Is Kmax = hf − φ on the HSC data sheet?

Yes. The 2019 data sheet gives Kmax = hf − φ, E = hf, v = fλ (use c for light) and Wien's law λmax = b ÷ T, plus h, c and the value of 1 eV. E = hc ÷ λ and Kmax = qV0 are not printed, but you get them from E = hf with c = fλ, and from W = qV.

What is the difference between the work function and the threshold frequency?

The work function φ is an energy: the least energy needed to free an electron from the metal's surface. The threshold frequency f0 is the lowest light frequency that can do it. They are linked by φ = hf0, and both are fixed for a given metal.

Why doesn't the intensity of light affect the kinetic energy of photoelectrons?

Because each electron is freed by one photon, and every photon at a given frequency has the same energy hf. Brighter light means more photons per second, so more electrons and a bigger current, but each electron still gets the same energy.

What does the gradient of a Kmax versus frequency graph give?

Planck's constant h, when Kmax is in joules. If Kmax is in eV, or you plot stopping voltage in volts, the gradient is h ÷ e, about 4.14 × 10^−15 eV s or V s. Every metal gives the same gradient.

Why is it the maximum kinetic energy?

The work function is the energy needed for the most loosely held electrons at the surface. Electrons from deeper inside lose extra energy getting out, so they leave with less. Kmax = hf − φ describes the fastest ones.

Light above the threshold frequency falls on a metal. The intensity is doubled and the frequency kept the same. What happens?

The number of electrons per second doubles and Kmax stays the same.

Kmax doubles and the number of electrons per second stays the same.

Both Kmax and the number of electrons per second double.

Nothing changes, because only frequency matters.

On a Kmax against frequency graph, metal B's line is parallel to metal A's but crosses the frequency axis further to the right. Which statement is correct?

B has the larger work function, and both lines have gradient h.

B has the smaller work function, and both lines have gradient h.

B has the larger work function and a larger value of h.

B's electrons always have more kinetic energy than A's at the same frequency.

HSC Physics · Module 7 · The Nature of Light · 10 min read

The photoelectric effect

Shine ultraviolet light on a clean zinc plate and electrons fly off it. Shine bright red light on the same plate and nothing happens, however bright you make it. That is the photoelectric effect, and the wave model of light cannot explain it. Einstein's answer, that light arrives in packets of energy called photons, is one of the ideas the HSC asks about most.

This guide covers the black body background that led Planck to E = hf, what the photoelectric experiments showed, how K_max = hf − φ explains them, and how to read the K_max–frequency graph. Then you get the common mistakes, a quick quiz and a full worked example.

What you need to know

Analyse the evidence from black body radiation experiments, including Wien's law (λ_max = b ÷ T), and how it led to Planck's role in changing how light is modelled.
Investigate the photoelectric effect results that a wave picture of light could not explain.
Use the photon model of light and conservation of energy to analyse the photoelectric effect in metals, using K_max = hf − φ.

The equations (from the 2019 HSC data sheet unless marked):

E = hfThe energy of one photon. h = 6.626 × 10⁻³⁴ J s, f in hertz, E in joules.

c = fλOn the sheet as v = fλ. Use it to turn a wavelength into a frequency before E = hf.

E = hc ÷ λ not on the data sheetE = hf and c = fλ combined. A longer wavelength means a lower photon energy.

K_max = hf − φPhoton energy in, work function used to escape, the rest is the fastest electron's kinetic energy.

K_max = qV₀ not on the data sheetFrom W = qV on the sheet: the stopping voltage V₀ just stops the fastest electrons. In eV, K_max has the same number as V₀ in volts.

λ_max = b ÷ TWien's law for black bodies, b = 2.898 × 10⁻³ m K, T in kelvin.

Dot points: PH-M7-11, PH-M7-12, PH-M7-13 (2017 Physics syllabus, HSC 2026 and 2027), in our own words.

Mini lesson

Black body radiation: where quanta began

A black body absorbs all the radiation that hits it and gives out a spectrum that depends only on its temperature. The curve rises to a peak and falls away. A hotter body gives out more at every wavelength, and its peak moves to a shorter wavelength: λ_max = b ÷ T. The Sun's surface, at about 5800 K, peaks near 2.898 × 10⁻³ ÷ 5800 = 5.0 × 10⁻⁷ m, or 500 nm.

