Particles and radiation: A-level Physics AQA revision notes

3.2.1

Particles

Constituents of the atom 3.2.1.1

Definitions

Nucleon: a constituent of the nucleus — a proton or a neutron.
Specific charge: charge per unit mass, $\frac{q}{m}$ , useful for comparing how strongly different particles respond to the same electric or magnetic field.

Key results

Proton: mass $1.673 \times 1 0^{- 27} kg$ , charge $+ 1.602 \times 1 0^{- 19} C$ . Neutron: mass $1.675 \times 1 0^{- 27} kg$ , charge $0$ . Electron: mass $9.109 \times 1 0^{- 31} kg$ , charge $- 1.602 \times 1 0^{- 19} C$ .
Nucleon number (mass number), $A$ = protons + neutrons. Proton number (atomic number), $Z$ = protons. Neutron number $= A - Z$ . Written $_{Z}^{A} X$ .
Specific charge: proton $\approx 9.58 \times 1 0^{7} C kg^{- 1}$ ; electron $\approx - 1.76 \times 1 0^{11} C kg^{- 1}$ — about 1836 times larger in magnitude, since the electron is far lighter for the same size of charge.

Notes

Isotopes share the same $Z$ (and so the same chemical element) but differ in neutron number, and therefore in $A$ .
For a nucleus, total charge is $q = Z e$ ; for an ion with $N_{e}$ electrons still attached, $q = (Z - N_{e}) e$ — use the mass of the whole nucleus or ion, not just a single particle, when finding specific charge for a bound system.
The proton and neutron have almost identical mass (the neutron is about 0.14% heavier), which is why nucleon number alone is a good first approximation to a nucleus's mass in atomic mass units.

Stable and unstable nuclei 3.2.1.2

Key results

Alpha decay: $_{Z}^{A} X \to_{Z - 2}^{A - 4} Y +_{2}^{4} He$ — nucleon number falls by 4, proton number by 2.

Notes

The strong nuclear force is attractive at separations of roughly $0.5$ – $3 fm$ but becomes strongly repulsive at very short range (below about $0.5 fm$ ), which is what stops a nucleus collapsing in on itself.
Beyond about $3 fm$ the strong force has essentially no effect, which is why a nucleus cannot grow arbitrarily large without extra neutrons to dilute the cumulative proton-proton electrostatic repulsion (see Nuclear physics for how this shapes the band of stability).
An alpha particle is emitted as a single bound unit (a helium-4 nucleus) rather than four separate nucleons, since the alpha particle is itself an unusually tightly bound, low-energy configuration.

Particles, antiparticles and photons 3.2.1.3

Definitions

Antiparticle: a particle of equal mass and rest energy but opposite charge (and opposite other additive quantum numbers) to its corresponding particle — a neutral antiparticle is still neutral.

Key results

Rest energy: $E_{0} = m c^{2}$ .
Pair production: a photon converts into a particle-antiparticle pair, requiring $E_{γ} \geq 2 m c^{2}$ for the pair's combined rest energy (e.g. $\geq 1.022 MeV$ for an electron-positron pair) — occurring near a nucleus so the nucleus can absorb any recoil momentum.
Annihilation: a particle and its antiparticle meet and convert entirely into energy, typically as two photons emitted in opposite directions to conserve momentum.

Notes

Particle/antiparticle pairs: $e^{-} / e^{+}$ , $μ^{-} / μ^{+}$ , $p / \overset{p}{ˉ}$ , $n / \overset{n}{ˉ}$ , $ν_{e} / \overset{ν}{ˉ}_{e}$ , $ν_{μ} / \overset{ν}{ˉ}_{μ}$ — every particle needs a distinct antiparticle even when (as for the neutron) it carries no charge, since other quantum numbers like baryon number still differ.
A photon is its own antiparticle: it has zero rest mass and no charge, so there is nothing left to distinguish an 'antiphoton'.
Pair production near a nucleus (rather than in free space) is essential: an isolated photon converting into a pair could not simultaneously conserve both energy and momentum, but the nucleus can absorb the small recoil needed to satisfy both.
Two photons, not one, are required for two-body annihilation at rest, since a single photon could not carry away the initial zero net momentum while also carrying the full energy released.

Particle interactions 3.2.1.4

Definitions

Exchange particle: a virtual particle exchanged between interacting particles that carries the force between them, in the exchange-particle (Feynman) model of fundamental interactions.

Key results

Four fundamental interactions and their exchange particles: strong (binding of nucleons — pion exchange, or gluons at the quark level), electromagnetic (charged particles — virtual photon), weak (changes a quark or lepton's type — $W^{+}$ , $W^{-}$ or $Z^{0}$ ), gravitational (all mass-energy — no exchange particle yet observed).

