 E.2 Quantum physics cards
- DefinitionHL only
Define the photoelectric effect.
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The photoelectric effect is the emission of electrons from the surface of a metal when electromagnetic radiation of sufficiently high frequency is incident on that surface. Exam tip: the mark is for "electrons emitted from a metal (surface) when EM radiation/light is incident on it" — write ELECTRONS, not "particles" or "energy", and say the radiation must be above the threshold frequency, never "bright enough". The emitted electrons are called photoelectrons and are ordinary electrons of charge −1.60 × 10⁻¹⁹ C. The effect is a phenomenon, so it has no unit; emission is essentially instantaneous (< 10⁻⁹ s) even at very low intensity provided f > f₀.
- DefinitionHL only
Define the work function of a metal.
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The work function Φ is the minimum energy required to remove an electron from the surface of a metal. Exam tip: the mark scheme wants "minimum energy needed to release/remove an electron from the (surface of the) metal" — omitting "minimum" or "surface" loses the mark, and "energy to remove an electron from an atom" is ionisation energy, not the work function. Unit: joule (J), but almost always quoted in electronvolts (typical metals 2–5 eV: caesium 2.1 eV, sodium 2.3 eV, zinc 4.3 eV, platinum 5.6 eV). Φ is a scalar and is a property of the metal only, independent of the incident light.
- DefinitionHL only
Define the threshold frequency for photoelectric emission.
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The threshold frequency f₀ is the minimum frequency of incident electromagnetic radiation that will cause photoelectrons to be emitted from a given metal surface. Exam tip: state "minimum frequency for emission of electrons" — answers such as "the frequency at which the metal starts to glow" score zero. Below f₀ no electrons are emitted no matter how intense the light or how long you wait; above f₀ emission is immediate however dim the source. Unit: hertz (Hz = s⁻¹); scalar. The corresponding threshold (cut-off) wavelength is λ₀ = c/f₀, and radiation of longer wavelength than λ₀ causes no emission.
- DefinitionHL only
Define the stopping potential in a photoelectric experiment.
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The stopping potential V_s is the minimum reverse (retarding) potential difference that must be applied between the collecting electrode and the emitting photosurface to reduce the photoelectric current to exactly zero. Exam tip: the mark is for "pd needed to reduce the photocurrent to zero", i.e. it stops even the MOST energetic photoelectrons; describing it as "the pd that stops the electrons" without "the fastest/most energetic" is usually not enough at HL. Unit: volt (V); scalar. V_s depends only on the frequency of the incident light and on Φ — it is completely independent of the intensity of the light.
- DefinitionHL only
Define a photon.
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A photon is a discrete quantum (packet) of electromagnetic radiation carrying energy E = hf, where f is the frequency of the radiation. Exam tip: mark-scheme wording is "a quantum/packet of EM energy" — do not write "a small particle of light with mass". A photon has zero rest mass, travels at c = 3.00 × 10⁸ m s⁻¹ in a vacuum, and carries momentum p = E/c = h/λ. Energy is in joules (J) or eV; a photon is uncharged. In the photoelectric effect one photon interacts with one electron in a one-to-one interaction and gives up ALL of its energy.
- DefinitionHL only
Define the electronvolt and state its conversion to joules.
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One electronvolt is the energy transferred when an electron (charge magnitude e = 1.60 × 10⁻¹⁹ C) is accelerated through a potential difference of exactly 1 volt: 1 eV = 1.60 × 10⁻¹⁹ J. Exam tip: the definition mark needs "charge of one electron" AND "through a pd of 1 V"; the eV is a unit of ENERGY, never of potential difference or charge. To convert eV → J multiply by 1.60 × 10⁻¹⁹; J → eV divide by it. Useful shortcuts: 1 MeV = 1.60 × 10⁻¹³ J, and hc = 1240 eV nm so a 620 nm photon has energy 2.0 eV.
- DefinitionHL only
Define the maximum kinetic energy of photoelectrons and explain why there is a spread of energies.
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E_max is the largest kinetic energy that any emitted photoelectron can have for a given incident frequency, given by E_max = hf − Φ. Exam tip: electrons emitted from the very surface need only the minimum energy Φ and so leave with E_max; electrons liberated from deeper inside the metal lose extra energy in collisions on the way out, so photoelectron energies range continuously from 0 up to E_max. Students who state that "all photoelectrons have energy hf − Φ" lose the mark. Unit: joule (J) or eV; kinetic energy is a scalar. Measured experimentally from eV_s = E_max.
- DefinitionHL only
Define the saturation current in a photoelectric cell and state what it depends on.
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The saturation current is the constant maximum photoelectric current reached when the accelerating pd is large enough that every photoelectron emitted from the cathode reaches the anode. Exam tip: the required point is that saturation current ∝ intensity (rate of arrival of photons) at fixed frequency, and is INDEPENDENT of the accelerating pd once saturation is reached; it does NOT depend on how energetic the electrons are. Unit: ampere (A), often μA in practice. Doubling the intensity at constant frequency doubles the photon rate, so it doubles the electron emission rate and doubles the saturation current, but leaves V_s unchanged.
- DefinitionHL only
Define wave–particle duality.
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Wave–particle duality is the principle that all matter and all radiation exhibit both wave-like and particle-like properties, with the behaviour observed depending on the experiment performed. Exam tip: mark schemes accept "radiation/matter behaves as a wave in some experiments (interference, diffraction) and as a particle in others (photoelectric effect, Compton scattering)", and reward naming one experiment of each kind — never say something "is both at the same time". Evidence for the wave nature of electrons is electron diffraction; evidence for the particle nature of light is the photoelectric effect and Compton scattering. It is a principle, so no unit.
- DefinitionHL only
State the de Broglie hypothesis and define the de Broglie wavelength.
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De Broglie proposed that every particle with momentum p has an associated matter wave of wavelength λ = h/p. Exam tip: the mark is for "all moving particles/matter have an associated wavelength given by h divided by momentum" — it applies to every particle, not only to electrons, and λ is inversely proportional to p, so heavier or faster particles have shorter wavelengths. Unit of λ: metre (m); λ is a scalar while p is a vector. A 1 kg ball at 1 m s⁻¹ has λ ≈ 6.6 × 10⁻³⁴ m, far too small to diffract — this is why duality is invisible in everyday life.
- DefinitionHL only
Define Compton scattering.
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Compton scattering is the inelastic scattering of a photon (typically an X-ray or γ-ray) by a free or loosely bound electron, in which the photon is deflected through an angle θ and emerges with a longer wavelength, the lost energy being transferred to the recoiling electron. Exam tip: the mark scheme wants "photon collides with a (free) electron; photon loses energy so its wavelength increases; electron recoils". Both energy and momentum are conserved in the collision. The wavelength shift Δλ = λ_f − λ_i is measured in metres and depends ONLY on the scattering angle, never on the incident wavelength or on the target material.
- DefinitionHL only
Define the Compton wavelength of the electron and give its value.
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The Compton wavelength λ_C = h/(m_ec) is the wavelength shift a photon undergoes when it is Compton-scattered through exactly 90°. Numerically λ_C = 6.63 × 10⁻³⁴ / (9.11 × 10⁻³¹ × 3.00 × 10⁸) = 2.43 × 10⁻¹² m (2.43 pm). Exam tip: it is a fixed constant of the electron, not of the radiation, and it sets the scale of the whole effect — the maximum possible shift, at θ = 180°, is 2λ_C = 4.85 × 10⁻¹² m. This is why the Compton shift is negligible for visible light (λ ≈ 500 nm, shift ≈ 0.0005 %) but easily measured for X-rays.