Classical wave physics, which let energy take any value, predicted that the intensity would keep climbing as the wavelength got shorter. That runaway at short wavelengths (the 'ultraviolet catastrophe') is not what experiments found. In 1900 Planck matched the real curve by assuming energy is emitted in whole packets of size E = hf. He treated it as a mathematical trick; Einstein later took it seriously for light itself.

The hotter body is higher at every wavelength and peaks (λ_max) at a shorter wavelength. The dashed classical prediction runs away at short wavelengths (the 'ultraviolet catastrophe'); Planck's E = hf fixed that.

What the experiments showed

In a photocell, light falls on a metal plate in a vacuum tube and any electrons knocked out (photoelectrons) are collected and measured as a current. Four results did not fit the idea of light as a continuous wave, whose energy depends on its intensity:

A threshold frequency. Below a certain frequency f₀ no electrons come out, however bright the light. A wave model says bright enough light should always work.
No delay. Above f₀, electrons come out straight away, even in very dim light. A wave would need time to build up enough energy in one electron.
Intensity changes the number, not the energy. Brighter light gives more electrons per second (a bigger current) but the same maximum kinetic energy.
Frequency changes the energy. A higher frequency gives a higher K_max, in a straight-line relationship.

Einstein's photon model

Einstein (1905) said light itself is made of photons, each with energy E = hf, and that one photon gives all its energy to one electron. Conservation of energy then gives K_max = hf − φ. The work function φ is the least energy needed to free an electron from that metal's surface. Electrons from deeper down lose extra energy on the way out, which is why the equation gives the maximum kinetic energy.

Every observation now follows. If hf is less than φ, no single photon can free an electron, so there is a threshold frequency f₀ = φ ÷ h. One photon acts at once, so there is no delay. Brighter light means more photons, so more electrons, but each photon still has the same hf, so K_max does not change.

Photon energies are tiny in joules, so they are often given in electronvolts: 1 eV = 1.602 × 10⁻¹⁹ J. A 400 nm violet photon has E = hc ÷ λ = (6.626 × 10⁻³⁴ × 3.00 × 10⁸) ÷ (400 × 10⁻⁹) = 4.97 × 10⁻¹⁹ J, or 3.10 eV.

The K_max–frequency graph

K_max = hf − φ has the form y = mx + c. Plot K_max against f and you get a straight line. The gradient is h, the same for every metal. The x-intercept is the threshold frequency f₀. Extending the line back to f = 0 gives a y-intercept of −φ. If K_max is plotted in eV (or the stopping voltage in V), the gradient is h ÷ e = 4.14 × 10⁻¹⁵ eV s instead.

A different metal gives a parallel line, shifted along the axis: a larger work function means a larger threshold frequency, so the line sits further right.

Gradient h, x-intercept f₀, y-intercept −φ (extrapolated, dashed). Metal B has a larger work function, so its parallel line starts further right.

Stopping voltage, and intensity versus frequency

Make the collector negative and the photoelectrons are pushed back. The smallest reverse voltage that stops even the fastest electron, so the current drops to zero, is the stopping voltage V₀. The electric field does work qV₀ on the electron, so K_max = qV₀. A stopping voltage of 1.5 V means K_max = 1.5 eV.

On a current–voltage graph this separates the two variables cleanly. Brighter light at the same frequency raises the current (more photons, more electrons) but leaves V₀ the same. Higher frequency at the same brightness makes V₀ bigger, because each photon carries more energy.

Same frequency, two brightnesses: different maximum current, the same stopping voltage −V₀. Only a change of frequency moves V₀.

Common mistakes

Saying brighter light gives the electrons more kinetic energy.Brightness sets how many photons arrive per second, so how many electrons come out. Each photon's energy, and so K_max, depends only on frequency.
Thinking very bright light below the threshold frequency will eventually eject electrons.One photon frees one electron. If hf < φ, no photon has enough energy, so there is no emission at any brightness.
Saying a longer wavelength photon has more energy.E = hc ÷ λ, so a longer wavelength means a lower frequency and a lower photon energy. Red photons carry less energy than ultraviolet ones.
Mixing eV and joules: subtracting φ in eV from hf in joules.Pick one unit for the whole line. Convert with 1 eV = 1.602 × 10⁻¹⁹ J before you subtract.
Reading the work function off the frequency axis, or calling the y-intercept +φ.The x-intercept is the threshold frequency f₀. The extrapolated y-intercept is −φ, so the work function is the size of that intercept.
Joining plotted points dot to dot, or not extending the line to the axes.Rule one straight line of best fit with points balanced either side, then extend it to read f₀ and −φ.

