What is the photoelectric effect and how does it relate to Edexcel International A Levels Physics?

The photoelectric effect is a phenomenon where electrons are emitted from a material when it is exposed to light of a certain frequency. This concept is crucial in Edexcel International A Levels Physics as it demonstrates the particle nature of light and supports the quantum theory. Understanding this effect helps students grasp how energy is quantized and how it interacts with matter, which is a fundamental aspect of the Waves and Electricity topic.

How does the photoelectric effect support the particle theory of light?

The photoelectric effect supports the particle theory of light by showing that light can be thought of as a stream of particles called photons. When light of a sufficient frequency strikes a material, it transfers energy to electrons in discrete packets, or quanta, causing them to be emitted. This observation contradicts the wave theory of light, which cannot explain why light below a certain frequency fails to eject electrons, regardless of its intensity.

What is the significance of atomic spectra in understanding atomic structure?

Atomic spectra are significant because they provide evidence for the quantized nature of energy levels within an atom. When electrons transition between these levels, they emit or absorb light at specific wavelengths, creating a spectrum. This concept is crucial in Edexcel International A Levels Physics as it helps students understand the electronic structure of atoms and the principles of quantum mechanics, which are essential for explaining chemical properties and reactions.

How do you calculate the energy of a photon in the context of the photoelectric effect?

In the context of the photoelectric effect, the energy of a photon can be calculated using the equation E = hf, where E is the energy, h is Planck's constant (6.63 x 10^-34 Js), and f is the frequency of the light. This formula is fundamental in Edexcel International A Levels Physics as it links the frequency of light to its energy, helping students understand how different frequencies can affect the emission of electrons from a material.

Why are certain frequencies of light unable to cause the photoelectric effect, regardless of intensity?

Certain frequencies of light are unable to cause the photoelectric effect because they do not have enough energy per photon to overcome the work function of the material, which is the minimum energy required to eject an electron. This threshold frequency is a key concept in Edexcel International A Levels Physics, as it highlights the quantized nature of energy transfer and the limitations of classical wave theory in explaining the behavior of light and electrons.

Why is "Thinking that increasing the intensity (brightness) of the light increases the kinetic energy of the photoelectrons" a common mistake in the Photoelectric Effect and Atomic Spectra?

Why it happens: Students carry over the everyday idea that 'brighter = more energetic', and confuse the number of photons with the energy of each photon. How to avoid it: Separate the two clearly: **intensity controls the NUMBER of photons per second (hence the rate of emission / photocurrent); frequency controls the ENERGY of each photon (hence KE_max = hf - ).** More intense light frees more electrons, not faster ones. Source: Edexcel IAL Physics Unit 2 examiner reports

Why is "In a levels-based answer, failing to explain WHY the wave model fails — just describing the photoelectric effect instead" a common mistake in the Photoelectric Effect and Atomic Spectra?

Why it happens: Students learn the observations but not the contrast between what each model predicts, so they list facts rather than argue. How to avoid it: **Always contrast the prediction with the observation.** The wave model predicts emission at any frequency given enough intensity/time and a time delay for dim light; experiment shows a sharp threshold frequency and instantaneous emission. The photon model (E = hf, one-to-one) explains both. Write linked prose, not a list. Source: Edexcel IAL Physics Unit 2 examiner reports

Why is "Mixing electronvolts and joules in the same equation (e.g. $KE_{\text{max}} = hf - \phi$ with $hf$ in joules but $\phi$ in eV)" a common mistake in the Photoelectric Effect and Atomic Spectra?

Why it happens: Work functions and energy levels are often quoted in eV, while h, c and e are in SI units, so the two unit systems collide. How to avoid it: **Convert everything to joules first** (multiply eV by 1.60 × 10^-19) before adding or subtracting, then convert back to eV at the end only if asked. Never subtract an energy in eV from one in joules. Source: Edexcel IAL Physics Unit 2 examiner reports

Why is "Treating $hf - \phi$ as the kinetic energy of every emitted electron, rather than the MAXIMUM kinetic energy" a common mistake in the Photoelectric Effect and Atomic Spectra?

Why it happens: The equation KE = hf - is memorised without the subscript 'max', so the special meaning is lost. How to avoid it: Remember the equation gives **KE_max only**. Electrons released from below the surface use up extra energy escaping the metal, so they emerge with LESS than hf - . The stopping potential measures this maximum: eV_s = KE_max. Source: Edexcel IAL Physics Unit 2 examiner reports

Why is "Getting the sign or order wrong when finding the photon energy from energy levels (e.g. $E_{\text{lower}} - E_{\text{upper}}$ giving a negative answer)" a common mistake in the Photoelectric Effect and Atomic Spectra?

Why it happens: Energy levels are negative numbers, and subtracting negatives is error-prone under exam pressure. How to avoid it: Use E = E_upper - E_lower and check the **photon energy is positive** for an emitted photon. For a fall from -1.51\ eV to -3.40\ eV: E = (-1.51) - (-3.40) = +1.89\ eV. Source: Edexcel IAL Physics Unit 2 examiner reports

Why is "Suggesting that an atom can emit or absorb photons of any energy, rather than only those matching the gaps between discrete levels" a common mistake in the Photoelectric Effect and Atomic Spectra?

Why it happens: Students picture energy as continuous (like a ramp) instead of quantised (like a ladder), so they miss why spectra are LINE spectra. How to avoid it: Stress that energy levels are **discrete**, so only photons whose energy exactly equals a gap ( E = hf) can be emitted or absorbed. This is precisely why we see sharp **lines** at specific frequencies, not a continuous band. Source: Edexcel IAL Physics Unit 2 examiner reports

Edexcel International A Levels Physics (XPH11-YPH11)

the Photoelectric Effect and Atomic Spectra

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Short Notes - the Photoelectric Effect and Atomic Spectra

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Detailed Study Notes

Detailed notes on Waves and Electricity for Edexcel International A Levels Physics, covering key concepts, explanations, examples, and exam-focused revision points.

The Photoelectric Effect and Atomic Spectra — Pearson Edexcel International A Level Physics (XPH11/YPH11) Study Notes (Unit 2)

Light as photons of energy $E=hf$ ; the photoelectric effect and the one-photon-one-electron model; work function, threshold frequency and Einstein's photoelectric equation; why the wave model fails and the photon model succeeds; the stopping potential; and discrete atomic energy levels producing emission and absorption line spectra.

At a glance

Light is quantised: it arrives as photons, each carrying energy $E = hf = \dfrac{hc}{\lambda}$ , where $h = 6.63\times10^{-34}\ \text{J s}$ .
Photoelectric effect: photons of high enough frequency eject electrons from a metal. One photon interacts with one electron, giving up all its energy at once.
Work function $\phi$ = the minimum energy to release an electron. Threshold frequency $f_0$ satisfies $\phi = hf_0$ — below it, no emission however bright the light.
Einstein's equation: $hf = \phi + KE_{\text{max}}$ . The maximum kinetic energy is $KE_{\text{max}} = hf - \phi$ , and the stopping potential gives it: $eV_s = KE_{\text{max}}$ .
Intensity vs frequency: intensity sets the number of photons (rate of emission); frequency sets the energy of each photon ( $KE_{\text{max}}$ ). Brighter light ≠ faster electrons.
Why the wave model fails: it predicts emission at any frequency given enough intensity/time and a time delay for dim light — both contradicted by the threshold frequency and instant emission.
Atomic energy levels are discrete: an electron dropping from level $E_2$ to $E_1$ emits a photon of energy $\Delta E = E_2 - E_1 = hf$ . Levels are usually negative.
Emission spectrum = bright lines (electrons fall, photons emitted); absorption spectrum = dark lines at the same frequencies (photons absorbed, electrons jump up).

What you’ll learn

Mapped to the Pearson Edexcel International A Levels XPH11-YPH11 syllabus (2018-onwards).

Understand that electromagnetic radiation can be modelled as a stream of photons, each of energy $E = hf = \dfrac{hc}{\lambda}$ .
Describe the photoelectric effect as evidence for the particle (photon) nature of light, and explain it as a one-photon-one-electron interaction.
Define the work function $\phi$ and threshold frequency $f_0$ , and use $\phi = hf_0$ .
Use Einstein's photoelectric equation $hf = \phi + KE_{\text{max}}$ , and relate the maximum kinetic energy to the stopping potential via $eV_s = KE_{\text{max}}$ .
Explain why the wave model of light cannot account for the threshold frequency and instantaneous emission, whereas the photon model can.
Explain the existence of discrete atomic energy levels and use $\Delta E = hf$ to relate transitions to the frequency/wavelength of emitted or absorbed photons (emission and absorption line spectra).
Use the electronvolt as a unit of energy and convert between electronvolts and joules ( $1\ \text{eV} = 1.60\times10^{-19}\ \text{J}$ ).

