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The photoelectric effect (SACE Stage 2 Physics)

Syllabus dot point

“Explain the photoelectric effect using the photon model, including threshold frequency, work function and maximum kinetic energy.”

9 min read

Quick answer

When light above a threshold frequency hits a metal, electrons are ejected immediately. The wave model cannot explain this. Einstein's photon model can: each photon carries energy hfhf; an electron needs at least the work function WW to escape, so the maximum kinetic energy of an ejected electron is Ek,max=hf−WE_{k,\text{max}} = hf - W. Below the threshold frequency f0=W/hf_0 = W/h, no electrons are emitted no matter how bright the light.

Jump to a section
  1. What this dot point is asking
  2. What is observed
  3. Why the wave model fails
  4. Einstein's photon explanation
  5. Stopping voltage
  6. How SACE assesses this
  7. Exam-style questions

What this dot point is asking

You need to explain why the photoelectric effect contradicts the wave model and use the photoelectric equation, including threshold frequency, work function and maximum kinetic energy.

Maximum kinetic energy and threshold

Light of frequency 7.0×10147.0\times10^{14} Hz hits a metal of work function 2.02.0 eV

Find the maximum kinetic energy of the ejected electrons and the threshold frequency. Use h=6.63×10−34h = 6.63\times10^{-34} J s, 1 eV=1.60×10−191\ \text{eV} = 1.60\times10^{-19} J.

Step 1: Compute the photon energy

Each photon carries energy E=hfE = hf. Multiplying Planck's constant by the given frequency gives the energy of one photon in joules.

hf=(6.63×10−34)(7.0×1014)=4.64×10−19 J.hf = (6.63\times10^{-34})(7.0\times10^{14}) = 4.64\times10^{-19}\ \text{J}.

Step 2: Convert the work function to joules

The work function is quoted in electronvolts, but the photoelectric equation requires consistent units. Multiplying by the conversion factor gives the work function in joules.

W=2.0×1.60×10−19=3.20×10−19 J.W = 2.0\times1.60\times10^{-19} = 3.20\times10^{-19}\ \text{J}.

Step 3: Apply the photoelectric equation to find Ek,maxE_{k,\text{max}}

The photoelectric equation Ek,max=hf−WE_{k,\text{max}} = hf - W expresses that the photon energy goes partly into freeing the electron (the work function) and the remainder becomes kinetic energy.

Ek,max=hf−W=4.64×10−19−3.20×10−19=1.44×10−19 J.E_{k,\text{max}} = hf - W = 4.64\times10^{-19} - 3.20\times10^{-19} = 1.44\times10^{-19}\ \text{J}.

Step 4: Find the threshold frequency

The threshold frequency is the minimum frequency for emission, found by setting the kinetic energy to zero so that the entire photon energy just equals the work function: f0=W/hf_0 = W/h.

f0=Wh=3.20×10−196.63×10−34=4.83×1014 Hz.f_0 = \frac{W}{h} = \frac{3.20\times10^{-19}}{6.63\times10^{-34}} = 4.83\times10^{14}\ \text{Hz}.

Final answer: Ek,max=1.44×10−19 JE_{k,\text{max}} = 1.44\times10^{-19}\ \text{J}; threshold frequency f0=4.83×1014 Hzf_0 = 4.83\times10^{14}\ \text{Hz}.

What is observed

When light shines on a metal surface, electrons (photoelectrons) can be ejected. The key experimental facts are:

  • Emission happens only above a threshold frequency f0f_0. Below it, no electrons are emitted however intense the light.
  • Above f0f_0, electrons are emitted immediately, with no time delay even for very dim light.
  • The maximum kinetic energy of the electrons depends on the frequency, not the intensity.
  • Increasing the intensity (above f0f_0) increases the number of electrons per second, not their energy.

Why the wave model fails

The wave model predicts that energy is delivered continuously and spread over the surface, so even low-frequency light should eventually eject electrons if you wait, and brighter light should give more energetic electrons. Both predictions are wrong: there is a sharp frequency threshold and intensity has no effect on electron energy. The wave model cannot account for these facts.

Einstein's photon explanation

Einstein proposed that light arrives as discrete photons, each of energy hfhf, and that one electron absorbs one photon.

The photoelectric equation

Ek,max=hf−WE_{k,\text{max}} = hf - W

Ek,maxE_{k,\text{max}} is the maximum kinetic energy of an ejected electron, hfhf the photon energy, and WW the work function (the minimum energy needed to free an electron from the metal). Emission occurs only when hf≥Whf \ge W, i.e. above the threshold frequency f0=W/hf_0 = W/h.

This explains every observation: below f0f_0 a single photon lacks the energy WW, so no electron escapes regardless of how many photons arrive (intensity); above f0f_0 each photon has surplus energy that becomes the electron's kinetic energy, and a brighter beam simply delivers more photons, freeing more electrons.