- DefinitionHL only
Define the momentum of a photon.
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A photon of energy E and wavelength λ carries linear momentum of magnitude p = E/c = hf/c = h/λ, directed along the photon's direction of travel. Exam tip: students object that p = mv requires mass — state clearly that a photon has zero rest mass and that its momentum comes from the relativistic relation E = pc for a massless particle. Unit: kg m s⁻¹ (equivalently N s); momentum is a VECTOR, which is why Compton scattering requires a vector momentum triangle. A 0.10 nm X-ray photon has p = 6.63 × 10⁻³⁴ / 1.0 × 10⁻¹⁰ = 6.6 × 10⁻²⁴ kg m s⁻¹.
- DefinitionHL only
Define the intensity of a monochromatic beam in terms of photons.
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Intensity I is the power per unit area delivered by the beam, I = P/A = nhf/A, where n is the number of photons striking the surface per second. Exam tip: the required idea is that at a fixed frequency the intensity is proportional to the RATE of arrival of photons, and each photon still carries the same energy hf. This is the key to the photoelectric effect: raising the intensity supplies more photons per second (more photoelectrons per second, larger current) but never a more energetic photon, so E_max and V_s are unchanged. Unit of I: W m⁻²; scalar. n has unit s⁻¹.
- EquationHL onlyData booklet: Yes
State Einstein's photon energy equation and explain how to use it.
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E = hf (data booklet), and since c = fλ also E = hc/λ. E = photon energy in joules (J); h = Planck constant = 6.63 × 10⁻³⁴ J s; f = frequency in hertz (Hz); λ = wavelength in metres (m); c = 3.00 × 10⁸ m s⁻¹. Valid for a photon in a vacuum. Common misuse: substituting λ in nanometres — always convert to metres. Sanity check: green light of λ = 500 nm gives E = (6.63 × 10⁻³⁴ × 3.00 × 10⁸)/(5.00 × 10⁻⁷) = 3.98 × 10⁻¹⁹ J = 2.5 eV. Useful memorised form: E(eV) = 1240/λ(nm).
- EquationHL onlyData booklet: Yes
State Einstein's photoelectric equation and define every symbol.
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E_max = hf − Φ (data booklet). E_max = maximum kinetic energy of the emitted photoelectrons (J); h = 6.63 × 10⁻³⁴ J s; f = frequency of the incident radiation (Hz); Φ = work function of the metal (J). Valid only for f ≥ f₀; it is a statement of energy conservation for a ONE-photon-one-electron interaction. Common misuse: using it with f < f₀ and reporting a negative kinetic energy — the correct answer there is "no emission". Rearranged forms you must know: Φ = hf − E_max, f = (E_max + Φ)/h, and in eV: E_max(eV) = 1240/λ(nm) − Φ(eV).
- EquationHL onlyData booklet: No – derive
Give the relation between work function and threshold frequency.
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Φ = hf₀, so f₀ = Φ/h and the threshold wavelength is λ₀ = c/f₀ = hc/Φ. Φ = work function (J); h = 6.63 × 10⁻³⁴ J s; f₀ = threshold frequency (Hz); λ₀ = threshold wavelength (m); c = 3.00 × 10⁸ m s⁻¹. Data-booklet status: not printed separately — it is the special case of E_max = hf − Φ with E_max = 0, so derive it in one line. Common misuse: forgetting to convert Φ from eV to J before dividing by h. Sanity check: sodium, Φ = 2.3 eV = 3.68 × 10⁻¹⁹ J → f₀ = 5.6 × 10¹⁴ Hz, λ₀ = 540 nm (green).
- EquationHL onlyData booklet: No – derive
State the equation linking stopping potential to maximum kinetic energy.
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eV_s = E_max = hf − Φ, so V_s = (h/e)f − Φ/e. e = elementary charge = 1.60 × 10⁻¹⁹ C; V_s = stopping potential (V); E_max in joules (J); h in J s; f in Hz; Φ in J. Valid when the fastest photoelectron is just brought to rest at the collector, i.e. the retarding pd does work eV_s equal to its whole kinetic energy. Common misuse: writing E_max = V_s (mixing energy with pd) or forgetting the factor e. Sanity check: if E_max = 1.6 eV then V_s = 1.6 V exactly — in electronvolts the numbers are identical.
- EquationHL onlyData booklet: Yes
State the de Broglie equation and its useful rearrangements.
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λ = h/p (data booklet). λ = de Broglie wavelength (m); h = 6.63 × 10⁻³⁴ J s; p = momentum of the particle (kg m s⁻¹). For a non-relativistic particle p = mv, so λ = h/(mv); and since E_k = p²/(2m), p = √(2mE_k) and λ = h/√(2mE_k). Valid for v ≪ c. Common misuse: substituting the SPEED for the momentum, or using E_k in eV without converting to joules. Sanity check: an electron with E_k = 100 eV = 1.60 × 10⁻¹⁷ J gives p = 5.4 × 10⁻²⁴ kg m s⁻¹ and λ = 1.2 × 10⁻¹⁰ m — atomic scale, so it diffracts from crystals.
- EquationHL onlyData booklet: No – derive
Derive and state the de Broglie wavelength of an electron accelerated from rest through a pd V.
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Work done eV = ½m_ev² = p²/(2m_e), so p = √(2m_eeV) and λ = h/√(2m_eeV). λ in metres (m); h = 6.63 × 10⁻³⁴ J s; m_e = 9.11 × 10⁻³¹ kg; e = 1.60 × 10⁻¹⁹ C; V = accelerating pd in volts (V). Valid for electrons with V up to a few kV (non-relativistic). Data-booklet status: derive — the booklet gives only λ = h/p. Common misuse: forgetting the square root, or the factor 2. Sanity check: V = 100 V → λ = 1.23 × 10⁻¹⁰ m. Handy shortcut: λ(nm) ≈ 1.23/√(V in volts).
- EquationHL onlyData booklet: Yes
State the Compton scattering equation and define every symbol.
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λ_f − λ_i = (h/(m_ec))(1 − cosθ) (data booklet). λ_i = wavelength of the incident photon (m); λ_f = wavelength of the scattered photon (m); h = 6.63 × 10⁻³⁴ J s; m_e = 9.11 × 10⁻³¹ kg; c = 3.00 × 10⁸ m s⁻¹; θ = angle between the incident and scattered photon directions (degrees or radians). Valid for scattering from a free or loosely bound electron. Common misuse: taking θ as the recoil angle of the ELECTRON — it is the photon's deflection. Sanity check: θ = 90° gives 1 − cos90° = 1, so Δλ = 2.43 × 10⁻¹² m exactly.
- EquationHL onlyData booklet: Yes
State the momentum–wavelength relations for a photon.
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p = E/c = hf/c = h/λ. p = photon momentum (kg m s⁻¹ = N s); E = photon energy (J); c = 3.00 × 10⁸ m s⁻¹; h = 6.63 × 10⁻³⁴ J s; f in Hz; λ in metres. Valid for any massless quantum; it follows from the relativistic energy–momentum relation E² = p²c² + m²c⁴ with m = 0. Data-booklet status: p = h/λ follows directly from the printed λ = h/p, so quote it as the de Broglie relation applied to a photon. Common misuse: attempting p = mv with a "photon mass". Sanity check: a 2.5 eV visible photon has p = 4.0 × 10⁻¹⁹/3.00 × 10⁸ = 1.3 × 10⁻²⁷ kg m s⁻¹.