Light as photons — the quantum idea

Light comes in packets of energy E = hf; the electronvolt is the handy energy unit.

Classically light is a wave. But to explain the photoelectric effect and line spectra, we treat light as a stream of discrete energy packets called photons. Each photon carries a fixed amount of energy set only by the frequency of the light:

$E = hf = \frac{hc}{\lambda}$

$h = 6.63\times10^{-34}\ \text{J s}$ is the Planck constant.
$f$ is the frequency (Hz), $\lambda$ the wavelength (m), and $c = 3.00\times10^{8}\ \text{m s}^{-1}$ the speed of light.

So higher frequency (shorter wavelength) = more energetic photons. A single ultraviolet photon carries far more energy than a single red-light photon, even though both travel at $c$ .

The electronvolt (eV). Atomic-scale energies in joules are tiny, so physicists use the electronvolt: the energy gained by an electron accelerated through a potential difference of 1 V.

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

To convert eV → J, multiply by $1.60\times10^{-19}$ ; to convert J → eV, divide by $1.60\times10^{-19}$ . This single conversion underlies almost every calculation in this topic, because work functions and energy levels are usually quoted in eV while $h$ , $c$ and $e$ are in SI units.

Light delivers energy in discrete packets (photons), each of energy E = hf. The photon model — not the continuous wave model — is what explains the photoelectric effect and atomic line spectra.

Photon energy $E = hf = \dfrac{hc}{\lambda}$ , with $h = 6.63\times10^{-34}\ \text{J s}$ .
Higher frequency / shorter wavelength = more energetic photons.
$1\ \text{eV} = 1.60\times10^{-19}\ \text{J}$ (eV → J: multiply; J → eV: divide).
Convert all energies to joules before combining with SI constants.

See the full worked example for the photoelectric effect and atomic spectra →

The photoelectric effect — work function and threshold frequency

One photon frees one electron only if hf ≥ φ; below the threshold frequency, nothing happens.

Shine light on a clean metal surface and, if the frequency is high enough, electrons are ejected. This is the photoelectric effect, and the ejected electrons are called photoelectrons.

The crucial experimental facts are:

There is a threshold frequency $f_0$ . Below it, no electrons are emitted at all, however intense the light or however long it shines.
Above $f_0$ , emission is instantaneous — even for very dim light there is no measurable time delay.
Increasing the intensity (above $f_0$ ) increases the number of electrons emitted per second, but not their kinetic energy.
Increasing the frequency increases the maximum kinetic energy of the electrons.

These facts are explained by the one-photon-one-electron model: a single photon interacts with a single electron and transfers all its energy in one go. To escape, the electron must be given at least the work function $\phi$ — the minimum energy needed to free an electron from the metal's surface.

At the threshold frequency, the photon energy exactly equals the work function and the electron just escapes with zero kinetic energy:

$\phi = hf_0 \qquad\Rightarrow\qquad f_0 = \frac{\phi}{h}$

Below $f_0$ , no single photon has enough energy, and turning up the intensity only sends more photons — not more energetic ones — so still nothing happens. This is why the wave picture (energy delivered continuously, so brightness should always eventually free electrons) fails, while the photon picture succeeds.

A single photon of energy hf strikes one electron. Part of the energy (the work function φ) is used to escape the metal surface; the rest becomes the electron's maximum kinetic energy: hf = φ + KE max.

Work function $\phi$ = minimum energy to release a surface electron.
Threshold frequency: $\phi = hf_0$ , so $f_0 = \phi/h$ ; below $f_0$ no emission.
One photon interacts with one electron, transferring all its energy at once.
Intensity → number of electrons; frequency → kinetic energy of electrons.

See the full worked example for the photoelectric effect and atomic spectra →

Einstein's photoelectric equation and the stopping potential

hf = φ + KEmax is energy conservation; the stopping potential measures KEmax.

Einstein applied conservation of energy to the one-photon-one-electron interaction. The photon's energy goes partly into freeing the electron (the work function) and partly into the electron's kinetic energy:

$hf = \phi + KE_{\text{max}}$

Rearranged, the maximum kinetic energy of a photoelectron is:

$KE_{\text{max}} = hf - \phi = \tfrac{1}{2}mv_{\text{max}}^2$

Measuring $KE_{\text{max}}$ — the stopping potential. Apply a reverse voltage to the photocell so the collector is negative. As you raise this voltage, fewer electrons reach the collector; the stopping potential $V_s$ is the voltage that just stops even the fastest electrons. The work done against the field then equals their maximum kinetic energy:

$eV_s = KE_{\text{max}}$

A neat consequence: the maximum kinetic energy in electronvolts is numerically equal to the stopping potential in volts (e.g. $KE_{\text{max}} = 0.90\ \text{eV} \Rightarrow V_s = 0.90\ \text{V}$ ).

The $KE_{\text{max}}$ against $f$ graph. Writing Einstein's equation as $KE_{\text{max}} = hf - \phi$ and comparing with $y = mx + c$ shows it is a straight line:

gradient $= h$ (the Planck constant — the same for every metal);
y-intercept $= -\phi$ (negative);
x-intercept $= f_0$ (the threshold frequency, where $KE_{\text{max}} = 0$ ).

This graph is a favourite "determine the Planck constant / work function" question. To find $h$ , take the gradient from a large triangle; to find $\phi$ , use $\phi = hf_0$ from the x-intercept.

Plotting KE max against frequency gives a straight line of gradient h (the Planck constant). It crosses the frequency axis at the threshold frequency f₀, and its back-extension meets the energy axis at −φ. Every metal gives a line of the same gradient but a different intercept.

Einstein: $hf = \phi + KE_{\text{max}}$ , so $KE_{\text{max}} = hf - \phi$ .
Stopping potential: $eV_s = KE_{\text{max}}$ .
$KE_{\text{max}}$ vs $f$ graph: gradient = $h$ , x-intercept = $f_0$ , y-intercept = $-\phi$ .
Use a large gradient triangle to find $h$ ; use $\phi = hf_0$ for the work function.

See the full worked example for the photoelectric effect and atomic spectra →

Why the wave model fails and the photon model wins

Threshold frequency + instant emission kill the wave model; E = hf saves the day.

The photoelectric effect is the classic evidence for the particle nature of light, and it is examined as a levels-based 6-marker. The trick is to contrast what each model predicts with what is actually observed.

What the wave model predicts (and why it is wrong):

A wave delivers energy continuously and spread over the whole wavefront, so a brighter beam should always free electrons eventually — at any frequency. ✗ Observed: below $f_0$ , no emission however bright.
A dim wave should take time to feed enough energy to an electron, so there should be a measurable time delay. ✗ Observed: emission is instantaneous above $f_0$ .
A more intense wave should give electrons more energy. ✗ Observed: intensity changes the rate of emission, not the maximum kinetic energy.

What the photon model predicts (and why it is right):

Light arrives as photons, each of energy $E = hf$ , and one photon frees one electron in a single interaction, transferring all its energy at once.
An electron is released only if a single photon supplies at least the work function: $hf \geq \phi$ , i.e. $f \geq f_0$ . This explains the threshold frequency exactly.
Increasing intensity sends more photons (more electrons per second) but does not make individual photons more energetic — so $KE_{\text{max}}$ depends on frequency, not intensity.
The first suitable photon to arrive frees an electron straight away, so emission is instantaneous, even in dim light.

Write your answer as continuous linked prose, applying $E = hf$ and the one-to-one interaction to each observation. A bullet list of facts, with no model contrast, caps the marks.

Wave model: predicts emission at any frequency, a time delay, and energy rising with intensity — all wrong.
Observations: a sharp threshold frequency, instantaneous emission, $KE_{\text{max}}$ set by frequency.
Photon model: $E = hf$ , one photon frees one electron, needs $hf \geq \phi$ .
Intensity → more photons (rate); frequency → photon energy (so $KE_{\text{max}}$ ).

See the full worked example for the photoelectric effect and atomic spectra →

Atomic energy levels and line spectra

Electrons sit on discrete energy rungs; ΔE = hf links transitions to spectral lines.

The photoelectric effect shows that light is quantised. Atomic line spectra show that an atom's electron energies are quantised too: an electron in an atom can only occupy certain discrete energy levels, like rungs on a ladder. The levels are usually written as negative numbers, measured from the ionised electron at $0\ \text{eV}$ , with the ground state the most negative (most tightly bound).