Stopping voltage

The maximum kinetic energy can be measured with a stopping voltage VsV_s - the reverse voltage that just stops the most energetic electrons:

eVs=Ek,max.eV_s = E_{k,\text{max}}.

A graph of Ek,maxE_{k,\text{max}} against frequency is a straight line of gradient hh and intercept −W-W, which is how Planck's constant can be measured from the photoelectric effect.

How SACE assesses this

SACE Stage 2 photoelectric questions are usually calculations using Ek,max=hf−WE_{k,\text{max}} = hf - W: find the maximum kinetic energy from a given frequency and work function, or find the stopping voltage from eVs=Ek,maxeV_s = E_{k,\text{max}}, or determine the threshold frequency f0=Whf_0 = \dfrac{W}{h}. The recurring trap is unit consistency, since the work function may be quoted in eV or in joules; convert it to joules before subtracting from hfhf. A common explanation part asks why the wave model fails, where the marking points are the existence of a sharp threshold frequency and the fact that intensity changes the number of electrons but not their energy. Quote the photoelectric equation, evaluate hfhf as a separate step, and state answers with units.

Common errors
Thinking brighter light gives more energetic electrons
Intensity changes the number of electrons, not their energy. Only frequency sets Ek,maxE_{k,\text{max}}.
Forgetting the threshold
Below f0=W/hf_0 = W/h no electrons are emitted at all, no matter how intense the light.
Mixing up units
Convert the work function between eV and joules consistently before subtracting from hfhf.
Confusing work function and kinetic energy
WW is the energy to escape; Ek,max=hf−WE_{k,\text{max}} = hf - W is what is left over as kinetic energy.

Exam-style questions

Questions in the style of SACE Board exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

SACE 2025-style3 marks
The work function of a metal target is 1.78 eV1.78\ \text{eV}. Monochromatic light of frequency 5.45×1014 Hz5.45\times10^{14}\ \text{Hz} is incident on it. Calculate the maximum kinetic energy of the emitted electrons. Use h=6.63×10−34 J sh = 6.63\times10^{-34}\ \text{J s} and 1 eV=1.60×10−19 J1\ \text{eV} = 1.60\times10^{-19}\ \text{J}.
Show worked answer →

Use Ek,max=hf−WE_{k,\text{max}} = hf - W.

Photon energy: hf=(6.63×10−34)(5.45×1014)=3.61×10−19 Jhf = (6.63\times10^{-34})(5.45\times10^{14}) = 3.61\times10^{-19}\ \text{J}.

Work function in joules: W=1.78×1.60×10−19=2.85×10−19 JW = 1.78\times1.60\times10^{-19} = 2.85\times10^{-19}\ \text{J}.

Ek,max=3.61×10−19−2.85×10−19=7.6×10−20 J.E_{k,\text{max}} = 3.61\times10^{-19} - 2.85\times10^{-19} = 7.6\times10^{-20}\ \text{J}.

1 mark for the photon energy, 1 mark for converting the work function to joules, 1 mark for the answer of about 7.6×10−20 J7.6\times10^{-20}\ \text{J}.

SACE 2024-style3 marks
The work function of a metal surface is 5.30×10−19 J5.30\times10^{-19}\ \text{J}. Light of frequency 1.50×1015 Hz1.50\times10^{15}\ \text{Hz} is incident on it. Calculate the stopping voltage. Use h=6.63×10−34 J sh = 6.63\times10^{-34}\ \text{J s} and e=1.60×10−19 Ce = 1.60\times10^{-19}\ \text{C}.
Show worked answer →

First the maximum kinetic energy, Ek,max=hf−WE_{k,\text{max}} = hf - W:

hf=(6.63×10−34)(1.50×1015)=9.945×10−19 J,hf = (6.63\times10^{-34})(1.50\times10^{15}) = 9.945\times10^{-19}\ \text{J},

Ek,max=9.945×10−19−5.30×10−19=4.645×10−19 J.E_{k,\text{max}} = 9.945\times10^{-19} - 5.30\times10^{-19} = 4.645\times10^{-19}\ \text{J}.

The stopping voltage just removes this kinetic energy, eVs=Ek,maxeV_s = E_{k,\text{max}}:

Vs=Ek,maxe=4.645×10−191.60×10−19=2.90 V.V_s = \dfrac{E_{k,\text{max}}}{e} = \dfrac{4.645\times10^{-19}}{1.60\times10^{-19}} = 2.90\ \text{V}.

1 mark for Ek,maxE_{k,\text{max}}, 1 mark for eVs=Ek,maxeV_s = E_{k,\text{max}}, 1 mark for the answer of about 2.90 V2.90\ \text{V}.

Practise this

Sources & how we know this

Based on the official SACE Board subject outline. Not yet individually reviewed by a person; always check SACE Board and your school's materials for current rules.

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