- EquationHL onlyData booklet: No – derive
Give the maximum Compton shift and the fractional energy loss it implies.
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Maximum shift occurs at θ = 180° (backscatter), where 1 − cosθ = 2: Δλ_max = 2h/(m_ec) = 4.85 × 10⁻¹² m. Symbols: h in J s, m_e = 9.11 × 10⁻³¹ kg, c = 3.00 × 10⁸ m s⁻¹. Fractional loss of photon energy ≈ Δλ/λ_f, so it is large only when λ_i is comparable with λ_C. Data-booklet status: derived from the printed Compton equation. Common misuse: assuming the photon can lose all its energy — it cannot; a photon is never absorbed in Compton scattering. Sanity check: for a 0.0100 nm X-ray backscattered, λ_f = 0.0149 nm, so about 33 % of the energy goes to the electron.
- EquationHL onlyData booklet: No – derive
State the energy and momentum conservation equations used in a Compton collision.
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Energy: hf_i = hf_f + E_k(electron), i.e. hc/λ_i = hc/λ_f + E_k. Momentum (vector): h/λ_i = (h/λ_f)cosθ + p_ecosφ along the incident direction and 0 = (h/λ_f)sinθ − p_esinφ perpendicular to it. Symbols: f in Hz, λ in m, E_k in J, p_e = electron recoil momentum (kg m s⁻¹), θ = photon scattering angle, φ = electron recoil angle (on the opposite side). Valid for a free stationary electron. Data-booklet status: derive. Common misuse: adding momenta as scalars. Sanity check: E_k of the electron always equals hc(1/λ_i − 1/λ_f), which is positive because λ_f > λ_i.
- EquationHL onlyData booklet: No – derive
Give the equation linking beam intensity to the rate of photon arrival.
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I = P/A = nhf/A, so n = P/(hf) = Pλ/(hc). I = intensity (W m⁻²); P = power of the beam (W); A = illuminated area (m²); n = number of photons per second (s⁻¹); h = 6.63 × 10⁻³⁴ J s; f in Hz; λ in m. Valid for monochromatic radiation. Data-booklet status: not printed — build it from E = hf and P = E/t. Common misuse: assuming doubling intensity doubles the electron energy; it doubles n and hence the photocurrent only. Sanity check: a 1.0 mW, 633 nm laser emits n = 1.0 × 10⁻³ × 6.33 × 10⁻⁷/(6.63 × 10⁻³⁴ × 3.00 × 10⁸) ≈ 3.2 × 10¹⁵ photons per second.
- EquationHL onlyData booklet: No – derive
State the equation for the maximum photocurrent and the electron emission rate in a photocell.
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I_sat = N e, where N is the number of photoelectrons released per second and e = 1.60 × 10⁻¹⁹ C. With quantum efficiency η (fraction of incident photons that release an electron), N = ηn = ηP/(hf), so I_sat = ηPe/(hf) = ηPeλ/(hc). Units: I_sat in amperes (A), N and n in s⁻¹, P in watts (W). Valid at saturation, when every emitted electron is collected. Data-booklet status: derive from Q = It. Common misuse: assuming η = 1 — real photocathodes have η of order 10⁻³ to 10⁻¹. Sanity check: N = 1.0 × 10¹³ s⁻¹ gives I_sat = 1.6 μA, a typical photocell reading.
- EquationHL onlyData booklet: No – derive
Give the relation between kinetic energy and momentum used to convert electron energies to de Broglie wavelengths.
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For a non-relativistic particle E_k = p²/(2m), hence p = √(2mE_k) and λ = h/√(2mE_k). E_k = kinetic energy (J); p = momentum (kg m s⁻¹); m = particle mass (kg); h = 6.63 × 10⁻³⁴ J s; λ in metres. Valid while v ≪ c (for electrons, E_k well below 511 keV). Data-booklet status: derive from E_k = ½mv² and p = mv. Common misuse: writing p = 2mE_k without the square root, or using E_k in eV. Sanity check: a thermal neutron (m = 1.67 × 10⁻²⁷ kg, E_k ≈ 0.025 eV = 4.0 × 10⁻²¹ J) has λ ≈ 1.8 × 10⁻¹⁰ m, ideal for neutron diffraction.
- Graph/diagramHL only
Describe the graph of maximum photoelectron kinetic energy against frequency of incident light.
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Plot E_max (y-axis, J or eV) against f (x-axis, Hz). The graph is a straight line that exists only for f ≥ f₀; below f₀ there is no emission, so the line is not extended there except as a dashed extrapolation. Gradient = h = 6.63 × 10⁻³⁴ J s (or 4.14 × 10⁻¹⁵ eV s). The x-intercept is the threshold frequency f₀; the extrapolated y-intercept is −Φ. To find Φ, read the magnitude of the negative intercept, or use Φ = hf₀. Changing the metal shifts the line sideways to a new f₀ and a new intercept but the gradient stays h, because h is a universal constant.
- Graph/diagramHL only
Describe the graph of stopping potential against frequency and how to obtain h from it.
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Plot V_s (y-axis, volts) against f (x-axis, Hz). The graph is a straight line V_s = (h/e)f − Φ/e. Gradient = h/e = 4.14 × 10⁻¹⁵ V s; multiply the gradient by e = 1.60 × 10⁻¹⁹ C to obtain Planck's constant. The x-intercept is f₀ and the y-intercept is −Φ/e, so Φ in electronvolts is the numerical magnitude of that intercept. This is the classic Millikan plot. Exam tip: read the gradient from two widely separated points on the LINE, not from data points, and quote it to the same significant figures as the data; use max/min gradients to estimate the uncertainty in h.
- Graph/diagramHL only
Describe photoelectric current against applied pd for the same frequency at two different intensities.
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Plot photocurrent I (y-axis, μA) against pd V (x-axis, V), with negative V representing retarding pd. Each curve rises from zero at V = −V_s, climbs steeply, then flattens to a horizontal saturation plateau at large positive V. Doubling the intensity doubles the height of the plateau because twice as many photons per second release twice as many electrons per second, but both curves cut the axis at exactly the SAME stopping potential −V_s, since E_max depends only on f. Exam tip: students who shift the intercept when intensity changes lose the mark. The curve for higher intensity lies entirely above the other.
- Graph/diagramHL only
Describe photoelectric current against applied pd for the same intensity at two different frequencies.
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Axes as before: photocurrent I (μA) against pd V (V). The higher-frequency light gives a more negative intercept, because V_s = (hf − Φ)/e increases with f. If the intensity (W m⁻²) is kept the same while f rises, each photon carries more energy hf, so there are FEWER photons per second and the saturation plateau is LOWER. Exam tip: the mark scheme distinguishes "same intensity" (fewer photons at higher f, lower plateau) from "same photon rate" (identical plateau); read the stem carefully. Both curves still saturate, and the extra kinetic energy shows only in the horizontal shift of the intercept, never in the plateau slope.
- Graph/diagramHL only
Describe how the Compton equation is linearised in a practical investigation.