When an electron moves between levels, the energy difference is carried by a single photon:

$\Delta E = E_2 - E_1 = hf = \frac{hc}{\lambda}$

Emission: an electron drops from a higher level $E_2$ to a lower level $E_1$ and a photon is emitted with energy $\Delta E = E_2 - E_1$ . Many excited atoms produce bright emission lines at definite frequencies.
Absorption: a photon of exactly the right energy is absorbed, lifting an electron up to a higher level. In white light passing through a cool gas, these frequencies are removed, leaving dark absorption lines at the same positions as the emission lines.

Because the levels are discrete, only photons whose energy exactly matches a gap can be emitted or absorbed — which is why spectra are lines at specific frequencies, not a continuous band. Always convert energies in eV to joules before using $h$ and $c$ , and take $\Delta E = E_{\text{upper}} - E_{\text{lower}}$ so the emitted photon energy comes out positive.

Electrons occupy discrete energy levels. A downward transition emits a photon (emission line); an upward transition absorbs a photon (absorption line). The photon energy equals the gap, ΔE = hf, so only specific frequencies appear — hence sharp spectral lines rather than a continuous band.

Atomic energy levels are discrete; levels are usually negative, ground state most negative.
Transition photon energy: $\Delta E = E_2 - E_1 = hf = \dfrac{hc}{\lambda}$ .
Emission = electron falls, photon emitted (bright lines).
Absorption = photon absorbed, electron rises (dark lines at the same frequencies).

See the full worked example for the photoelectric effect and atomic spectra →

Quick recap

Light is quantised into photons of energy $E = hf = \dfrac{hc}{\lambda}$ ; $1\ \text{eV} = 1.60\times10^{-19}\ \text{J}$ (convert eV → J before using SI constants).
The photoelectric effect is a one-photon-one-electron interaction: an electron escapes only if $hf \geq \phi$ , i.e. $f \geq f_0$ where $\phi = hf_0$ .
Einstein's equation $hf = \phi + KE_{\text{max}}$ gives the MAXIMUM kinetic energy; the stopping potential measures it via $eV_s = KE_{\text{max}}$ .
Intensity controls the number of photons (rate of emission); frequency controls each photon's energy (and so $KE_{\text{max}}$ ).
The wave model fails (it predicts emission at any frequency and a time delay); the photon model explains the threshold frequency and instant emission.
Atomic energy levels are discrete; a transition emits/absorbs a photon of energy $\Delta E = E_2 - E_1 = hf$ — giving line spectra.
Emission = bright lines (electron falls); absorption = dark lines at the same frequencies (electron rises).

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

Photon energy: $E = hf = \dfrac{hc}{\lambda}$ , with $h = 6.63\times10^{-34}\ \text{J s}$ .
Einstein's photoelectric equation: $hf = \phi + KE_{\text{max}}$ , so $KE_{\text{max}} = hf - \phi = \tfrac{1}{2}mv_{\text{max}}^2$ .
Work function and threshold frequency: $\phi = hf_0$ , so $f_0 = \dfrac{\phi}{h}$ (no emission below $f_0$ ).
Stopping potential: $eV_s = KE_{\text{max}}$ .
$KE_{\text{max}}$ vs $f$ graph: gradient $= h$ ; x-intercept $= f_0$ ; y-intercept $= -\phi$ .
Energy-level transition: $\Delta E = E_2 - E_1 = hf = \dfrac{hc}{\lambda}$ (levels usually negative).
Intensity → number of photons (rate); frequency → energy per photon (so $KE_{\text{max}}$ ).
$1\ \text{eV} = 1.60\times10^{-19}\ \text{J}$ — convert eV → J before combining with $h$ , $c$ , $e$ .

How it’s examined

This subtopic sits in Unit 2 (WPH12, Waves and Electricity) and is examined on the Unit 2 paper as short calculations, a 'show that', a graph question, and a levels-based 6-marker. Typical formats: define photon / work function / threshold frequency (1–2 marks, AO1); calculate a photon energy, threshold frequency, $KE_{\text{max}}$ or stopping potential, with full $E=hf$ / $hf=\phi+KE_{\text{max}}$ working and an eV ↔ J conversion (2–4 marks, M and A marks); determine the Planck constant or work function from a $KE_{\text{max}}$ -against- $f$ graph using a large gradient triangle and the intercepts (3–4 marks, AO3); the classic 'explain why the wave model cannot account for the photoelectric effect but the photon model can' (6 marks, levels-based, continuous prose contrasting both models); and energy-level calculations of the frequency/wavelength of an emitted or absorbed photon (2–4 marks). ALWAYS show the equation, substitution with units, and a 3 s.f. answer — 'no working = risk zero'; for a 'show that' work through to one extra s.f.; keep all energies in the same unit before comparing.

Sources: Pearson Edexcel International Advanced Level Physics (YPH11/XPH11) specification (Issue 3, July 2021); Edexcel IAL Physics Unit 2 examiner reports; Edexcel IAL Physics Unit 2 sample assessment materials and mark schemes. Last reviewed 2026-06-26.

Take this whole topic with you

Step-by-step worked examples — the Photoelectric Effect and Atomic Spectra

Step-by-step solutions to past-paper-style questions on the photoelectric effect and atomic spectra, written exactly the way a tutor would explain them at the board.

Question type:

