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Physics (IAL)

The Photoelectric Effect & Atomic Spectra

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Edexcel IAL Physics · Quantum Physics

The Photoelectric Effect
& Atomic Spectra

Big idea: Light doesn't just travel as a smooth wave — it also arrives in individual, indivisible "packets" of energy called photons. Whether it's knocking an electron clean out of a metal, or getting swallowed whole by an atom to boost an electron to a higher energy level, energy in these interactions always moves in one discrete lump at a time, never in a continuous trickle.

📋 Summary — The Whole Chapter at a Glance
  • The photoelectric effect is when electrons ("photoelectrons") are knocked out of a metal's surface by light — and it only works above a minimum ("threshold") frequency, no matter how bright the light is.
  • This proves light is quantised — delivered in discrete packets called photons, each carrying energy E = hf, and each photon can only be absorbed by one electron.
  • The photoelectric equation hf = Φ + KEmax is just energy conservation: photon energy in = energy needed to escape (work function) + leftover kinetic energy.
  • The electronvolt (eV) is a tiny, convenient energy unit for quantum-scale energies: 1 eV = 1.6 × 10⁻¹⁹ J.
  • The gold-leaf electroscope experiment gives real, observable evidence for photons — intensity changes speed of emission, frequency changes energy of electrons, and there's a hard threshold frequency cutoff.
  • Atomic line spectra happen when excited electrons drop down between fixed energy levels inside atoms, releasing a photon whose energy exactly matches the energy gap — creating a unique "fingerprint" of coloured lines for each element.
⚡ 1. The Photoelectric Effect

Definition: The photoelectric effect is the phenomenon where electrons are emitted from the surface of a metal when it absorbs electromagnetic radiation (usually UV or visible light). These ejected electrons are called photoelectrons — same particle as a normal electron, just given a special name because of how it was released.

Analogy: Imagine electrons are trapped at the bottom of a well inside the metal, like people stuck in a pit. To climb out, a person needs a single, strong enough boost of energy — a friend throwing down one big rope isn't enough if it's too short (not enough energy per throw); it doesn't matter how many short ropes you throw down together (that's "intensity" — more photons per second, but each one still too weak). Only when a single rope is long enough (a high-enough photon) can any one person actually climb out.

This is exactly why the photoelectric effect provides some of the best evidence that light is quantised — carried in discrete packets (photons) rather than as a smooth, continuous wave:

  • Each photoelectron can only absorb one whole photon — never a little bit from many photons added together.
  • This means only light above a certain threshold frequency f₀ will ever release an electron — dimmer or brighter doesn't matter, only the "size" of each individual packet does.
The Gold-Leaf Electroscope Experiment

This classic setup lets you literally the photoelectric effect happen:

UV LAMP ZINC PLATE (charged –) | | | photons --> e⁻ e⁻ e⁻ e⁻ ← photoelectrons knocked off v | ________ ______|______ | | | metal | | UV | |______________| |________| | GOLD-LEAF ELECTROSCOPE (leaf falls as charge is lost from the leaf)
  1. A zinc plate is attached to a gold leaf, which is negatively charged — so it repels away from a central charged rod and stands up at an angle.
  2. UV light shines on the plate. If the UV frequency is above the threshold, electrons are knocked off the plate.
  3. As negative charge (electrons) leaves the system, the leaf becomes less negatively charged, repels less, and falls back down towards the rod.
What the Experiment Shows Us
You change...You observe...What it tells us
UV lamp moved closer (↑ intensity)Leaf falls fasterMore photons/second → more photoelectrons/second. Intensity affects , not energy.
Use higher-frequency lightLeaf falls at the same rateKEmax of each electron increases, but rate of emission is unaffected by frequency alone.
Use a filament (low-frequency) lamp instead of UVNo change at all — leaf stays upBelow threshold frequency, electrons are ever emitted, no matter how bright.
Charge the plate positively insteadNo change — leaf stays downEmitted electrons get pulled straight back by the positive charge before escaping.
Any moment UV (above threshold) hits the plateEmission is instantOne photon → one electron, immediately. No "charging up" delay like wave theory predicted.
Q1.A student shines a very bright red light and then a very dim violet light onto the same piece of sodium metal. Only the violet light causes photoelectrons to be emitted. Explain why, using the idea of photon energy.
Q2.Explain why photoelectric emission happens instantaneously, and why this was surprising evidence against the classical wave theory of light.
🧮 2. The Photoelectric Equation

This equation is just a statement of conservation of energy. The energy carried in by one photon has to go — some of it is "spent" pulling the electron free of the metal (the work function), and whatever energy is left over becomes the electron's kinetic energy as it flies off.