Worked example

Ultraviolet light of wavelength 365 nm from a mercury lamp falls on a clean metal plate in a photocell. The photocurrent drops to zero when the reverse (stopping) voltage reaches 0.95 V.

Calculate the work function of the metal in eV, and use it to decide whether a very bright green lamp of wavelength 546 nm would eject photoelectrons from the same plate. Justify your answer.

(5 marks)

I'll think out loud, the way you should in the exam, using KNOW-UD: Knowns, Not known, Organise, Work it out, Units, Direction.

Prediction first. A 365 nm photon is about 3.4 eV (ultraviolet is a few eV). The stopping voltage says the fastest electrons have 0.95 eV, so the work function is about 3.4 − 0.95 ≈ 2.45 eV. Green light is 546 ÷ 365 ≈ 1.5 times the wavelength, so its photons have about 3.4 ÷ 1.5 ≈ 2.3 eV. I predict that is just short of the work function, so no electrons, however bright the lamp.

K · Knowns

I sketch the photocell with the values on it, then write them as symbols in SI units. Nanometres become metres straight away, and the stopping voltage gives me K_max directly.

λ₁ = 365 nm = 365 × 10⁻⁹ m (UV), V₀ = 0.95 V
λ₂ = 546 nm = 546 × 10⁻⁹ m (green)
h = 6.626 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m s⁻¹, 1 eV = 1.602 × 10⁻¹⁹ J

Knowns on the sketch: λ = 365 nm, stopping voltage V₀ = 0.95 V (current just zero). The green lamp, λ = 546 nm, is the second case to test.

N · Not known

I need φ in eV first. Then the decision: I'll compare the green photon's energy with φ (or, the same thing, its frequency with the threshold frequency).

φ = ? eV, E_green = ? eV (compare with φ)

O · Organise

Conservation of energy for one photon and one electron. I write the equations in symbols before any numbers, and I'll work in joules then convert, so I never subtract eV from J.

E = hc ÷ λ (from E = hf and c = fλ)
K_max = qV₀
φ = hf − K_max = hc ÷ λ₁ − qV₀
Emission only if hc ÷ λ₂ > φ

W · Work it out

Photon energy for the UV first, keeping the full calculator value. Then the electron's energy from the stopping voltage, then subtract. The threshold frequency is a useful extra line for the justification.

E_UV = (6.626 × 10⁻³⁴ × 3.00 × 10⁸) ÷ (365 × 10⁻⁹) = 5.446 × 10⁻¹⁹ J
K_max = 1.602 × 10⁻¹⁹ × 0.95 = 1.522 × 10⁻¹⁹ J
φ = 5.446 × 10⁻¹⁹ − 1.522 × 10⁻¹⁹ = 3.924 × 10⁻¹⁹ J
φ = (3.924 × 10⁻¹⁹) ÷ (1.602 × 10⁻¹⁹) = 2.450 eV
f₀ = φ ÷ h = (3.924 × 10⁻¹⁹) ÷ (6.626 × 10⁻³⁴) = 5.92 × 10¹⁴ Hz
Green: f = c ÷ λ₂ = (3.00 × 10⁸) ÷ (546 × 10⁻⁹) = 5.49 × 10¹⁴ Hz
E_green = hf = 6.626 × 10⁻³⁴ × 5.495 × 10¹⁴ = 3.641 × 10⁻¹⁹ J = 2.27 eV

U · Units

The question asks for the work function in eV, so I give that, to two significant figures like the stopping voltage. The comparison is energy with energy, both in eV.

φ ≈ 2.4 eV (2.45 eV)
E_green = 2.27 eV < φ = 2.45 eV; f_green = 5.49 × 10¹⁴ Hz < f₀ = 5.92 × 10¹⁴ Hz

D · Direction and decision

Energy is a scalar, so there's no direction; instead I finish the 'justify' with the physics reason, not just the numbers. The key idea is one photon per electron, so brightness can't make up the shortfall.