1Energy of a single photon (2 marks)
Getting startedDirect calculation• photon energy, AO2, Unit 2
Question
Light of frequency $5.0 \times 10^{14}\ \text{Hz}$ is emitted by a source. Take $h = 6.63 \times 10^{-34}\ \text{J s}$ . Calculate the energy of a single photon of this light. (2 marks)
Step-by-step solution
1. Step 1
  The energy of one photon is $E = hf$ , where $h$ is the Planck constant and $f$ is the frequency.
  $E = hf$
2. Step 2
  Substitute with units and evaluate.
  $E = 6.63 \times 10^{-34} \times 5.0 \times 10^{14} = 3.315 \times 10^{-19}\ \text{J}$
Answer
$E = 3.32 \times 10^{-19}\ \text{J}$ (to 3 s.f.).
Examiner tip
Mark scheme: (1) correct use of $E=hf$ with substitution (M); (2) $3.32 \times 10^{-19}\ \text{J}$ (A). A frequent slip is mismanaging the powers of ten ( $10^{-34} \times 10^{14} = 10^{-20}$ , then $6.63 \times 5.0 = 33.15$ gives $33.15 \times 10^{-20} = 3.32 \times 10^{-19}$ ).
2Converting an energy from electronvolts to joules (2 marks)
Getting startedDirect calculation• electronvolt, AO2, Unit 2
Question
The work function of a certain metal is $2.30\ \text{eV}$ . Using $1\ \text{eV} = 1.60 \times 10^{-19}\ \text{J}$ , express the work function in joules. (2 marks)
Step-by-step solution
1. Step 1
  One electronvolt is the energy gained by an electron accelerated through a potential difference of one volt, equal to $1.60 \times 10^{-19}\ \text{J}$ . Multiply the number of eV by this conversion factor.
  $\phi = 2.30 \times (1.60 \times 10^{-19})$
2. Step 2
  Evaluate the product.
  $\phi = 3.68 \times 10^{-19}\ \text{J}$
Answer
$\phi = 3.68 \times 10^{-19}\ \text{J}$ .
Examiner tip
Mark scheme: (1) multiply by $1.60 \times 10^{-19}$ (M); (2) $3.68 \times 10^{-19}\ \text{J}$ (A). To go the other way (J → eV) you DIVIDE by $1.60 \times 10^{-19}$ . Mixing eV and J in the same equation (e.g. $hf - \phi$ with $hf$ in joules but $\phi$ in eV) is one of the most penalised errors in this topic.
3Threshold frequency from the work function (3 marks)
Getting startedDirect calculation• threshold frequency, work function, Unit 2
Question
Caesium has a work function of $3.04 \times 10^{-19}\ \text{J}$ . Take $h = 6.63 \times 10^{-34}\ \text{J s}$ . Calculate the threshold frequency for photoemission from caesium. (3 marks)
Step-by-step solution
1. Step 1
  At the threshold frequency $f_0$ , a photon carries exactly enough energy to release an electron with no kinetic energy left over, so $\phi = hf_0$ .
  $\phi = hf_0$
2. Step 2
  Rearrange for the threshold frequency $f_0$ .
  $f_0 = \frac{\phi}{h} = \frac{3.04 \times 10^{-19}}{6.63 \times 10^{-34}}$
3. Step 3
  Evaluate.
  $f_0 = 4.586 \times 10^{14}\ \text{Hz}$
Answer
$f_0 = 4.59 \times 10^{14}\ \text{Hz}$ (to 3 s.f.).
Examiner tip
Mark scheme: (1) $\phi = hf_0$ stated; (2) rearranged and substituted (M); (3) $4.59 \times 10^{14}\ \text{Hz}$ (A). Photons of frequency below $f_0$ never cause emission, however bright the light — a key conceptual point examined alongside the calculation.
4Maximum kinetic energy from Einstein's equation (3 marks)
Building confidenceDirect calculation• Adapted from Edexcel IAL Physics Unit 2• photoelectric equation, AO2, Unit 2
Question
Light of frequency $7.0 \times 10^{14}\ \text{Hz}$ is shone onto a metal of work function $3.2 \times 10^{-19}\ \text{J}$ . Take $h = 6.63 \times 10^{-34}\ \text{J s}$ . Calculate the maximum kinetic energy of the emitted photoelectrons. (3 marks)
Step-by-step solution
1. Step 1
  Einstein's photoelectric equation is $hf = \phi + KE_{\text{max}}$ , so $KE_{\text{max}} = hf - \phi$ . First find the photon energy $hf$ .
  $hf = 6.63 \times 10^{-34} \times 7.0 \times 10^{14} = 4.641 \times 10^{-19}\ \text{J}$
2. Step 2
  Subtract the work function (both energies are in joules, so they are compatible).
  $KE_{\text{max}} = hf - \phi = 4.641 \times 10^{-19} - 3.2 \times 10^{-19}$
3. Step 3
  Evaluate.
  $KE_{\text{max}} = 1.441 \times 10^{-19}\ \text{J}$
Answer
$KE_{\text{max}} = 1.44 \times 10^{-19}\ \text{J}$ (to 3 s.f.).
Examiner tip
Mark scheme: (1) photon energy $hf = 4.64 \times 10^{-19}\ \text{J}$ ; (2) subtract $\phi$ (M); (3) $1.44 \times 10^{-19}\ \text{J}$ (A). This is the MAXIMUM kinetic energy — electrons deeper in the metal lose more energy escaping and emerge with less. The equation gives the maximum, not a typical, value.
5Stopping potential from maximum kinetic energy (3 marks)
Building confidenceDirect calculation• stopping potential, AO2, Unit 2
Question
In a photocell the maximum kinetic energy of the emitted photoelectrons is $1.44 \times 10^{-19}\ \text{J}$ . Take $e = 1.60 \times 10^{-19}\ \text{C}$ . Calculate the stopping potential needed to just prevent any electrons from reaching the collector. (3 marks)
Step-by-step solution
1. Step 1
  The stopping potential $V_s$ is the reverse voltage that just stops the fastest electrons. Work done against it equals their maximum kinetic energy: $eV_s = KE_{\text{max}}$ .
  $eV_s = KE_{\text{max}}$
2. Step 2
  Rearrange for $V_s$ .
  $V_s = \frac{KE_{\text{max}}}{e} = \frac{1.44 \times 10^{-19}}{1.60 \times 10^{-19}}$
3. Step 3
  Evaluate.
  $V_s = 0.900\ \text{V}$
Answer
$V_s = 0.900\ \text{V}$ .
Examiner tip
Mark scheme: (1) $eV_s = KE_{\text{max}}$ stated; (2) substitution (M); (3) $0.900\ \text{V}$ (A). Because $eV_s = KE_{\text{max}}$ , the maximum kinetic energy in electronvolts is numerically equal to the stopping potential in volts — a handy check ( $1.44 \times 10^{-19}\ \text{J} = 0.900\ \text{eV} \Rightarrow V_s = 0.900\ \text{V}$ ).
6Wavelength of an emitted spectral line (4 marks)
Building confidenceMulti-step problem• Adapted from Edexcel IAL Physics Unit 2• atomic spectra, energy levels, AO2, Unit 2
Question
In a hydrogen atom an electron falls from an energy level at $-1.51\ \text{eV}$ to a level at $-3.40\ \text{eV}$ . Take $h = 6.63 \times 10^{-34}\ \text{J s}$ , $c = 3.00 \times 10^{8}\ \text{m s}^{-1}$ and $1\ \text{eV} = 1.60 \times 10^{-19}\ \text{J}$ . Calculate the wavelength of the emitted photon. (4 marks)
Step-by-step solution
1. Step 1
  The photon energy equals the difference between the two levels: $\Delta E = E_{\text{upper}} - E_{\text{lower}} = (-1.51) - (-3.40) = 1.89\ \text{eV}$ .
  $\Delta E = (-1.51) - (-3.40) = 1.89\ \text{eV}$
2. Step 2
  Convert this energy to joules before using SI constants.
  $\Delta E = 1.89 \times 1.60 \times 10^{-19} = 3.024 \times 10^{-19}\ \text{J}$
3. Step 3
  The photon energy is $\Delta E = hf = \dfrac{hc}{\lambda}$ , so rearrange for the wavelength $\lambda = \dfrac{hc}{\Delta E}$ .
  $\lambda = \frac{hc}{\Delta E} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^{8}}{3.024 \times 10^{-19}}$
4. Step 4
  Evaluate.
  $\lambda = 6.58 \times 10^{-7}\ \text{m} = 658\ \text{nm}$
Answer
$\lambda = 6.58 \times 10^{-7}\ \text{m}$ (about $658\ \text{nm}$ , red light — the hydrogen $H_\alpha$ line).
Examiner tip
Mark scheme: (1) $\Delta E = 1.89\ \text{eV}$ (subtracting the negative levels correctly); (2) convert to $3.02 \times 10^{-19}\ \text{J}$ ; (3) $\lambda = hc/\Delta E$ substitution (M); (4) $6.58 \times 10^{-7}\ \text{m}$ (A). Forgetting to convert eV to J before using $h$ and $c$ in SI units is the dominant error.
7Why the wave model cannot explain a threshold frequency (6 marks)
StretchWord problem• Adapted from Edexcel IAL Physics Unit 2• photon model, wave model, AO1, Unit 2
Question
When monochromatic light below a certain (threshold) frequency is shone onto a clean metal surface, no electrons are emitted, no matter how intense the light or how long it is left. Above the threshold frequency, electrons are emitted instantaneously even for very dim light. Explain why these observations cannot be accounted for by the wave model of light but are explained by the photon (particle) model. (6 marks)
Step-by-step solution
1. Step 1
  State the wave-model prediction. In the wave model, energy is delivered continuously and is spread over the whole wavefront, so a more intense beam delivers more energy per second. The wave model therefore predicts that ANY frequency should eventually free electrons if the light is bright enough or left long enough.
2. Step 2
  Contrast with observation 1 — the threshold frequency. Experimentally, below the threshold frequency no electrons are emitted however intense or prolonged the light. This directly contradicts the wave model, because intensity should make no difference to whether emission happens, only how quickly.
3. Step 3
  Contrast with observation 2 — instant emission. The wave model predicts a time delay while a weak wave gradually delivers enough energy to an electron. In fact emission is instantaneous, even in very dim light above $f_0$ .
4. Step 4
  Introduce the photon model. Light arrives as discrete packets (photons) each of energy $E = hf$ . One photon interacts with one electron in a single event (a one-to-one interaction), giving up all its energy at once.
5. Step 5
  Explain the threshold with the photon model. An electron is only released if a single photon supplies at least the work function $\phi$ . Since $E = hf$ , this requires $hf \geq \phi$ , i.e. $f \geq f_0 = \phi/h$ . Below $f_0$ no single photon has enough energy, and increasing intensity only sends MORE photons, not more energetic ones — so still no emission.
  $hf_0 = \phi$
6. Step 6
  Explain instant emission. Above $f_0$ , the very first photon to strike an electron frees it immediately, so emission is instantaneous even for dim light (fewer photons, but each still energetic enough). This is why the photon model succeeds where the wave model fails.
Answer
The wave model fails because it predicts emission at any frequency given enough intensity/time, and a time delay for dim light — both contradicted by experiment. The photon model (one photon, energy $hf$ , frees one electron only if $hf \geq \phi$ ) explains the threshold frequency and the instantaneous emission.
Examiner tip
Mark scheme (levels-based, up to 6): wave model predicts intensity/time would free electrons at any frequency (1); contradicted by the threshold frequency (1); wave model predicts a delay, but emission is instant (1); photon model — discrete quanta $E=hf$ , one-to-one interaction (1); threshold explained $hf \geq \phi$ , intensity gives more photons not more energy (1); instant emission because the first suitable photon frees the electron (1). Write CONTINUOUS LINKED PROSE applying the physics — a bullet list of facts caps the marks.
8Determining the Planck constant from a KEmax–f graph (6 marks)
StretchGraph or diagram• Adapted from Edexcel IAL Physics Unit 2• graph, Planck constant, AO3, Unit 2
Question
A graph of the maximum kinetic energy $KE_{\text{max}}$ of photoelectrons (y-axis, in J) against the frequency $f$ of the incident light (x-axis, in Hz) is a straight line. A large gradient triangle gives $\Delta(KE_{\text{max}}) = 3.32 \times 10^{-19}\ \text{J}$ for $\Delta f = 5.00 \times 10^{14}\ \text{Hz}$ , and the line crosses the frequency axis at $f_0 = 4.60 \times 10^{14}\ \text{Hz}$ . Take $h_{\text{accepted}} = 6.63 \times 10^{-34}\ \text{J s}$ . (a) Determine a value for the Planck constant. (b) Determine the work function of the metal. (6 marks)
Step-by-step solution
1. Step 1
  Rearrange Einstein's equation into $y = mx + c$ form: $KE_{\text{max}} = hf - \phi$ . Comparing with $y = mx + c$ (with $y = KE_{\text{max}}$ , $x = f$ ), the GRADIENT equals $h$ and the y-intercept equals $-\phi$ .
  $KE_{\text{max}} = hf - \phi \;\Rightarrow\; \text{gradient} = h$
2. Step 2
  Calculate the gradient from the triangle, including units — this gives $h$ directly.
  $h = \frac{\Delta(KE_{\text{max}})}{\Delta f} = \frac{3.32 \times 10^{-19}}{5.00 \times 10^{14}}$
3. Step 3
  Evaluate the gradient.
  $h = 6.64 \times 10^{-34}\ \text{J s}$
4. Step 4
  (b) The work function is most reliably found from the x-intercept (the threshold frequency), where $KE_{\text{max}} = 0$ so $\phi = hf_0$ .
  $\phi = hf_0 = 6.63 \times 10^{-34} \times 4.60 \times 10^{14}$
5. Step 5
  Evaluate the work function.
  $\phi = 3.05 \times 10^{-19}\ \text{J}$
Answer
(a) $h = 6.64 \times 10^{-34}\ \text{J s}$ (very close to the accepted $6.63 \times 10^{-34}\ \text{J s}$ ). (b) $\phi = hf_0 = 3.05 \times 10^{-19}\ \text{J}$ ( $\approx 1.90\ \text{eV}$ ).
Examiner tip
Mark scheme: (a) recognise gradient $= h$ (1); gradient from a large triangle (1); $h = 6.64 \times 10^{-34}\ \text{J s}$ (1). (b) $\phi = hf_0$ from the x-intercept (1); substitution (1); $3.05 \times 10^{-19}\ \text{J}$ (1). Note the y-intercept is $-\phi$ (negative) and the x-intercept is $f_0$ — confusing the two is common. Every metal gives a line of the SAME gradient $h$ but a different intercept.
9Absorption line and the emitted photon's frequency (6 marks)
StretchMulti-step problem• atomic spectra, absorption, emission, AO2, Unit 2
Question
An atom has energy levels at $E_1 = -10.4\ \text{eV}$ (ground state), $E_2 = -5.5\ \text{eV}$ and $E_3 = -3.0\ \text{eV}$ . Take $h = 6.63 \times 10^{-34}\ \text{J s}$ and $1\ \text{eV} = 1.60 \times 10^{-19}\ \text{J}$ . (a) Calculate the frequency of a photon that can be absorbed to excite the atom from the ground state to the level $E_2$ . (b) An electron then drops from $E_3$ to $E_2$ . Calculate the frequency of the photon emitted. (c) Explain why an absorption spectrum shows dark lines at exactly the frequencies seen as bright lines in the emission spectrum. (6 marks)
Step-by-step solution
1. Step 1
  (a) For absorption from $E_1$ to $E_2$ the photon must supply exactly the energy gap $\Delta E = E_2 - E_1 = (-5.5) - (-10.4) = 4.9\ \text{eV}$ .
  $\Delta E = (-5.5) - (-10.4) = 4.9\ \text{eV}$
2. Step 2
  Convert to joules, then use $\Delta E = hf$ rearranged for $f$ .
  $\Delta E = 4.9 \times 1.60 \times 10^{-19} = 7.84 \times 10^{-19}\ \text{J}$
3. Step 3
  Find the absorbed photon's frequency.
  $f = \frac{\Delta E}{h} = \frac{7.84 \times 10^{-19}}{6.63 \times 10^{-34}} = 1.18 \times 10^{15}\ \text{Hz}$
4. Step 4
  (b) For emission $E_3 \to E_2$ the photon energy is $\Delta E = E_3 - E_2 = (-3.0) - (-5.5) = 2.5\ \text{eV} = 4.00 \times 10^{-19}\ \text{J}$ .
  $\Delta E = (-3.0) - (-5.5) = 2.5\ \text{eV} = 4.00 \times 10^{-19}\ \text{J}$
5. Step 5
  Find the emitted photon's frequency.
  $f = \frac{4.00 \times 10^{-19}}{6.63 \times 10^{-34}} = 6.03 \times 10^{14}\ \text{Hz}$
6. Step 6
  (c) The energy levels are discrete, so the same energy gaps exist whether the atom absorbs (electron up) or emits (electron down) a photon. Absorption removes photons of those exact energies from a continuous spectrum (dark lines), while emission produces photons of exactly those energies (bright lines) — so they coincide in frequency.
Answer
(a) $f = 1.18 \times 10^{15}\ \text{Hz}$ . (b) $f = 6.03 \times 10^{14}\ \text{Hz}$ . (c) Because the energy levels are discrete and fixed, the same $\Delta E$ gaps govern both absorption (upward) and emission (downward), so the dark absorption lines fall at the same frequencies as the bright emission lines.
Examiner tip
Mark scheme: (a) $\Delta E = 4.9\ \text{eV}$ , convert, $f = 1.18 \times 10^{15}\ \text{Hz}$ (M+A). (b) $\Delta E = 2.5\ \text{eV}$ , $f = 6.03 \times 10^{14}\ \text{Hz}$ (M+A). (c) discrete levels → identical energy gaps → same photon frequencies for absorption and emission (2). Subtracting the negative level energies in the correct order ( $E_{\text{upper}} - E_{\text{lower}}$ ) is the key discriminator.