E = hf = Φ + ½mv²max
photon energy = work function + maximum kinetic energy of photoelectron
  • h = Planck's constant = 6.63 × 10⁻³⁴ J s
  • f = frequency of the incident radiation (Hz)
  • Φ = work function of the metal (J) — minimum energy needed to free an electron
  • ½mv²max (= KEmax) = maximum kinetic energy of the emitted photoelectron (J)
Analogy: Think of the work function like the cover charge to get into a club, paid in energy. A photon arrives with a certain amount of "cash" (hf). It pays the cover charge (Φ) to let the electron leave the metal — whatever cash is left over becomes the electron's kinetic energy as it walks away (KEmax). If the photon doesn't have enough cash to cover Φ in the first place, the electron simply isn't let out — no matter how many other broke photons show up at the same time.
Threshold Frequency

The threshold frequency f₀ is the minimum frequency of EM radiation needed to just barely release a photoelectron — at this exact frequency, the photon has energy to cover the work function, with nothing left over for kinetic energy:

hf₀ = Φ
at threshold, KEmax = 0
Three things this equation tells you
  • If hf < Φ, no electrons are emitted at all — regardless of intensity.
  • KEmax depends only on frequency of the incident light, on intensity.
  • Most emitted photoelectrons actually have less than KEmax — some energy is lost as electrons collide with atoms on their way out of the metal. KEmax is the energy of electrons released right at the very surface, with no energy lost on the way out.
Graphical Representation

Rearranging the equation into the familiar straight-line form y = mx + c is one of the most useful things you can do with it:

KEmax = hf − Φ
KE_max /J ^ | / | / | / | / ← gradient = h (Planck's constant) | / |________________/________________> f /Hz | f0 (threshold frequency = x-intercept) -Φ ┄┄┄┄┄┄┄┄┄┄┄┄┄┄/ | / | / (y-intercept = -Φ)
  • y-intercept = −Φ (the work function, read off as a negative value)
  • x-intercept = f₀ (the threshold frequency)
  • Gradient = h (Planck's constant) — this is actually how Planck's constant was first measured experimentally!
  • Below f₀, the line simply doesn't exist — zero electrons are emitted, so the graph sits flat on the x-axis until f₀ is reached.
Worked Example — Finding the Work Function from a Graph

A graph of KEmax (eV) against frequency f for sodium metal shows a straight line crossing the x-axis at f₀ = 4 × 10¹⁴ Hz. Calculate the work function of sodium in eV.

Step 1 — Rearrange the photoelectric equation into straight-line form:
KE_max = hf − Φ, matching y = mx + c
Step 2 — At the threshold frequency, KE_max = 0, so:
Φ = hf₀
Step 3 — Substitute in values:
Φ = (6.63 × 10⁻³⁴) × (4 × 10¹⁴) = 2.652 × 10⁻¹⁹ J
Step 4 — Convert Joules → eV by dividing by 1.6 × 10⁻¹⁹:
Φ = 2.652 × 10⁻¹⁹ ÷ 1.6 × 10⁻¹⁹ = 1.66 eV
Q3.Light of frequency 7.5 × 10¹⁴ Hz is shone onto a metal with a work function of 3.0 × 10⁻¹⁹ J. Calculate the maximum kinetic energy of the emitted photoelectrons, in Joules. (h = 6.63 × 10⁻³⁴ J s)
🔋 3. The Electronvolt

Quantum-scale energies (like photon energies or work functions) come out as tiny numbers in Joules — around 10⁻¹⁹ J. That's awkward to write and compare, so physicists invented a more convenient unit built specifically for this scale: the electronvolt (eV).

Definition: One electronvolt is the energy gained by a single electron when it is accelerated, from rest, through a potential difference of exactly 1 volt.

1 eV = 1.6 × 10⁻¹⁹ J

This comes directly from the definition of potential difference, V = E/Q, rearranged to E = QV. Since an electron's charge is 1.6 × 10⁻¹⁹ C, pushing it through 1 V transfers exactly 1.6 × 10⁻¹⁹ J of energy to it — and we just call that amount "1 eV".