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Plot the measured wavelength shift Δλ = λ_f − λ_i (y-axis, metres, typically 10⁻¹² m) against (1 − cosθ) (x-axis, no unit, running 0 to 2). The result is a straight line through the origin. Gradient = h/(m_ec) = 2.43 × 10⁻¹² m, the Compton wavelength; multiplying the gradient by m_ec gives an experimental value of Planck's constant. Exam tip: plotting Δλ against θ itself gives a curve and earns no linearisation mark. A line that does not pass through the origin indicates a systematic error in the zero of the spectrometer angle scale or in the reference wavelength λ_i.
- Graph/diagramHL only
Describe the spectrum of X-rays scattered from a graphite target at a fixed angle in the Compton experiment.
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Plot scattered intensity (y-axis, arbitrary units) against wavelength (x-axis, pm). Two peaks appear: an unmodified peak at the incident wavelength λ_i, from photons scattered by tightly bound inner electrons (which recoil with the whole atom, so the shift is negligible), and a modified peak at the longer wavelength λ_f, from photons scattered by loosely bound outer electrons. The separation of the peaks is Δλ, read directly off the wavelength axis. As the detector angle θ is increased from 0° to 180°, the modified peak moves further to the right and grows relative to the unmodified peak; at θ = 0 the two peaks coincide.
- Graph/diagramHL only
Describe how de Broglie's relation is linearised for an electron diffraction experiment.
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For electrons accelerated through pd V, λ = h/√(2m_eeV), so λ ∝ V^(−½). Plot λ (y-axis, m) against 1/√V (x-axis, V^(−½)): a straight line through the origin of gradient h/√(2m_ee) = 1.23 × 10⁻⁹ m V^½, from which h can be found. In the graphite tube the ring radius r ∝ λ, so plotting r against 1/√V is equivalent and gives d, the layer spacing, from r = 2Lλ/d. Exam tip: plotting r against V gives a curve — no marks. Increasing V decreases λ and so shrinks the rings; a line not through the origin suggests a systematic error in V.
- Graph/diagramHL only
Describe the graph of de Broglie wavelength against momentum and its linearised form.
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A direct plot of λ (y-axis, m) against p (x-axis, kg m s⁻¹) is a rectangular hyperbola: λ falls steeply at small p and tends asymptotically to zero as p → ∞, never touching either axis. Because λ ∝ 1/p, the linearised version plots λ against 1/p, giving a straight line through the origin whose gradient is Planck's constant h = 6.63 × 10⁻³⁴ J s. Equivalently, a log–log plot of lnλ against lnp is a straight line of gradient −1 and intercept lnh. Exam tip: state "gradient = h" with the unit J s; students often quote the gradient of the hyperbola at a point, which has no physical meaning.
- Concept/explainHL only
Outline the four experimental observations of the photoelectric effect that cannot be explained by the classical wave model of light.
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- A threshold frequency f₀ exists: below it no electrons are emitted however intense the light or however long it shines, whereas a wave of any frequency should eventually supply enough energy
- Emission is instantaneous (< 10⁻⁹ s) even for very dim light, but a wave spread over the whole metal surface would need seconds to minutes to accumulate the work function at one electron
- The maximum kinetic energy of the emitted electrons depends only on the frequency, not on the intensity — classically a more intense wave has a larger amplitude and should give faster electrons
- The photocurrent (rate of emission) is proportional to intensity, showing intensity controls how many electrons leave, not how energetic each one is. Exam tip: many students list only the threshold frequency; a 4-mark 'outline' question expects the observation AND the classical prediction it contradicts.
- Concept/explainHL only
Explain how Einstein's photon model accounts for the threshold frequency and for the instantaneous emission of photoelectrons.
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- Light of frequency f is treated as a stream of discrete quanta (photons) each carrying energy E = hf, which is delivered whole or not at all
- One photon interacts with one electron in a single, localised event, transferring all of its energy to that electron
- The electron must do at least the work function Φ against the attraction of the lattice to escape the surface, so E_max = hf − Φ
- If hf < Φ the electron cannot escape no matter how many photons arrive, giving a sharp threshold frequency f₀ = Φ/h
- Because the interaction is one-to-one and instantaneous, the first photon to strike a suitable electron ejects it immediately — no energy accumulation time is needed
- Nature of Science: Einstein reinterpreted Planck's mathematical quantisation as a physical property of light, a paradigm shift that Planck himself initially resisted. Exam tip: writing 'the photon gives some of its energy' loses the mark — the whole quantum is absorbed by a single electron.
- Concept/explainHL only
Explain why increasing the intensity of monochromatic light incident on a metal surface increases the photocurrent but leaves the maximum kinetic energy of the photoelectrons unchanged.
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- Intensity is the power per unit area, and for monochromatic light of frequency f the power is delivered as n photons per second each of fixed energy hf
- Doubling the intensity doubles n but does not change hf, because photon energy is fixed by frequency alone
- Each emission is a one-photon–one-electron event, so twice as many photons eject twice as many electrons per second and the saturation current doubles
- The energy available to any single electron is still hf, so E_max = hf − Φ is unaltered and the stopping potential is unchanged
- Classically the opposite is predicted: greater amplitude means greater energy per electron, so this result is direct evidence for quantisation. Exam tip: incomplete answers say 'more intense light has more energy so electrons move faster' — energy per photon, not total energy, sets E_max.
- Concept/explainHL only
Explain what is meant by the stopping potential in a photoelectric experiment and why it depends on frequency but not on intensity.
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- The collecting electrode is made negative with respect to the emitting surface, so electrons leaving the metal must do work against the retarding electric field
- An electron of kinetic energy E arriving with charge e loses energy eV in crossing a potential difference V
- The stopping potential V_s is the minimum retarding pd at which the photocurrent falls exactly to zero, so even the fastest electrons just fail to reach the collector: eV_s = E_max
- Combining with Einstein's equation gives eV_s = hf − Φ, so V_s rises linearly with f and is zero at f₀
- Intensity changes only the number of electrons, so the current at any given pd scales up but the pd at which it reaches zero is unchanged. Exam tip: students often write eV_s = hf; the work function must be subtracted first.
- Concept/explainHL only
Explain, using an order-of-magnitude argument, why the classical wave model predicts a measurable delay before photoemission and why this is not observed.
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- Classically the energy of a light wave is spread uniformly over the wavefront, so an electron can only absorb the fraction falling on its own effective area, roughly that of an atom, ≈ 10⁻²⁰ m²
- For a dim source giving an intensity of about 10⁻⁶ W m⁻² the power absorbed by one electron would be ≈ 10⁻²⁶ W
- A typical work function of 2–5 eV is about 5 × 10⁻¹⁹ J, so the accumulation time would be ≈ 5 × 10⁻¹⁹ / 10⁻²⁶ ≈ 5 × 10⁷ s, i.e. years
- Experiment shows emission begins within nanoseconds of illumination, independent of how weak the beam is
- Only a localised quantum delivering hf in one event explains this. Exam tip: 'the delay is small' is not an answer — the mark scheme wants the contrast between a predicted macroscopic delay and an observed immediate current.
- Concept/explainHL only
Describe and explain the shape of a graph of photocurrent against collecting-plate potential difference for a fixed frequency, and how the curve changes when the intensity is doubled.