Model Answers — the Photoelectric Effect and Atomic Spectra

High-scoring sample answers for the photoelectric effect and atomic spectra on the Cambridge IGCSE paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
2 marks
Q (2 marks). State what is meant by a photon, and write down the equation relating a photon's energy to the frequency of the light.
Model answer
A photon is a discrete quantum (packet) of electromagnetic energy; light is delivered in these indivisible packets rather than continuously.

The energy of a photon is given by:

$E = hf$

where $h$ is the Planck constant and $f$ is the frequency of the radiation.
Why this scores
Why this scores 2/2. (1) photon = a quantum/discrete packet of electromagnetic energy; (2) $E = hf$ with the symbols identified. Describing a photon merely as 'a particle of light' without the idea of a discrete energy quantum is weak; the word quantum/packet is what examiners credit.
Question 2
2 marks
Q (2 marks). Define the work function of a metal, and state the relationship between the work function and the threshold frequency.
Model answer
The work function $\phi$ is the minimum energy needed to release (emit) an electron from the surface of the metal.

It is related to the threshold frequency $f_0$ by:

$\phi = hf_0$

where $f_0$ is the minimum frequency of light that can cause photoemission and $h$ is the Planck constant.
Why this scores
Why this scores 2/2. (1) work function = minimum energy to remove/release an electron from the metal surface; (2) $\phi = hf_0$ . The word minimum is essential — 'the energy to release an electron' without it is incomplete, because a typical photon may carry more than $\phi$ .
Question 3
3 marks
Q (3 marks). Explain the difference between an emission line spectrum and an absorption line spectrum in terms of electron transitions between energy levels.
Model answer
Both arise because an atom's electrons can only occupy discrete (fixed) energy levels.

In an emission spectrum, an electron drops from a higher level to a lower level and the energy lost is carried away as a photon. The atom emits photons of definite energies $\Delta E = hf$ , seen as bright coloured lines on a dark background.