Converting between J and eV
  • eV → J: multiply by 1.6 × 10⁻¹⁹
  • J → eV: divide by 1.6 × 10⁻¹⁹
Relation to Kinetic Energy

When a charged particle accelerates from rest through a potential difference V, all the electrical potential energy it loses is converted into kinetic energy:

eV = ½mv²
rearranged: v = √(2eV / m)
Worked Example — Photon Energy in eV

Show that the photon energy of light with wavelength 700 nm is about 1.8 eV.

Step 1 — Combine wave speed and photon energy equations:
c = fλ → f = c/λ, and E = hf → E = hc/λ
Step 2 — Calculate E in Joules:
E = (6.63×10⁻³⁴ × 3.0×10⁸) / (700×10⁻⁹) = 2.84 × 10⁻¹⁹ J
Step 3 — Convert to eV (divide by 1.6×10⁻¹⁹):
E = 2.84×10⁻¹⁹ / 1.6×10⁻¹⁹ = 1.78 eV ≈ 1.8 eV
Q4.An electron is accelerated from rest through a potential difference of 500 V. Calculate the speed it gains. (mass of electron = 9.11 × 10⁻³¹ kg, e = 1.6 × 10⁻¹⁹ C)
🌊 4. The Particle Nature of EM Radiation

Here's the deep conflict this whole chapter is built around: classical physics treated light purely as a wave — and that model works great for explaining diffraction and interference (light bending round obstacles, and beams interfering to make patterns of light and dark). But the photoelectric effect be explained by a pure wave model. It only makes sense if light also behaves as a stream of individual particles — photons — each carrying a fixed, discrete amount of energy E = hf.

Wave vs Particle — pick your evidence
  • Wave behaviour: diffraction, interference (light spreading and combining continuously)
  • Particle behaviour: photoelectric effect (instant emission, threshold frequency, one photon = one electron)

This is the famous wave-particle duality of light — it isn't "really" one or the other; it shows whichever behaviour the experiment is set up to reveal.

This is why the photoelectric effect is considered such powerful evidence: the observations from the gold-leaf electroscope experiment (see Section 1) simply don't add up under wave theory, but every single one falls out naturally once you assume light arrives in discrete photon packets that interact one-to-one with individual electrons.

🌈 5. Atomic Line Spectra

Electrons inside an atom aren't free to have amount of energy — they can only occupy specific, fixed energy levels (like rungs on a ladder — you can stand on any rung, but never hover in between them). When an atom is heated or otherwise "excited", electrons absorb energy and jump up to higher energy levels. But electrons don't like staying excited — they quickly fall back down to lower levels, and every time they do, they release the energy difference as a single photon.

Definition: An emission line spectrum is produced when an excited electron moves from a higher to a lower energy level and emits a photon with energy exactly equal to the difference between those two levels.

Analogy: Think of energy levels like floors in a building connected only by a single elevator that can't stop between floors. If an electron on floor 4 wants to go to floor 1, it must "pay" out an amount of energy exactly equal to the height dropped — no more, no less — and that exact amount gets released as a single photon. Different buildings (different elements) have differently-spaced floors, so the "receipts" (photon energies/wavelengths) they produce are unique to each building. That's why every element has its own unmistakable spectral fingerprint.
ENERGY LEVELS (Hydrogen, simplified) n=4 ───────────────────────── 0 eV (ionisation) n=3 ───────────────────────── -1.51 eV | | | | (photon released n=2 ─────────────|───|──────── -3.40 eV as electron drops) \ \ \ \ n=1 ─────────────────\───\──── -13.6 eV (ground state) Electron drops n=4 → n=2: releases a LOWER energy photon (longer λ) Electron drops n=4 → n=1: releases a HIGHER energy photon (shorter λ)
The Key Equations
ΔE = E₁ − E₂ = hf = hc/λ
energy released = energy of level left − energy of level arrived at = photon energy

Because E = hf and c = fλ, a bigger energy jump between levels always means a higher frequency and therefore shorter wavelength photon — and vice versa. This is a really common exam trap, so let's nail it down clearly:

Bigger energy gap ΔESmaller energy gap ΔE
Higher frequency fLower frequency f
Shorter wavelength λLonger wavelength λ
e.g. drops to ground state (n=1) — often UVe.g. drops between high levels — often infrared

In hydrogen specifically, transitions ending at n = 2 happen to fall in the visible range — which is why hydrogen's famous rainbow-line spectrum (violet, blue, light blue, red) is such a classic textbook image. Transitions ending at n = 1 are higher energy (ultraviolet), and those ending at n = 3 or higher are lower energy (infrared).