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- For large positive pd the curve is flat at the saturation current, because every electron emitted per second is collected and no more are available
- As the pd is reduced towards zero the current falls only slightly; at V = 0 there is still a current since electrons leave with kinetic energy of their own
- For negative (retarding) pd the current falls progressively as slower electrons are turned back, reaching zero at V = −V_s
- Doubling the intensity at the same frequency doubles the saturation current and doubles the current at every pd, but the curve still cuts the axis at exactly the same V_s
- Raising the frequency instead moves the intercept further left without changing the saturation current, provided photon rate is fixed. Exam tip: sketches are frequently drawn cutting the axis at different points for different intensities — that is the classic error.
- Concept/explainHL only
Outline Millikan's photoelectric experiment and explain its significance in establishing the photon model.
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- Millikan set out to disprove Einstein's 1905 photon equation, which he regarded as 'reckless', and worked on it for a decade
- Alkali metal surfaces oxidise rapidly, so he cut a fresh surface with a knife inside an evacuated tube to obtain a clean, reproducible work function
- Monochromatic light of several known frequencies was used and the stopping potential measured for each
- Plotting V_s against f gave a straight line of gradient h/e for every metal tested, with only the intercept −Φ/e differing
- The value of h obtained agreed with Planck's value from black-body radiation to about 0.5%, an independent confirmation from a completely different phenomenon
- Nature of Science: a scientist's attempt to falsify a theory produced its strongest confirmation, and both men received Nobel Prizes. Exam tip: state that the gradient is h/e (not h) when V_s is plotted.
- Concept/explainHL only
Explain the de Broglie hypothesis and what the wavelength associated with a moving particle physically represents.
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- De Broglie proposed a symmetry with light: if photons of momentum p behave as particles, then particles of momentum p should show wave behaviour with λ = h/p
- For a non-relativistic particle p = mv, so λ = h/(mv), and the wavelength is inversely proportional to both mass and speed
- The wave is not a physical oscillation of the particle; the wave amplitude squared gives the probability per unit volume of detecting the particle at that point
- Wave behaviour is only observable when λ is comparable to the size of the aperture or spacing the particle encounters, which for electrons requires atomic-scale spacings ≈ 10⁻¹⁰ m
- Nature of Science: a bold theoretical symmetry argument preceded and predicted the experimental evidence found by Davisson and Germer. Exam tip: students say the particle 'turns into a wave'; the correct statement is that it exhibits wave-like behaviour in its detection probabilities.
- Concept/explainHL only
Explain how electron diffraction through a thin graphite film provides evidence for the wave nature of matter.
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- Electrons are accelerated through a known pd of a few kilovolts, giving them a de Broglie wavelength of order 10⁻¹¹–10⁻¹⁰ m, comparable to the spacing of carbon atom planes in graphite
- The regularly spaced atomic planes act as a natural diffraction grating; the polycrystalline film has crystallites at all orientations, so the maxima appear as concentric rings on a fluorescent screen
- Diffraction and interference are wave phenomena only — particles travelling in straight lines would give a single bright spot
- Increasing the accelerating pd increases p, decreases λ, and the rings shrink in radius, exactly as λ = h/p predicts
- Davisson and Germer obtained the same result by scattering electrons from a nickel crystal and measuring the angle of the intensity maximum. Exam tip: the ring pattern alone is not the full answer — you must state that ring radius varies with V in the way λ = h/p predicts.
- Concept/explainHL only
Explain why wave properties are not observed for everyday macroscopic objects even though de Broglie's relation applies to them.
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- λ = h/p and Planck's constant is only 6.63 × 10⁻³⁴ J s, so any object with an everyday momentum has an extraordinarily small wavelength
- A 60 kg person walking at 1 m s⁻¹ has p = 60 kg m s⁻¹ and λ ≈ 1 × 10⁻³⁵ m, some twenty orders of magnitude smaller than a nucleus
- Diffraction and interference are only detectable when the wavelength is of the same order as the aperture or obstacle; no aperture of 10⁻³⁵ m exists or could exist
- The angular spread of any diffraction, θ ≈ λ/b, is therefore immeasurably small and the object travels in an apparently straight line
- This is a correspondence-principle result: quantum predictions reduce to classical behaviour in the limit of large momentum. Exam tip: saying 'large objects are not quantum' earns nothing; a numerical estimate of λ and a comparison with a realistic aperture size is what is credited.
- Concept/explainHL only
Explain, using conservation of energy and momentum, why the wavelength of an X-ray photon increases when it is Compton-scattered by a free electron.
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- The photon is treated as a particle carrying energy E = hf = hc/λ and momentum p = E/c = h/λ
- In the collision with an essentially free, initially stationary electron both total energy and total momentum must be conserved, as in an elastic collision between two particles
- Unless the photon is undeviated, the electron must recoil to conserve momentum, and the recoiling electron carries away kinetic energy
- The photon therefore leaves with less energy, so its frequency falls and its wavelength rises: λ_f > λ_i always
- The shift λ_f − λ_i = (h/(m_e c))(1 − cosθ) depends only on the scattering angle θ, not on the incident wavelength or the material
- Nature of Science: this 1923 result convinced the remaining sceptics that photons carry momentum and are genuine particles. Exam tip: do not say the photon 'slows down' — photons always travel at c; it is the frequency that changes.
- Concept/explainHL only
Explain why the Compton shift is easily detected with X-rays but effectively invisible with visible light, and state what the Compton wavelength represents.
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- The absolute shift Δλ = (h/(m_e c))(1 − cosθ) has a maximum value of 2h/(m_e c) ≈ 4.85 × 10⁻¹² m, the same for every incident wavelength
- What is measurable in a spectrometer is the fractional shift Δλ/λ, so a small λ gives a large fractional change
- For 0.050 nm X-rays the maximum fractional shift is about 10%, easily resolved; for 500 nm visible light it is about 10⁻⁵, far below the resolution of ordinary spectrometers
- X-ray photon energies (tens of keV) also greatly exceed the binding energy of outer electrons, so the electrons behave as free — a requirement of the derivation
- The Compton wavelength h/(m_e c) = 2.43 × 10⁻¹² m is the shift produced at 90° scattering and is a fixed property of the electron. Exam tip: 'X-rays have more energy' is too vague; argue from the fractional shift.
- Concept/explainHL only
Compare and contrast the photoelectric effect and Compton scattering as evidence for the particle nature of electromagnetic radiation.
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- Both treat radiation as discrete photons of energy hf that interact one-to-one with a single electron, and both fail completely under a wave description
- Photoelectric effect: the photon is completely absorbed by an electron bound in a metal, the energy is shared with the lattice through the work function, and no photon survives
- Compton scattering: the photon is scattered, not absorbed, by an essentially free electron and survives with reduced energy and changed direction
- Photoelectric evidence rests on energy quantisation (E_max = hf − Φ, threshold frequency); Compton evidence rests on photon momentum p = h/λ, a stronger and more complete particle property
- Photon energies differ greatly: a few eV (UV/visible) for photoelectric, tens of keV (X-ray) for Compton, because the photon energy must greatly exceed the electron's binding energy for the electron to behave as free
- Absorption requires a third body (the lattice) to conserve momentum, which is why a truly free electron cannot absorb a photon. Exam tip: a 'compare' answer needs both similarities and differences, in linked pairs.
- Concept/explainHL only
Outline the probability interpretation of matter waves and how it resolves the apparent contradiction between particle and wave behaviour.