In an absorption spectrum, white light passes through the (cooler) gas and electrons absorb photons of exactly the right energy to jump up to a higher level. Those specific frequencies are removed from the continuous spectrum, leaving dark lines at the same positions as the emission lines.
Why this scores
Why this scores 3/3. (1) discrete energy levels; (2) emission = electron falls to lower level, photon emitted, bright lines; (3) absorption = photon absorbed, electron jumps up, dark lines at the same frequencies. The fact that the lines coincide (same energy gaps) is the high-value detail.
Question 4
4 marks
Q (4 marks). Light of wavelength $4.50 \times 10^{-7}\ \text{m}$ falls on a metal of work function $1.95\ \text{eV}$ . Take $h = 6.63 \times 10^{-34}\ \text{J s}$ , $c = 3.00 \times 10^{8}\ \text{m s}^{-1}$ and $1\ \text{eV} = 1.60 \times 10^{-19}\ \text{J}$ . Calculate the maximum kinetic energy of the emitted photoelectrons, in joules.
Model answer
First find the photon energy from the wavelength:

$hf = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^{8}}{4.50 \times 10^{-7}} = 4.42 \times 10^{-19}\ \text{J}$

Convert the work function to joules so the energies are compatible:

$\phi = 1.95 \times 1.60 \times 10^{-19} = 3.12 \times 10^{-19}\ \text{J}$

Apply Einstein's photoelectric equation, $KE_{\text{max}} = hf - \phi$ :

$KE_{\text{max}} = 4.42 \times 10^{-19} - 3.12 \times 10^{-19} = 1.30 \times 10^{-19}\ \text{J}$

The maximum kinetic energy is $1.30 \times 10^{-19}\ \text{J}$ (about $0.81\ \text{eV}$ ).
Why this scores
Why this scores 4/4. (1) $hf = hc/\lambda$ method; (2) photon energy $4.42 \times 10^{-19}\ \text{J}$ ; (3) convert $\phi$ to joules $3.12 \times 10^{-19}\ \text{J}$ ; (4) $KE_{\text{max}} = 1.30 \times 10^{-19}\ \text{J}$ . The marks are lost most often by subtracting a work function still in eV from a photon energy in joules — convert first.
Question 5
4 marks
Q (4 marks). A clean metal surface is illuminated with monochromatic light above its threshold frequency. (a) State and explain the effect on the rate of electron emission of increasing the intensity of the light (frequency unchanged). (b) State and explain the effect on the maximum kinetic energy of the electrons of increasing the frequency of the light (intensity unchanged).
Model answer
(a) Increasing the intensity (at fixed frequency) means more photons per second strike the surface. Since each photon can free at most one electron in a one-to-one interaction, more electrons are emitted per second — the photocurrent increases. The intensity does not change each photon's energy, so the maximum kinetic energy is unaffected.

(b) Increasing the frequency increases each photon's energy, $E = hf$ . From $KE_{\text{max}} = hf - \phi$ , with the work function $\phi$ fixed, a larger $hf$ gives a larger maximum kinetic energy. The number of photons (and hence the rate of emission) does not change if the intensity is held constant.
Why this scores
Why this scores 4/4. (a) 2 marks: more intensity → more photons per second → more electrons per second (rate up), via one-photon-one-electron. (b) 2 marks: higher frequency → more energetic photons → larger $KE_{\text{max}}$ via $KE_{\text{max}} = hf - \phi$ . The classic trap is claiming intensity raises the electrons' energy — it raises the NUMBER, not the energy.
Question 6
4 marks
Q (4 marks). An electron in an atom falls from an energy level of $-2.4 \times 10^{-19}\ \text{J}$ to a level of $-8.6 \times 10^{-19}\ \text{J}$ . Take $h = 6.63 \times 10^{-34}\ \text{J s}$ and $c = 3.00 \times 10^{8}\ \text{m s}^{-1}$ . Calculate (a) the energy of the emitted photon and (b) its wavelength.
Model answer
(a) The photon energy is the difference between the two levels (upper minus lower):

$\Delta E = E_{\text{upper}} - E_{\text{lower}} = (-2.4 \times 10^{-19}) - (-8.6 \times 10^{-19}) = 6.2 \times 10^{-19}\ \text{J}$

(b) Using $\Delta E = \dfrac{hc}{\lambda}$ , rearranged for the wavelength:

$\lambda = \frac{hc}{\Delta E} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^{8}}{6.2 \times 10^{-19}} = 3.21 \times 10^{-7}\ \text{m}$

The emitted photon has energy $6.2 \times 10^{-19}\ \text{J}$ and wavelength $3.21 \times 10^{-7}\ \text{m}$ (about $321\ \text{nm}$ , in the ultraviolet).
Why this scores
Why this scores 4/4. (a) 2 marks: subtract the levels in the right order → $6.2 \times 10^{-19}\ \text{J}$ (the answer is positive for an emitted photon). (b) 2 marks: $\lambda = hc/\Delta E$ → $3.21 \times 10^{-7}\ \text{m}$ . A sign slip — getting $-6.2 \times 10^{-19}$ — signals the subtraction order was reversed; the emitted photon energy must be positive.
Question 7
4 marks
Q (4 marks). A metal has a work function of $4.30\ \text{eV}$ . Take $h = 6.63 \times 10^{-34}\ \text{J s}$ and $1\ \text{eV} = 1.60 \times 10^{-19}\ \text{J}$ . Show that the threshold frequency for this metal is about $1 \times 10^{15}\ \text{Hz}$ .
Model answer
Convert the work function to joules first:

$\phi = 4.30 \times 1.60 \times 10^{-19} = 6.88 \times 10^{-19}\ \text{J}$

At the threshold frequency a photon supplies exactly the work function, so $\phi = hf_0$ . Rearranging:

$f_0 = \frac{\phi}{h} = \frac{6.88 \times 10^{-19}}{6.63 \times 10^{-34}} = 1.04 \times 10^{15}\ \text{Hz}$

This is about $1 \times 10^{15}\ \text{Hz}$ , as required.
Why this scores
Why this scores 4/4. (1) convert $\phi$ to joules $6.88 \times 10^{-19}\ \text{J}$ ; (2) state $\phi = hf_0$ ; (3) substitution (M); (4) $f_0 = 1.04 \times 10^{15}\ \text{Hz}$ , quoted to one more s.f. than the target value, confirming 'about $1 \times 10^{15}$ '. For a 'show that' you must work through to the value and not simply assert it.
Question 8
6 marks
Q (6 marks). When ultraviolet light is shone on a negatively charged zinc plate connected to an electroscope, the leaf falls (the plate loses charge). When the same plate is illuminated with very bright visible (red) light, the leaf does not fall, no matter how long the light is left on. Use the photon model of light to explain these observations fully.
Model answer
These observations are explained by treating light as a stream of photons, each of energy $E = hf$ , where one photon interacts with one electron in a single event and gives up all its energy at once.

For an electron to be released from the zinc, a single photon must supply at least the work function $\phi$ — the minimum energy to free an electron. This requires the photon's energy to satisfy $hf \geq \phi$ , i.e. the frequency must be at least the threshold frequency $f_0 = \phi/h$ .

Ultraviolet light has a high frequency, so each UV photon has a large energy $hf$ that exceeds the work function. Each suitable photon frees an electron from the (negatively charged) plate, so the plate loses negative charge and the leaf falls.

Red (visible) light has a lower frequency, so each red photon has energy below the work function ( $hf < \phi$ ). No single photon can free an electron. Increasing the intensity simply delivers more photons per second, not more energetic ones, so still no electrons are emitted — however bright the light or however long it is left. This is exactly why a threshold frequency exists, and why the wave model (which predicts that bright enough light of any colour would eventually work) fails.
Why this scores
Why this scores 6/6. Light as photons, $E = hf$ , one-to-one interaction (1); need $hf \geq \phi$ / threshold frequency (1); UV is high frequency so $hf > \phi$ (1); electrons released, plate discharges, leaf falls (1); red light has $hf < \phi$ so no emission (1); higher intensity = more photons not more energy, so still nothing (1). Continuous, applied prose using the given context (zinc plate, electroscope) is required — a generic photoelectric essay that ignores the scenario is capped.
Question 9
6 marks
Q (6 marks). In a photoelectric experiment the maximum kinetic energy of photoelectrons is measured for several frequencies of incident light, giving a straight-line graph of $KE_{\text{max}}$ (y-axis) against frequency $f$ (x-axis). (a) Use Einstein's photoelectric equation to explain why the graph is a straight line, and state what the gradient and the two intercepts represent. (b) The line crosses the frequency axis at $5.00 \times 10^{14}\ \text{Hz}$ . Taking $h = 6.63 \times 10^{-34}\ \text{J s}$ , calculate the work function of the metal.
Model answer
(a) Einstein's equation $hf = \phi + KE_{\text{max}}$ can be rearranged as:

$KE_{\text{max}} = hf - \phi$

Comparing with the straight-line form $y = mx + c$ (with $y = KE_{\text{max}}$ and $x = f$ ):

the equation is linear in $f$ , so the graph is a straight line;

the gradient equals the Planck constant $h$ (the same for every metal);

the y-intercept is $-\phi$ (negative), the value extrapolated to $f = 0$ ;

the x-intercept is the threshold frequency $f_0$ , where $KE_{\text{max}} = 0$ .

