SACE Board SACE 2025-style practice: The work function of a metal target is 1.78\ eV. Monochromatic light of frequency 5.4510^14\ Hz is incident on it. Calculate the maximum kinetic energy of the emitted electrons. Use h = 6.6310^-34\ J s and 1\ eV = 1.6010^-19\ J.

Use E_k,max = hf - W. Photon energy: hf = (6.6310^-34)(5.4510^14) = 3.6110^-19\ J. Work function in joules: W = 1.781.6010^-19 = 2.8510^-19\ J. $E_k,max = 3.6110^-19 - 2.8510^-19 = 7.610^-20\ J.$ 1 mark for the photon energy, 1 mark for converting the work function to joules, 1 mark for the answer of about 7.610^-20\ J.

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Why can't the wave model explain the photoelectric effect, and what does it reveal about the nature of light?

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

Why the photoelectric effect contradicts the wave model, Einstein's photon explanation, and the photoelectric equation linking photon energy, work function and electron kinetic energy.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
What is observed
Why the wave model fails
Einstein's photon explanation
Stopping voltage
How SACE assesses this

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 $7.0\times10^{14}$ Hz hits a metal of work function $2.0$ eV

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

Step 1: Compute the photon energy

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

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\times1.60\times10^{-19} = 3.20\times10^{-19}\ \text{J}.

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

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

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: $f_0 = W/h$ .

f_0 = \frac{W}{h} = \frac{3.20\times10^{-19}}{6.63\times10^{-34}} = 4.83\times10^{14}\ \text{Hz}.

Final answer: $E_{k,\text{max}} = 1.44\times10^{-19}\ \text{J}$ ; threshold frequency $f_0 = 4.83\times10^{14}\ \text{Hz}$ .

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 $f_0$ . Below it, no electrons are emitted however intense the light.
Above $f_0$ , 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 $f_0$ ) 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 $hf$ , and that one electron absorbs one photon.

This explains every observation: below $f_0$ a single photon lacks the energy $W$ , so no electron escapes regardless of how many photons arrive (intensity); above $f_0$ 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 $V_s$ - the reverse voltage that just stops the most energetic electrons:

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

A graph of

E_{k,\text{max}}

against frequency is a straight line of gradient

h

and intercept

-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 $E_{k,\text{max}} = hf - W$ : find the maximum kinetic energy from a given frequency and work function, or find the stopping voltage from $eV_s = E_{k,\text{max}}$ , or determine the threshold frequency $f_0 = \dfrac{W}{h}$ . 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 $hf$ . 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 $hf$ as a separate step, and state answers with units.

Exam-style practice questions

Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SACE 20253 marksThe work function of a metal target is

1.78\ \text{eV}

. Monochromatic light of frequency

5.45\times10^{14}\ \text{Hz}

is incident on it. Calculate the maximum kinetic energy of the emitted electrons. Use

h = 6.63\times10^{-34}\ \text{J s}

and

1\ \text{eV} = 1.60\times10^{-19}\ \text{J}

Show worked answer →

Use $E_{k,\text{max}} = hf - W$ .

Photon energy: $hf = (6.63\times10^{-34})(5.45\times10^{14}) = 3.61\times10^{-19}\ \text{J}$ .

Work function in joules: $W = 1.78\times1.60\times10^{-19} = 2.85\times10^{-19}\ \text{J}$ .

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\times10^{-20}\ \text{J}

SACE 20243 marksThe work function of a metal surface is

5.30\times10^{-19}\ \text{J}

. Light of frequency

1.50\times10^{15}\ \text{Hz}

is incident on it. Calculate the stopping voltage. Use

h = 6.63\times10^{-34}\ \text{J s}

and

e = 1.60\times10^{-19}\ \text{C}

Show worked answer →

First the maximum kinetic energy, $E_{k,\text{max}} = hf - W$ :

hf = (6.63\times10^{-34})(1.50\times10^{15}) = 9.945\times10^{-19}\ \text{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, $eV_s = E_{k,\text{max}}$ :

V_s = \dfrac{E_{k,\text{max}}}{e} = \dfrac{4.645\times10^{-19}}{1.60\times10^{-19}} = 2.90\ \text{V}.

1 mark for

E_{k,\text{max}}

, 1 mark for

eV_s = E_{k,\text{max}}

, 1 mark for the answer of about

2.90\ \text{V}