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- Each particle is described by a wavefunction whose amplitude varies in space and time and which obeys the superposition principle, so it can interfere and diffract
- The particle itself is always detected as a single localised event — one dot on a screen, one click in a detector — never as a spread-out wave
- The square of the wavefunction amplitude at a point gives the probability per unit volume of detecting the particle there, so the wave predicts where detections are likely, not where the particle 'is'
- In a double-slit experiment with electrons fired one at a time, individual dots appear at random positions but the accumulated pattern reproduces the interference fringes predicted by λ = h/p
- Determining which slit the electron passed through destroys the fringes: wave and particle descriptions are complementary, never used simultaneously. Exam tip: the wave is a probability amplitude, not a description of the particle spreading out or vibrating.
- Worked problemHL onlyData booklet: Yes
Light of frequency 7.5 × 10¹⁴ Hz is incident on a sodium surface of work function 2.3 eV. Determine the maximum kinetic energy of the emitted photoelectrons, in eV and in J.
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Photon energy E = hf = 6.63 × 10⁻³⁴ × 7.5 × 10¹⁴ = 4.97 × 10⁻¹⁹ J. Convert to eV: 4.97 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 3.11 eV. Einstein's photoelectric equation (booklet): E_max = hf − Φ = 3.11 − 2.3 = 0.81 eV. In joules: E_max = 0.81 × 1.60 × 10⁻¹⁹ = 1.3 × 10⁻¹⁹ J (2 s.f., limited by Φ). Since hf > Φ emission does occur, so the answer is physically valid. Check/Trap: never subtract a work function in eV from a photon energy in joules — convert one of them first; and if the answer comes out negative it means f < f₀ and E_max = 0, not a negative kinetic energy.
- Worked problemHL onlyData booklet: Yes
A metal has a work function of 4.2 eV. Determine its threshold frequency and the longest wavelength of light that will cause photoemission, and state the region of the spectrum.
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At threshold E_max = 0, so hf₀ = Φ. Convert: Φ = 4.2 × 1.60 × 10⁻¹⁹ = 6.72 × 10⁻¹⁹ J. f₀ = Φ/h = 6.72 × 10⁻¹⁹ / 6.63 × 10⁻³⁴ = 1.0 × 10¹⁵ Hz (2 s.f.). Longest wavelength λ₀ = c/f₀ = 3.00 × 10⁸ / 1.014 × 10¹⁵ = 2.96 × 10⁻⁷ m ≈ 3.0 × 10⁻⁷ m = 296 nm. Visible light spans about 400–700 nm, so 296 nm lies in the ultraviolet: visible light of any intensity will not eject electrons from this metal. Check/Trap: λ₀ is the LONGEST wavelength, because longer λ means lower f; students routinely call it the minimum wavelength. A useful shortcut is E(eV) = 1240/λ(nm), giving λ₀ = 1240/4.2 ≈ 295 nm.
- Worked problemHL onlyData booklet: Yes
Ultraviolet light of wavelength 250 nm falls on a metal of work function 3.0 eV. Calculate the stopping potential.
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Photon energy E = hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (250 × 10⁻⁹) = 1.989 × 10⁻²⁵ / 2.50 × 10⁻⁷ = 7.96 × 10⁻¹⁹ J. In eV: 7.96 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 4.97 eV. E_max = hf − Φ = 4.97 − 3.0 = 1.97 eV = 3.15 × 10⁻¹⁹ J. Stopping potential from eV_s = E_max: V_s = E_max(in eV) / 1 = 2.0 V (2 s.f.). Check/Trap: when E_max is expressed in eV the stopping potential in volts is numerically equal to it — a fast check worth using. Do not divide 3.15 × 10⁻¹⁹ by 1.60 × 10⁻¹⁹ twice, and note that changing the lamp brightness leaves V_s at 2.0 V.
- Worked problemHL onlyData booklet: Yes
A laser emits light of wavelength 640 nm with an output power of 3.0 mW. Calculate the number of photons emitted per second.
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Energy of one photon: E = hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (640 × 10⁻⁹) = 1.989 × 10⁻²⁵ / 6.40 × 10⁻⁷ = 3.11 × 10⁻¹⁹ J (equivalently 1.94 eV). Power is energy per second, so the photon rate n = P/E = 3.0 × 10⁻³ / 3.11 × 10⁻¹⁹ = 9.7 × 10¹⁵ photons per second (2 s.f.). Check/Trap: convert mW to W and nm to m before substituting — a factor of 10⁻³ or 10⁻⁹ lost here is the single commonest arithmetic slip in this style of question. The enormous rate explains why a laser beam appears continuous even though it is quantised, and if the wavelength were halved the photon energy would double and the rate would halve for the same power.
- Worked problemHL onlyData booklet: Yes
Photoelectrons of maximum kinetic energy 0.81 eV are emitted from a sodium surface. Calculate their maximum speed and justify the use of the non-relativistic formula.
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Convert: E_max = 0.81 × 1.60 × 10⁻¹⁹ = 1.30 × 10⁻¹⁹ J. Using E_max = ½m_e v² with m_e = 9.11 × 10⁻³¹ kg: v = √(2E_max/m_e) = √(2 × 1.30 × 10⁻¹⁹ / 9.11 × 10⁻³¹) = √(2.85 × 10¹¹) = 5.3 × 10⁵ m s⁻¹ (2 s.f.). Justification: v/c = 5.3 × 10⁵ / 3.00 × 10⁸ = 1.8 × 10⁻³, so γ ≈ 1 + ½(v/c)² differs from 1 by about 2 × 10⁻⁶ and the classical expression is accurate well beyond 2 s.f. Check/Trap: this is the MAXIMUM speed; electrons from below the surface lose extra energy and emerge more slowly, so the beam has a continuous range of speeds from zero up to 5.3 × 10⁵ m s⁻¹.
- Worked problemHL onlyData booklet: No – derive
Show that an electron accelerated from rest through a potential difference V has de Broglie wavelength λ = h/√(2m_eeV), and evaluate λ for V = 250 V.
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Work–energy: the electron gains kinetic energy E_k = eV, so ½m_e v² = eV and v = √(2eV/m_e). Momentum p = m_e v = m_e√(2eV/m_e) = √(2m_e eV). Substituting into de Broglie's relation λ = h/p gives λ = h/√(2m_e eV) as required. Evaluate for V = 250 V: 2m_e eV = 2 × 9.11 × 10⁻³¹ × 1.60 × 10⁻¹⁹ × 250 = 7.29 × 10⁻⁴⁷, so p = √(7.29 × 10⁻⁴⁷) = 8.54 × 10⁻²⁴ N s. λ = 6.63 × 10⁻³⁴ / 8.54 × 10⁻²⁴ = 7.8 × 10⁻¹¹ m (2 s.f.). Check/Trap: λ ∝ 1/√V, so quadrupling the pd only halves the wavelength — a common error is to assume λ ∝ 1/V. The value 0.078 nm is comparable to atomic spacings, which is why diffraction is observable.
- Worked problemHL onlyData booklet: Yes
A cricket ball of mass 0.16 kg is bowled at 30 m s⁻¹. Calculate its de Broglie wavelength and comment on whether wave behaviour could be observed.