(b) At the x-intercept $KE_{\text{max}} = 0$ , so $\phi = hf_0$ :

$\phi = hf_0 = 6.63 \times 10^{-34} \times 5.00 \times 10^{14} = 3.32 \times 10^{-19}\ \text{J}$

The work function is $3.32 \times 10^{-19}\ \text{J}$ (about $2.07\ \text{eV}$ ).
Why this scores
Why this scores 6/6. (a) rearranged to $KE_{\text{max}} = hf - \phi$ (1); compared with $y = mx + c$ → straight line (1); gradient $= h$ (1); intercepts: y-intercept $= -\phi$ and x-intercept $= f_0$ (1). (b) $\phi = hf_0$ stated (1); $3.32 \times 10^{-19}\ \text{J}$ (1). Stating that the y-intercept is $-\phi$ (not $+\phi$ ) and that the x-intercept is $f_0$ is the discriminating detail.
Question 10
6 marks
Q (6 marks). A hot gas of atomic hydrogen emits light. Some of this light is then shone onto a metal surface in a photocell. The hydrogen energy levels include $-3.40\ \text{eV}$ ( $n=2$ ) and $-0.85\ \text{eV}$ ( $n=4$ ), and the metal has a work function of $1.90\ \text{eV}$ . Take $h = 6.63 \times 10^{-34}\ \text{J s}$ and $1\ \text{eV} = 1.60 \times 10^{-19}\ \text{J}$ . (a) Calculate the energy, in joules, of the photon emitted when an electron in hydrogen falls from $n=4$ to $n=2$ . (b) Determine whether this photon can cause photoemission from the metal, justifying your answer with a calculation.
Model answer
(a) The emitted photon energy is the gap between the two levels (upper minus lower):

$\Delta E = E_4 - E_2 = (-0.85) - (-3.40) = 2.55\ \text{eV}$

Convert to joules:

$\Delta E = 2.55 \times 1.60 \times 10^{-19} = 4.08 \times 10^{-19}\ \text{J}$

(b) Convert the work function to joules for a fair comparison:

$\phi = 1.90 \times 1.60 \times 10^{-19} = 3.04 \times 10^{-19}\ \text{J}$

Compare the photon energy with the work function:

$hf = 4.08 \times 10^{-19}\ \text{J} \;>\; \phi = 3.04 \times 10^{-19}\ \text{J}$

Since the photon energy exceeds the work function, the photon can cause photoemission. The maximum kinetic energy of the emitted electron would be:

$KE_{\text{max}} = hf - \phi = 4.08 \times 10^{-19} - 3.04 \times 10^{-19} = 1.04 \times 10^{-19}\ \text{J}$
Why this scores
Why this scores 6/6. (a) 3 marks: $\Delta E = 2.55\ \text{eV}$ (correct subtraction of negative levels), convert to $4.08 \times 10^{-19}\ \text{J}$ . (b) 3 marks: convert $\phi$ to $3.04 \times 10^{-19}\ \text{J}$ , compare $hf > \phi$ with a clear conclusion that emission occurs, and (credit) $KE_{\text{max}} = 1.04 \times 10^{-19}\ \text{J}$ . A synoptic question linking spectra to the photoelectric effect — keeping all energies in the SAME unit before comparing is essential.

Key Formulae — the Photoelectric Effect and Atomic Spectra

The formulae you need to memorise for the photoelectric effect and atomic spectra on the Cambridge IGCSE paper, with every variable defined in plain English and a note on when to use it.

Photon energy
$E = hf = \frac{hc}{\lambda}$
$E$
energy of one photon / J
$h$
Planck constant $= 6.63 \times 10^{-34}\ \text{J s}$
$f$
frequency of the light / Hz
$c$
speed of light $= 3.00 \times 10^{8}\ \text{m s}^{-1}$
$\lambda$
wavelength / m
When to use
Finding the energy carried by a single quantum of light from its frequency or wavelength. In the Edexcel formulae booklet (Appendix 7).
Example
$f = 5.0 \times 10^{14}\ \text{Hz}$ : $E = 6.63 \times 10^{-34} \times 5.0 \times 10^{14} = 3.32 \times 10^{-19}\ \text{J}$ .
Einstein's photoelectric equation
$hf = \phi + KE_{\text{max}}$
$hf$
energy of the incident photon / J
$\phi$
work function of the metal / J
$KE_{\text{max}}$
maximum kinetic energy of an emitted electron $= \tfrac{1}{2}mv_{\text{max}}^2$ / J
When to use
Linking photon energy, work function and the maximum kinetic energy of photoelectrons. Rearrange to $KE_{\text{max}} = hf - \phi$ . In the formulae booklet.
Example
$hf = 4.64 \times 10^{-19}$ , $\phi = 3.2 \times 10^{-19}$ : $KE_{\text{max}} = 1.44 \times 10^{-19}\ \text{J}$ .
Work function and threshold frequency
$\phi = hf_0$
$\phi$
work function / J
$h$
Planck constant
$f_0$
threshold frequency / Hz
When to use
The threshold frequency is the minimum frequency that causes emission; at $f_0$ the photon energy exactly equals the work function and $KE_{\text{max}} = 0$ . Must be recalled (follows from setting $KE_{\text{max}} = 0$ ).
Example
$\phi = 3.04 \times 10^{-19}\ \text{J}$ : $f_0 = \phi/h = 4.59 \times 10^{14}\ \text{Hz}$ .
Stopping potential
$eV_s = KE_{\text{max}} = \tfrac{1}{2}mv_{\text{max}}^2$
$e$
elementary charge $= 1.60 \times 10^{-19}\ \text{C}$
$V_s$
stopping potential / V
$KE_{\text{max}}$
maximum kinetic energy of a photoelectron / J
When to use
Finding the reverse voltage that just stops the fastest photoelectrons, or converting between $KE_{\text{max}}$ and a measured stopping potential. Must be recalled.
Example
$KE_{\text{max}} = 1.44 \times 10^{-19}\ \text{J}$ : $V_s = KE_{\text{max}}/e = 0.900\ \text{V}$ .
Energy-level transition (photon energy)
$\Delta E = E_2 - E_1 = hf$
$\Delta E$
photon energy = difference between two energy levels / J
$E_2,\ E_1$
the upper and lower energy levels (often negative) / J or eV
$f$
frequency of the emitted or absorbed photon / Hz
When to use
Finding the frequency (or, with $c = f\lambda$ , the wavelength) of a photon emitted or absorbed in a transition between discrete energy levels. Convert eV → J before using SI constants. In the formulae booklet.
Example
$\Delta E = 1.89\ \text{eV} = 3.02 \times 10^{-19}\ \text{J}$ : $f = \Delta E/h = 4.57 \times 10^{14}\ \text{Hz}$ .
Electronvolt–joule conversion
$E_{\text{(J)}} = E_{\text{(eV)}} \times 1.60 \times 10^{-19}$
$E_{\text{(J)}}$
energy in joules
$E_{\text{(eV)}}$
energy in electronvolts
$1\ \text{eV}$
$1.60 \times 10^{-19}\ \text{J}$ (energy gained by an electron through 1 V)
When to use
Whenever energies are quoted in eV but $h$ , $c$ and $e$ are in SI units. Multiply eV by $1.60 \times 10^{-19}$ to get joules; divide joules by $1.60 \times 10^{-19}$ to get eV. Must be recalled.
Example
$2.30\ \text{eV} = 2.30 \times 1.60 \times 10^{-19} = 3.68 \times 10^{-19}\ \text{J}$ .

Key Definitions and Keywords — the Photoelectric Effect and Atomic Spectra

Definitions to memorise and the exact keywords mark schemes credit for the photoelectric effect and atomic spectra answers — sharpened from recent examiner reports for the 2026 Cambridge IGCSE sitting.