No photoelectrons: each green photon (2.27 eV) has less energy than the work function (2.45 eV), and one photon frees at most one electron.
A brighter lamp only sends more of these photons per second; none of them can free an electron.

The work function is about 2.4 eV (2.45 eV). The 546 nm green lamp will not eject photoelectrons, however bright it is: its photons have only 2.27 eV, below the work function, and its frequency (5.49 × 10¹⁴ Hz) is below the threshold frequency (5.92 × 10¹⁴ Hz).

Check against the prediction. I predicted a work function of about 2.45 eV and green photons of about 2.3 eV, so no emission. The working gives 2.45 eV and 2.27 eV: it matches. As a further check, the threshold wavelength hc ÷ φ = 507 nm, and 546 nm is longer than that, so it is too red to work.

HSC advice

What markers keep saying

Plot points so they can be seen (small crosses, not dots), rule a straight line of best fit, and extend it to the frequency axis to read the threshold frequency, reading the axis scale carefully. If the graph shows stopping voltage against frequency, its gradient is h ÷ e, not h.NESA marking feedback, 2025, summarised
If a question asks you to justify using the gradient, give reasons: a line uses every data point, so it reduces the effect of errors in any one reading. Working out h again is not a justification.NESA marking feedback, 2021, summarised
Rule the line of best fit with the points balanced evenly either side, then extend it back to f = 0: the intercept there is −φ, so its size is the work function. Watch the scales and powers of ten on each axis.NESA marking feedback, 2019, summarised
When explaining why the photoelectric effect supports a particle model, describe the one-photon-to-one-electron transfer and the threshold frequency, and use the numbers in the question to show it, rather than just naming the effect.NESA marking feedback, 2021, summarised
Remember which way photon energy goes with wavelength: a longer wavelength means less energy per photon. Use that to pick the right photon (for example, the longest wavelength a detector responds to) before you calculate.NESA marking feedback, 2023, summarised
When comparing the classical and quantum models of black body radiation, state the assumption each one makes: energy can take any value (classical waves), or energy comes in packets of size hf (Planck). A consequence of the model is not the assumption.NESA marking feedback, 2020, summarised
λ_max in Wien's law is the wavelength at the peak of the curve, not the longest wavelength emitted. Read the peak off a graph carefully, and use T in kelvin.NESA marking feedback, 2023, summarised

Summaries in our own words. Read NESA's own feedback with the exam packs: 2019 · 2020 · 2021 · 2023 · 2025.

How often it is examined

The photoelectric effect questions were worth 53 marks across the seven HSC Physics exams from 2019 to 2025 (about 7.6 marks a paper), and appeared in 7 of 7 papers.

Marks per paper (Module 7, Light: Quantum Model). Source: CortexHSC's own tagging of the NESA HSC Physics exams 2019–2025, each question counted under its main topic, so treat the numbers as close, not exact.

FAQ

Is K_max = hf − φ on the HSC data sheet?: Yes. The 2019 data sheet gives K_max = hf − φ, E = hf, v = fλ (use c for light) and Wien's law λ_max = b ÷ T, plus h, c and the value of 1 eV. E = hc ÷ λ and K_max = qV₀ are not printed, but you get them from E = hf with c = fλ, and from W = qV.
What is the difference between the work function and the threshold frequency?: The work function φ is an energy: the least energy needed to free an electron from the metal's surface. The threshold frequency f₀ is the lowest light frequency that can do it. They are linked by φ = hf₀, and both are fixed for a given metal.
Why doesn't the intensity of light affect the kinetic energy of photoelectrons?: Because each electron is freed by one photon, and every photon at a given frequency has the same energy hf. Brighter light means more photons per second, so more electrons and a bigger current, but each electron still gets the same energy.
What does the gradient of a K_max versus frequency graph give?: Planck's constant h, when K_max is in joules. If K_max is in eV, or you plot stopping voltage in volts, the gradient is h ÷ e, about 4.14 × 10⁻¹⁵ eV s or V s. Every metal gives the same gradient.
Why is it the maximum kinetic energy?: The work function is the energy needed for the most loosely held electrons at the surface. Electrons from deeper inside lose extra energy getting out, so they leave with less. K_max = hf − φ describes the fastest ones.

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