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Momentum p = mv = 0.16 × 30 = 4.8 kg m s⁻¹ (= 4.8 N s). de Broglie wavelength λ = h/p = 6.63 × 10⁻³⁴ / 4.8 = 1.4 × 10⁻³⁴ m (2 s.f.). Comment: appreciable diffraction requires an aperture comparable to λ. The smallest structures available are nuclei, about 10⁻¹⁵ m, which is 10¹⁹ times larger than this wavelength, so the diffraction angle θ ≈ λ/b would be of order 10⁻¹⁹ rad — utterly unmeasurable. The ball therefore follows a classical trajectory. Check/Trap: the wave nature has not 'switched off'; it is simply unobservable because h is so small. Also note the units of p are kg m s⁻¹ ≡ N s, and h has units J s = kg m² s⁻¹, so h/p correctly gives metres — a useful unit check.
- Worked problemHL onlyData booklet: Yes
Electrons accelerated through 5.0 kV are directed at a thin graphite film in which the atomic plane spacing is 2.1 × 10⁻¹⁰ m. Estimate the angle of the first-order diffraction maximum.
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Step 1, wavelength: λ = h/√(2m_e eV) with V = 5.0 × 10³ V. 2m_e eV = 2 × 9.11 × 10⁻³¹ × 1.60 × 10⁻¹⁹ × 5.0 × 10³ = 1.458 × 10⁻⁴⁵, so p = 3.82 × 10⁻²³ N s. λ = 6.63 × 10⁻³⁴ / 3.82 × 10⁻²³ = 1.74 × 10⁻¹¹ m. Step 2, diffraction: treating the planes as a grating, d sinθ = nλ with n = 1 gives sinθ = λ/d = 1.74 × 10⁻¹¹ / 2.1 × 10⁻¹⁰ = 0.0827. Step 3: θ = sin⁻¹(0.0827) = 4.7° (2 s.f.). Check/Trap: raising the accelerating pd raises p, lowers λ and so shrinks θ — the rings contract, which is the observation that confirms λ = h/p. Make sure the calculator is in degrees, and note the estimate assumes the incident beam is normal to the planes.
- Worked problemHL onlyData booklet: Yes
An X-ray photon of wavelength 5.00 × 10⁻¹¹ m is Compton-scattered through 90° by a free electron. Calculate the scattered wavelength and the kinetic energy given to the electron.
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Compton wavelength: h/(m_e c) = 6.63 × 10⁻³⁴ / (9.11 × 10⁻³¹ × 3.00 × 10⁸) = 6.63 × 10⁻³⁴ / 2.733 × 10⁻²² = 2.43 × 10⁻¹² m. At θ = 90°, cosθ = 0 so Δλ = (h/m_e c)(1 − 0) = 2.43 × 10⁻¹² m. Scattered wavelength λ_f = 5.00 × 10⁻¹¹ + 0.243 × 10⁻¹¹ = 5.24 × 10⁻¹¹ m. Energies: E_i = hc/λ_i = 1.989 × 10⁻²⁵ / 5.00 × 10⁻¹¹ = 3.98 × 10⁻¹⁵ J; E_f = 1.989 × 10⁻²⁵ / 5.24 × 10⁻¹¹ = 3.79 × 10⁻¹⁵ J. Electron kinetic energy = E_i − E_f = 1.84 × 10⁻¹⁶ J ≈ 1.2 keV. Check/Trap: keep at least 4 s.f. through the energy subtraction — rounding λ_f to 2 s.f. first destroys the small difference. Δλ is added, never subtracted.
- Worked problemHL onlyData booklet: Yes
A 0.0200 nm X-ray photon is backscattered (θ = 180°) from a stationary free electron. Determine the energy of the scattered photon and of the recoil electron, both in keV.
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At θ = 180°, cos180° = −1, so 1 − cosθ = 2 and the shift is maximal: Δλ = 2h/(m_e c) = 2 × 2.43 × 10⁻¹² = 4.85 × 10⁻¹² m. λ_i = 2.00 × 10⁻¹¹ m, so λ_f = 2.00 × 10⁻¹¹ + 0.485 × 10⁻¹¹ = 2.485 × 10⁻¹¹ m. Incident energy E_i = hc/λ_i = 1.989 × 10⁻²⁵ / 2.00 × 10⁻¹¹ = 9.945 × 10⁻¹⁵ J = 62.2 keV. Scattered energy E_f = 1.989 × 10⁻²⁵ / 2.485 × 10⁻¹¹ = 8.00 × 10⁻¹⁵ J = 50.0 keV. Recoil electron kinetic energy = 62.2 − 50.0 = 12.2 keV. Check/Trap: 180° gives the largest possible shift and the largest energy transfer, but the photon still survives — it can never lose all its energy in Compton scattering, unlike in photoelectric absorption. Convert J to keV by dividing by 1.60 × 10⁻¹⁶.
- Worked problemHL onlyData booklet: Yes
Calculate the momentum of a gamma-ray photon of energy 10.0 MeV and hence its wavelength.
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For a photon E = pc, so p = E/c. Convert the energy: E = 10.0 × 10⁶ × 1.60 × 10⁻¹⁹ = 1.60 × 10⁻¹² J. Momentum p = 1.60 × 10⁻¹² / 3.00 × 10⁸ = 5.33 × 10⁻²¹ N s (3 s.f.). Wavelength from p = h/λ: λ = h/p = 6.63 × 10⁻³⁴ / 5.33 × 10⁻²¹ = 1.24 × 10⁻¹³ m. Cross-check directly: λ = hc/E = 1.989 × 10⁻²⁵ / 1.60 × 10⁻¹² = 1.24 × 10⁻¹³ m — the two routes agree. Check/Trap: p = E/c applies only to massless particles; for an electron of the same 10 MeV you must use the relativistic relation, and using p = mv would be badly wrong. Note this λ is far smaller than an atom, which is why gamma rays are not diffracted by crystals.
- Worked problemHL onlyData booklet: Yes
In an experiment the stopping potential is 0.50 V at f = 6.0 × 10¹⁴ Hz and 1.33 V at f = 8.0 × 10¹⁴ Hz. Determine h, the work function in eV, the threshold frequency and the threshold wavelength.
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Gradient of the V_s–f line = (1.33 − 0.50)/(8.0 − 6.0) × 10¹⁴ = 0.83 / 2.0 × 10¹⁴ = 4.15 × 10⁻¹⁵ V s. Since gradient = h/e, h = 1.60 × 10⁻¹⁹ × 4.15 × 10⁻¹⁵ = 6.6 × 10⁻³⁴ J s. Work function: Φ/e = (h/e)f − V_s = 4.15 × 10⁻¹⁵ × 6.0 × 10¹⁴ − 0.50 = 2.49 − 0.50 = 1.99 V, so Φ = 2.0 eV = 3.18 × 10⁻¹⁹ J. Threshold: f₀ = Φ/h = 3.18 × 10⁻¹⁹ / 6.64 × 10⁻³⁴ = 4.8 × 10¹⁴ Hz. λ₀ = c/f₀ = 3.00 × 10⁸ / 4.79 × 10¹⁴ = 6.3 × 10⁻⁷ m = 630 nm, in the red, consistent with caesium-like metals. Check/Trap: the gradient of a V_s–f graph is h/e, not h, and the y-intercept is −Φ/e, a negative number, so a positive intercept means the axes have been mislabelled. Both given frequencies must exceed f₀ — they do — otherwise the data are inconsistent. Do not read the intercept off a badly extrapolated sketch when two points allow an exact calculation.