Why each element has a unique spectrum

Every element has a different arrangement of electrons and therefore a different, unique set of energy levels. Since the photon energies (and thus wavelengths/colours) emitted depend entirely on the gaps between levels, no two elements ever produce the same pattern of spectral lines — this is how astronomers identify which elements exist in distant stars, just from analysing the light that reaches us!

Worked Example — Photon Wavelength from an Energy Transition

An electron in a hydrogen atom drops from the n = 2 level (E₂ = −3.40 eV) to the n = 1 level (E₁ = −13.6 eV). Calculate the wavelength of the emitted photon.

Step 1 — Find the energy difference:
ΔE = E(n=2) − E(n=1) = −3.40 − (−13.6) = 10.2 eV
Step 2 — Convert to Joules:
ΔE = 10.2 × 1.6×10⁻¹⁹ = 1.632 × 10⁻¹⁸ J
Step 3 — Use ΔE = hc/λ, rearranged to λ = hc/ΔE:
λ = (6.63×10⁻³⁴ × 3.0×10⁸) / 1.632×10⁻¹⁸
Step 4 — Calculate:
λ ≈ 1.22 × 10⁻⁷ m = 122 nm (this is in the ultraviolet range — makes sense, since it's a transition down to the ground state!)
Q5.Two possible electron transitions in an atom release photons of wavelength 400 nm and 700 nm respectively. Which transition corresponds to the larger energy level gap? Explain your reasoning.
Q6.Explain, in terms of energy levels, why a hot gas produces a line spectrum rather than a continuous spectrum of all colours.
🧠 What to Memorise
Term / FormulaMeaning
Photoelectric effectElectrons emitted from a metal surface when EM radiation is absorbed
PhotoelectronAn electron released via the photoelectric effect
Work function, ΦMinimum energy needed to release an electron from a metal surface
Threshold frequency, f₀Minimum frequency of radiation that can cause photoelectric emission
E = hf = Φ + ½mv²maxThe photoelectric equation (energy conservation)
hf₀ = ΦAt threshold frequency, KE_max = 0
1 eV= 1.6 × 10⁻¹⁹ J — energy gained by an electron across a 1 V p.d.
eV = ½mv²Kinetic energy gained accelerating through p.d. V
Emission line spectrumSeries of bright lines produced when excited electrons drop energy levels
ΔE = E₁ − E₂ = hf = hc/λEnergy of photon emitted = energy level difference
h (Planck's constant)6.63 × 10⁻³⁴ J s
c (speed of light)3.0 × 10⁸ m s⁻¹
✅ Concepts Checklist
🎯 Exam Tips & Common Mistakes
Units mismatch is the #1 error. hf, Φ, and KE_max must ALL be in the same units (Joules) before you plug them into the photoelectric equation. If a question gives you Φ in eV, convert it to Joules first — don't mix eV and J in the same calculation.
Intensity ≠ energy per photon. A very common trap: students think brighter light means higher-energy photoelectrons. It doesn't! Intensity only changes the of photons per second (and therefore the number of photoelectrons per second). Only changes KE_max.
KE_max vs actual KE. Examiners love testing this: not every photoelectron comes out with KE_max — most lose some energy to collisions on the way out of the metal. Only electrons emitted right at the very surface reach the maximum.
Graph sign errors. On a KE_max vs f graph, the y-intercept is −Φ (negative), not +Φ. Students often forget the negative sign when reading values off graphs.
Energy vs wavelength direction. Remember energy and wavelength are related (E = hc/λ). A bigger energy jump between levels gives a shorter wavelength, not longer — this trips people up constantly under exam pressure.
Show full working with formulas first. Mark schemes almost always award a method mark for correctly writing out the equation (e.g. KE_max = hf − Φ) before you substitute numbers in — even if your final numerical answer is wrong, you can still pick up marks.
Don't confuse "no photoelectrons" with "light is absorbed but does nothing". Below the threshold frequency, photons ARE still absorbed by the metal — they just don't have enough individual energy to free an electron, so that energy typically ends up as heat instead.
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