Notes

A Feynman-style diagram (e.g. electron-electron scattering via a virtual photon) reads as time running left to right or bottom to top, with each vertex marking one particle emitting or absorbing the exchange particle.
The exchanged particle is 'virtual' — it exists only briefly during the interaction and is never directly detected, unlike a real, freely-propagating particle of the same type.
Pion exchange models the residual strong force that binds protons and neutrons together within a nucleus, distinct from (though ultimately arising from) the more fundamental gluon exchange between the quarks inside each nucleon.

Hadrons, leptons and quarks 3.2.1.5–3.2.1.6

Definitions

Hadron: any particle that experiences the strong interaction, built from quarks — baryons (three quarks, e.g. proton, neutron) or mesons (a quark-antiquark pair, e.g. pion, kaon).
Lepton: a fundamental particle that does not experience the strong interaction — the electron, muon and their neutrinos, plus their antiparticles.
Strangeness, $S$ : a quantum number carried by the strange quark ( $S = - 1$ ) and conserved in strong and electromagnetic interactions, but not always in weak interactions.

Key results

Quark properties: up ( $Q = + \frac{2}{3}$ , $B = + \frac{1}{3}$ , $S = 0$ ), down ( $Q = - \frac{1}{3}$ , $B = + \frac{1}{3}$ , $S = 0$ ), strange ( $Q = - \frac{1}{3}$ , $B = + \frac{1}{3}$ , $S = - 1$ ); each antiquark has the opposite sign of every one of these.
Proton $= uud$ ( $Q = + 1$ ), neutron $= udd$ ( $Q = 0$ ); antiproton $= \overset{u}{ˉ} \overset{u}{ˉ} \overset{ˉ}{d}$ , antineutron $= \overset{u}{ˉ} \overset{ˉ}{d} \overset{ˉ}{d}$ .
Baryon number: $B = + 1$ for a baryon (3 quarks), $B = - 1$ for an antibaryon (3 antiquarks), $B = 0$ for a meson (quark + antiquark).

Notes

A free proton is stable, but a free neutron is unstable (mean lifetime around 15 minutes) — inside a stable nucleus, though, a neutron can be perfectly stable, since the energetics of the decay depend on what nucleus it is part of.
Every lepton carries its own family's lepton number ( $L_{e}$ for the electron and its neutrino, $L_{μ}$ for the muon and its neutrino): $+ 1$ for a particle, $- 1$ for its antiparticle, $0$ for anything outside that family.
Mesons carry integer strangeness and zero baryon number, since they are one quark plus one antiquark — $π^{+} = u \overset{ˉ}{d}$ ( $S = 0$ ), $K^{+} = u \overset{s}{ˉ}$ ( $S = + 1$ ).
$π^{0}$ is not simply $u \overset{u}{ˉ}$ or $d \overset{ˉ}{d}$ alone but a genuine quantum superposition of both — a subtlety worth knowing exists even though the full treatment is beyond this course.

Weak interactions and beta decay 3.2.1.7

Key results

$β^{-}$ decay: a down quark converts to an up quark, $d \to u + W^{-}$ , with $W^{-} \to e^{-} + \overset{ν}{ˉ}_{e}$ — overall $n \to p + e^{-} + \overset{ν}{ˉ}_{e}$ , nucleon number unchanged, proton number up by 1.
$β^{+}$ decay: an up quark converts to a down quark, $u \to d + W^{+}$ , with $W^{+} \to e^{+} + ν_{e}$ — overall $p \to n + e^{+} + ν_{e}$ , only possible within a suitable unstable nucleus, since a free proton's rest energy is less than a free neutron's.
Electron capture: a bound atomic electron and a proton interact via a virtual $W^{+}$ to give a neutron and an electron neutrino, $p + e^{-} \to n + ν_{e}$ .

Notes

Both $β^{-}$ and $β^{+}$ decay are quark-level processes at heart: a single quark inside a nucleon changes flavour, mediated by a $W$ boson, not the whole nucleon transforming directly.
The (anti)neutrino is essential to these processes, not an afterthought — without it, the beta particle would always carry away a single fixed energy (from two-body decay kinematics), but measured beta-particle energies actually form a continuous spectrum up to a maximum, exactly because the energy released is shared, in a varying proportion, between the beta particle and the (anti)neutrino.

Conservation laws 3.2.1.7

Key results

Quantities conserved in every particle interaction: charge $Q$ , baryon number $B$ , lepton number (each family, $L_{e}$ / $L_{μ}$ ), energy and momentum.
Strangeness $S$ is conserved in strong and electromagnetic interactions, but a weak interaction may change it, and only by $0$ or $\pm 1$ .