- Exam technique/trapHL only
State the trap in photoelectric calculations involving electronvolts and joules, and how to avoid it.
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The trap: work functions and stopping potentials are quoted in eV or V, while hf comes out in joules, and students subtract mixed units to get answers wrong by a factor of 1.60 × 10⁻¹⁹. Why students fall for it: the numbers still 'look reasonable' on a calculator, so no alarm is raised. Correct approach: choose one system at the start and convert everything into it — multiply eV by 1.60 × 10⁻¹⁹ to get J, divide J by 1.60 × 10⁻¹⁹ to get eV — and write the unit after every line. A powerful shortcut for photons is E(eV) = 1240/λ(nm). Command-term guidance: 'determine' allows any valid route but demands working; 'calculate' expects a numerical answer with unit; 'state' expects the value only. Always give the unit — an unlabelled number scores zero even when correct.
- Exam technique/trapHL only
Explain the common misconception that all photoelectrons are emitted with kinetic energy hf − Φ.
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The trap: E_max = hf − Φ gives only the MAXIMUM kinetic energy, obtained by an electron at the very surface that is bound by exactly the work function. Why students fall for it: the booklet equation carries the subscript 'max' but it is easily ignored, and mark schemes penalise answers that describe 'the' energy of the photoelectrons. Correct approach: state that electrons originating below the surface, or more tightly bound, lose additional energy before escaping, so emitted electrons have a continuous spread of kinetic energies from zero to E_max. This is exactly why the photocurrent falls gradually rather than abruptly as the retarding pd is increased, and why only the very fastest electrons are stopped at V_s. Command-term guidance: 'explain' requires the reason, not just the statement — link the spread of energies to the shape of the current–pd curve.
- Exam technique/trapHL only
Identify the frequency-versus-intensity trap in multiple-choice photoelectric questions and give a reliable method for answering them.
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The trap: a stem changes the intensity, the frequency, or both, and asks what happens to the photocurrent, E_max or V_s; distractors are built from confusing the two variables. Why students fall for it: everyday experience says brighter light is 'stronger', so intensity feels like it should give faster electrons. Correct approach: memorise the two independent chains — frequency controls the energy per photon, hence E_max and V_s and whether emission happens at all; intensity controls the number of photons per second, hence the rate of emission and the saturation current. Then read the stem and change only what it changes. Beware the case where the intensity is held constant while f is raised: the photon rate falls (n = P/hf), so the current DROPS while V_s rises. Command-term guidance: for 'deduce' you must state the reasoning chain, not merely quote the outcome.
- Exam technique/trapHL only
Describe how to determine Planck's constant with an uncertainty from a graph of stopping potential against frequency.
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Plot V_s (y) against f (x) with error bars on V_s, and draw the best-fit straight line. The gradient is h/e, so h = e × gradient. For the uncertainty draw the maximum- and minimum-gradient lines that still pass through all the error bars, calculate h from each, and take Δh = (h_max − h_min)/2, quoting h = (value ± Δh) J s with Δh to 1 s.f. and h rounded to the same decimal place. Absolute uncertainty in a gradient is found this way; fractional uncertainty Δh/h expressed as a percentage lets you compare with the accepted 6.63 × 10⁻³⁴ J s. A systematic error such as a contact potential between the electrodes shifts every V_s by the same amount: this moves the intercept, and hence the value of Φ, but leaves the gradient and therefore h unaffected — a key discriminating point. Random errors scatter points about the line and are reduced by repeating readings.
- Exam technique/trapHL only
Outline an experiment to determine the work function of a metal photoelectrically, identifying the variables, the main limitations and two improvements.
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Apparatus: an evacuated photocell with a clean metal cathode, a set of narrow-band filters or LEDs of known frequency, a variable dc supply with reversible polarity, a digital voltmeter and a picoammeter. Method: illuminate the cathode with one frequency, increase the reverse pd until the current just reaches zero, record V_s; repeat for at least six frequencies and plot V_s against f; Φ = e × (magnitude of the y-intercept), or Φ = h f₀ from the x-intercept. Independent variable f, dependent variable V_s; controlled: the same cathode surface, same collecting geometry, constant temperature. Limitations: surface oxidation raises Φ over time, stray light and photoemission from the anode give a small reverse current, and the 'zero' current is hard to judge. Improvements: cut or sputter a fresh surface in vacuum (as Millikan did), shield the cell from ambient light, and use a log-scale picoammeter with extrapolation to locate the true cut-off.
- Exam technique/trapHL only
Outline an electron diffraction tube experiment to test λ = h/p, including method, variables and how the data are linearised.
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Apparatus: an evacuated electron diffraction tube with an electron gun, a thin polycrystalline graphite target and a fluorescent screen, an EHT supply (0–5 kV) with a safety resistor, a voltmeter and a ruler or travelling microscope. Method: set an accelerating pd V, measure the diameter D of the first-order ring on the screen, and repeat for at least six values of V over the full range, allowing the tube to cool between readings. Independent variable V, dependent variable D; controlled: the same target and ring order, and a fixed tube–screen distance L. Theory gives D ∝ λ ∝ 1/√V, so plot D against 1/√V: a straight line through the origin supports λ = h/p, and the gradient combined with the known geometry and plane spacing yields h. Limitations: the ring is broad so D is hard to fix (large percentage uncertainty), screen parallax, and heating of the graphite. Improvements: photograph the screen and measure digitally, and take D as an average of several diameters.
- Exam technique/trapHL only
State the traps in Compton scattering calculations and how to avoid them.
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Traps: (1) omitting the (1 − cosθ) factor and using Δλ = h/(m_e c) for every angle — that value applies only at 90°; (2) having the calculator in radians when θ is given in degrees; (3) subtracting Δλ, giving a decrease in wavelength, which conservation of energy forbids; (4) rounding λ_f to 2 or 3 s.f. before subtracting energies, which destroys the small difference that is the electron's kinetic energy. Why students fall for it: the shift is a tiny fraction of λ, so premature rounding feels harmless. Correct approach: compute the Compton wavelength h/(m_e c) = 2.43 × 10⁻¹² m once, multiply by (1 − cosθ), ADD it, then carry at least 4 s.f. into hc/λ for both photons before subtracting. Sanity checks: 1 − cosθ runs from 0 at 0° to 2 at 180°, so Δλ can never exceed 4.85 × 10⁻¹² m, and m_e is used even when the target is a solid.
- Exam technique/trapHL only
Give command-term guidance for the standard graph questions in this sub-topic and the sketching errors that cost marks.
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'Sketch' means draw axes with labels and quantities (units not always required), show the correct shape and mark any intercept, asymptote or threshold — it does not mean plot to scale, but it does mean get the intercepts right. For E_max (or V_s) against f: a straight line that starts on the x-axis at f₀, with a negative y-intercept of −Φ (or −Φ/e) shown by extrapolating the dashed line back — the line must NOT be drawn from the origin, and no line should be shown for f < f₀. For different metals: parallel lines of identical gradient h (or h/e) with different f₀ — drawing lines of different gradients is the classic error, since h is universal. For photocurrent against pd: curves of different intensity share the same negative intercept −V_s; curves of different frequency share the same saturation current only if the photon rate is fixed. 'Describe' asks what the graph shows; 'explain' asks why, using photon language.
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