Photon
Examiner keyword
A discrete quantum (packet) of electromagnetic energy. The energy of one photon is $E = hf = \dfrac{hc}{\lambda}$ , where $h$ is the Planck constant.
Photoelectric effect
Examiner keyword
The emission of electrons from the surface of a metal when electromagnetic radiation of sufficiently high frequency is incident on it.
Photoelectron
An electron that is emitted from a metal surface as a result of the photoelectric effect.
Work function (φ)
Examiner keyword
The minimum energy required to release (emit) an electron from the surface of a metal. Related to the threshold frequency by $\phi = hf_0$ .
Threshold frequency (f₀)
Examiner keyword
The minimum frequency of incident light that will cause photoemission from a given metal. Below $f_0$ no electrons are emitted, however intense the light. $f_0 = \dfrac{\phi}{h}$ .
Planck constant (h)
Examiner keyword
The fundamental constant relating a photon's energy to its frequency, $E = hf$ . Its value is $6.63 \times 10^{-34}\ \text{J s}$ ; it is the gradient of a $KE_{\text{max}}$ against $f$ graph.
Einstein's photoelectric equation
Examiner keyword
$hf = \phi + KE_{\text{max}}$ — the energy of the incident photon equals the work function plus the maximum kinetic energy of the emitted electron (energy conservation for a one-photon-one-electron interaction).
Maximum kinetic energy (KEmax)
Examiner keyword
The greatest kinetic energy a photoelectron can have, $KE_{\text{max}} = hf - \phi = \tfrac{1}{2}mv_{\text{max}}^2$ . Electrons released from deeper in the metal emerge with less energy.
Stopping potential (Vs)
Examiner keyword
The minimum reverse potential difference that just stops the most energetic photoelectrons from reaching the collector, so that $eV_s = KE_{\text{max}}$ .
Electronvolt (eV)
Examiner keyword
The energy gained by an electron accelerated through a potential difference of one volt; $1\ \text{eV} = 1.60 \times 10^{-19}\ \text{J}$ . A convenient unit for atomic-scale energies.
Intensity (of light)
The power delivered per unit area. For light above the threshold frequency, increasing intensity means more photons per second, so a greater rate of electron emission — but no change to each electron's kinetic energy.
Energy level
Examiner keyword
One of the discrete (quantised) energies an electron may have in an atom. Levels are usually quoted as negative values (the ground state being the most negative), measured relative to the ionised electron at zero.
Ground state
The lowest energy level of an atom, in which the electron is most tightly bound (the most negative energy). An atom is in its ground state unless it has been excited.
Excitation
The process by which an electron absorbs energy (e.g. from a photon or a collision) and moves to a higher energy level. It later de-excites, emitting a photon of energy $\Delta E = hf$ .
Emission line spectrum
Examiner keyword
A spectrum of bright coloured lines on a dark background, produced when excited atoms de-excite and emit photons of definite energies equal to the gaps between energy levels.
Absorption line spectrum
Examiner keyword
A continuous spectrum crossed by dark lines, produced when photons of specific energies are absorbed by atoms (lifting electrons to higher levels). The dark lines fall at the same frequencies as the emission lines.

Common Mistakes and Misconceptions — the Photoelectric Effect and Atomic Spectra

The traps other students keep falling into on the photoelectric effect and atomic spectra questions — taken from recent Cambridge IGCSE examiner reports and mark schemes — and how to avoid them.

✕Thinking that increasing the intensity (brightness) of the light increases the kinetic energy of the photoelectrons
Edexcel IAL Physics Unit 2 examiner reports
Why it happens
Students carry over the everyday idea that 'brighter = more energetic', and confuse the number of photons with the energy of each photon.
How to avoid it
Separate the two clearly: intensity controls the NUMBER of photons per second (hence the rate of emission / photocurrent); frequency controls the ENERGY of each photon (hence $KE_{\text{max}} = hf - \phi$ ). More intense light frees more electrons, not faster ones.
✕In a levels-based answer, failing to explain WHY the wave model fails — just describing the photoelectric effect instead
Edexcel IAL Physics Unit 2 examiner reports
Why it happens
Students learn the observations but not the contrast between what each model predicts, so they list facts rather than argue.
How to avoid it
Always contrast the prediction with the observation. The wave model predicts emission at any frequency given enough intensity/time and a time delay for dim light; experiment shows a sharp threshold frequency and instantaneous emission. The photon model ( $E = hf$ , one-to-one) explains both. Write linked prose, not a list.
✕Mixing electronvolts and joules in the same equation (e.g. $KE_{\text{max}} = hf - \phi$ with $hf$ in joules but $\phi$ in eV)
Edexcel IAL Physics Unit 2 examiner reports
Why it happens
Work functions and energy levels are often quoted in eV, while $h$ , $c$ and $e$ are in SI units, so the two unit systems collide.
How to avoid it
Convert everything to joules first (multiply eV by $1.60 \times 10^{-19}$ ) before adding or subtracting, then convert back to eV at the end only if asked. Never subtract an energy in eV from one in joules.
✕Treating $hf - \phi$ as the kinetic energy of every emitted electron, rather than the MAXIMUM kinetic energy
Edexcel IAL Physics Unit 2 examiner reports
Why it happens
The equation $KE = hf - \phi$ is memorised without the subscript 'max', so the special meaning is lost.
How to avoid it
Remember the equation gives $KE_{\text{max}}$ only. Electrons released from below the surface use up extra energy escaping the metal, so they emerge with LESS than $hf - \phi$ . The stopping potential measures this maximum: $eV_s = KE_{\text{max}}$ .
✕Getting the sign or order wrong when finding the photon energy from energy levels (e.g. $E_{\text{lower}} - E_{\text{upper}}$ giving a negative answer)
Edexcel IAL Physics Unit 2 examiner reports
Why it happens
Energy levels are negative numbers, and subtracting negatives is error-prone under exam pressure.
How to avoid it
Use $\Delta E = E_{\text{upper}} - E_{\text{lower}}$ and check the photon energy is positive for an emitted photon. For a fall from $-1.51\ \text{eV}$ to $-3.40\ \text{eV}$ : $\Delta E = (-1.51) - (-3.40) = +1.89\ \text{eV}$ .
✕Suggesting that an atom can emit or absorb photons of any energy, rather than only those matching the gaps between discrete levels
Edexcel IAL Physics Unit 2 examiner reports
Why it happens
Students picture energy as continuous (like a ramp) instead of quantised (like a ladder), so they miss why spectra are LINE spectra.
How to avoid it
Stress that energy levels are discrete, so only photons whose energy exactly equals a gap ( $\Delta E = hf$ ) can be emitted or absorbed. This is precisely why we see sharp lines at specific frequencies, not a continuous band.

the Photoelectric Effect and Atomic Spectra — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

the Photoelectric Effect and Atomic Spectra

Short Notes - the Photoelectric Effect and Atomic Spectra

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Energy of a single photon (2 marks)

2Converting an energy from electronvolts to joules (2 marks)

3Threshold frequency from the work function (3 marks)

4Maximum kinetic energy from Einstein's equation (3 marks)

5Stopping potential from maximum kinetic energy (3 marks)

6Wavelength of an emitted spectral line (4 marks)

7Why the wave model cannot explain a threshold frequency (6 marks)

8Determining the Planck constant from a KEmax–f graph (6 marks)

9Absorption line and the emitted photon's frequency (6 marks)

Photon energy

Einstein's photoelectric equation

Work function and threshold frequency

Stopping potential

Energy-level transition (photon energy)

Electronvolt–joule conversion

Photon

Photoelectric effect

Photoelectron

Work function (φ)

Threshold frequency (f₀)

Planck constant (h)

Einstein's photoelectric equation

Maximum kinetic energy (KEmax)

Stopping potential (Vs)

Electronvolt (eV)

Intensity (of light)

Energy level

Ground state

Excitation

Emission line spectrum

Absorption line spectrum

✕Thinking that increasing the intensity (brightness) of the light increases the kinetic energy of the photoelectrons

✕In a levels-based answer, failing to explain WHY the wave model fails — just describing the photoelectric effect instead

✕Mixing electronvolts and joules in the same equation (e.g. KEmax=hf−ϕKE_{\text{max}} = hf - \phiKEmax​=hf−ϕ with hfhfhf in joules but ϕ\phiϕ in eV)

✕Treating hf−ϕhf - \phihf−ϕ as the kinetic energy of every emitted electron, rather than the MAXIMUM kinetic energy

✕Getting the sign or order wrong when finding the photon energy from energy levels (e.g. Elower−EupperE_{\text{lower}} - E_{\text{upper}}Elower​−Eupper​ giving a negative answer)

✕Suggesting that an atom can emit or absorb photons of any energy, rather than only those matching the gaps between discrete levels

the Photoelectric Effect and Atomic Spectra — frequently asked questions

✕Mixing electronvolts and joules in the same equation (e.g. $KE_{\text{max}} = hf - \phi$ with $hf$ in joules but $\phi$ in eV)

✕Treating $hf - \phi$ as the kinetic energy of every emitted electron, rather than the MAXIMUM kinetic energy

✕Getting the sign or order wrong when finding the photon energy from energy levels (e.g. $E_{\text{lower}} - E_{\text{upper}}$ giving a negative answer)