Notes

Conservation laws are the fastest way to test whether a proposed particle decay or interaction is physically allowed — write down every conserved quantity on both sides, and if any one fails to balance, the process cannot happen as written, regardless of how much energy is available.
A balanced conservation check is necessary but not sufficient: energy must also actually be available (the products' total rest energy cannot exceed the reactants'), and the interaction responsible must actually permit that specific vertex.
Worked check, neutron beta decay ( $n \to p + e^{-} + \overset{ν}{ˉ}_{e}$ ): charge ( $0 \to + 1 - 1 + 0 = 0$ ✓), baryon number ( $1 \to 1 + 0 + 0 = 1$ ✓), electron lepton number ( $0 \to 0 + 1 - 1 = 0$ ✓) — all balance, confirming the decay is allowed.
A strong production process like $π^{-} + p \to K^{+} + K^{-} + n$ must conserve strangeness exactly ( $0 + 0 = + 1 - 1 + 0$ ✓), but the same kaon can later decay weakly (e.g. $K^{+} \to π^{+} + π^{0}$ ) with strangeness changing by $\pm 1$ ( $+ 1 \to 0$ ) — the same particle obeying a stricter rule when produced strongly than when it later decays weakly.

Worked examples

Worked example 3.2.1 · 4 marks

The proton has mass:

1.673 \times 1 0^{- 27} kg

and charge:

+ 1.602 \times 1 0^{- 19} C

An alpha particle has mass:

6.645 \times 1 0^{- 27} kg

and charge $+ 2 e$ .

Calculate the specific charge of each particle, and state which is larger and by roughly what factor.

Show worked solution

Proton:

\frac{q}{m} = \frac{1.602 \times 1 0 ^{- 19}}{1.673 \times 1 0 ^{- 27}} \approx 9.58 \times 1 0^{7} C kg^{- 1}

Alpha particle charge:

= 2 \times 1.602 \times 1 0^{- 19} = 3.204 \times 1 0^{- 19} C

so:

\frac{q}{m} = \frac{3.204 \times 1 0 ^{- 19}}{6.645 \times 1 0 ^{- 27}} \approx 4.82 \times 1 0^{7} C kg^{- 1}

The proton's specific charge is about twice the alpha particle's, since the alpha particle has only twice the charge but roughly four times the mass.

Mark scheme · 4 marks

Calculates the proton's specific charge as $\approx 9.58 \times 1 0^{7} C kg^{- 1}$ 1 mark
Calculates the alpha particle's charge as $+ 2 e = 3.204 \times 1 0^{- 19} C$ 1 mark
Calculates the alpha particle's specific charge as $\approx 4.82 \times 1 0^{7} C kg^{- 1}$ 1 mark
States the proton's specific charge is about twice (roughly a factor of 2) the alpha particle's 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.2.1 · 5 marks

A photon undergoes pair production near a nucleus, producing an electron and a positron, each of rest mass:

9.11 \times 1 0^{- 31} kg

Calculate

(a) the minimum possible energy of the photon, in MeV, and

(b) the corresponding minimum photon frequency.

Explain why this pair production must occur near a nucleus rather than in completely free space. (:

c = 3.00 \times 1 0^{8} m s^{- 1}

h = 6.63 \times 1 0^{- 34} J s

e = 1.60 \times 1 0^{- 19} C

)

Show worked solution

Rest energy of one electron (or positron):

m c^{2} = 9.11 \times 1 0^{- 31} \times (3.00 \times 1 0^{8})^{2} \approx 8.20 \times 1 0^{- 14} J

Minimum photon energy (threshold):

E_{γ} = 2 m c^{2} \approx 1.64 \times 1 0^{- 13} J

which is:

\frac{1.64 \times 1 0 ^{- 13}}{1.60 \times 1 0 ^{- 19}} \approx 1.02 \times 1 0^{6} eV \approx 1.02 MeV

Minimum frequency:

f = \frac{E _{γ}}{h} = \frac{1.64 \times 1 0 ^{- 13}}{6.63 \times 1 0 ^{- 34}} \approx 2.47 \times 1 0^{20} Hz

This must happen near a nucleus because an isolated photon converting into a pair could not simultaneously conserve both energy and momentum; the nucleus absorbs the small recoil needed to satisfy both.

Mark scheme · 5 marks

Calculates the rest energy of one electron (or positron) as $\approx 8.20 \times 1 0^{- 14} J$ 1 mark
Doubles this to give the threshold photon energy $\approx 1.64 \times 1 0^{- 13} J$ 1 mark
Converts this energy to $\approx 1.02 MeV$ 1 mark
Calculates the minimum frequency as $\approx 2.47 \times 1 0^{20} Hz$ 1 mark
Explains that a nucleus must be present to absorb recoil momentum, so both energy and momentum are conserved 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.2.1 · 6 marks

Particle	Quark composition
$Σ^{+}$	uus
$Ξ^{-}$	dss
$π^{-}$	d $\overset{u}{ˉ}$
$K^{0}$	d $\overset{s}{ˉ}$

The table shows the quark composition of four particles.

Given the quark properties up (:

Q = + \frac{2}{3}

B = + \frac{1}{3}

$S = 0$ ), down (:

Q = - \frac{1}{3}

B = + \frac{1}{3}

$S = 0$ ) and strange (:

Q = - \frac{1}{3}

B = + \frac{1}{3}

$S = - 1$ ), with each antiquark carrying the opposite sign of every property, calculate the charge $Q$ (in units of $e$ ), baryon number $B$ and strangeness $S$ of each particle, and state whether each is a baryon or a meson.

Show worked solution

$Σ^{+} = uus$ :

Q = \frac{2}{3} + \frac{2}{3} - \frac{1}{3} = + 1

B = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = + 1

S = 0 + 0 - 1 = - 1

— a baryon (three quarks).

$Ξ^{-} = dss$ :

Q = - \frac{1}{3} - \frac{1}{3} - \frac{1}{3} = - 1

$B = + 1$ ,

S = 0 - 1 - 1 = - 2

— a baryon.

$π^{-} = d \overset{u}{ˉ}$ :

Q = - \frac{1}{3} - \frac{2}{3} = - 1

B = \frac{1}{3} - \frac{1}{3} = 0

$S = 0$ — a meson (quark-antiquark pair).

$K^{0} = d \overset{s}{ˉ}$ :

Q = - \frac{1}{3} + \frac{1}{3} = 0

$B = 0$ , $S = 0 + 1 = + 1$ — a meson.

Mark scheme · 6 marks

Calculates $Σ^{+}$ : $Q = + 1$ , $B = + 1$ , $S = - 1$ 1 mark
Calculates $Ξ^{-}$ : $Q = - 1$ , $B = + 1$ , $S = - 2$ 1 mark
Calculates $π^{-}$ : $Q = - 1$ , $B = 0$ , $S = 0$ 1 mark
Calculates $K^{0}$ : $Q = 0$ , $B = 0$ , $S = + 1$ 1 mark
Identifies $Σ^{+}$ and $Ξ^{-}$ as baryons, being three-quark states 1 mark
Identifies $π^{-}$ and $K^{0}$ as mesons, being quark-antiquark pairs 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.2.1 · 7 marks

For each of the following proposed interactions, use conservation of charge, baryon number and lepton number (and strangeness where relevant) to determine whether it is allowed, identifying which law (if any) is violated in a disallowed case.

(a) $π^{-} + p \to K^{+} + Σ^{-}$ , where $K^{+}$ has:

Q = + 1, B = 0, S = + 1

and $Σ^{-}$ has:

Q = - 1, B = + 1, S = - 1

(b) $p \to e^{+} + γ$ .

Show worked solution

(a) LHS: $Q = - 1 + 1 = 0$ , $B = 0 + 1 = 1$ , $S = 0 + 0 = 0$ .

RHS: $Q = 1 - 1 = 0$ , $B = 0 + 1 = 1$ , $S = 1 - 1 = 0$ .

All three balance, so this strong interaction is allowed.

(b) LHS: $Q = + 1$ , $B = + 1$ .

RHS: $Q = + 1 + 0 = + 1$ (charge balances) but $B = 0 + 0 = 0 \neq = 1$ — baryon number is not conserved, so this proposed proton decay is forbidden.

RHS: $Q = 1 - 1 + 0 = 0$ and $B = 1 + 0 + 0 = 1$ both balance, but:

L_{e} = 0 + 1 + 1 = 2 \neq = 0

since both the $e^{-}$ and $ν_{e}$ carry $L_{e} = + 1$ — lepton number is not conserved, so this is forbidden.

Writing $\overset{ν}{ˉ}_{e}$ instead of $ν_{e}$ would give:

L_{e} = 0 + 1 - 1 = 0

on the right, making the decay allowed — this is the correct form of neutron beta decay.

Mark scheme · 7 marks

States (a) is allowed, since charge, baryon number and strangeness all balance (0, 1, 0) on both sides 1 mark
States (b) is forbidden, identifying baryon number as not conserved ( $1 \to 0$ ) 1 mark
States charge is conserved in (b) ( $+ 1 \to + 1$ ) despite the interaction being forbidden overall 1 mark
States charge and baryon number are both conserved in (c) 1 mark
Calculates the electron lepton number as $0$ on the left but $+ 2$ on the right in (c), since both $e^{-}$ and $ν_{e}$ carry $L_{e} = + 1$ 1 mark
States (c) is forbidden by lepton number conservation 1 mark
States (c) would be allowed if $ν_{e}$ were replaced by $\overset{ν}{ˉ}_{e}$ , giving $L_{e} = 0$ on the right 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.2.1 · 5 marks

Sodium-22 ( $_{11}^{22} Na$ ) decays by $β^{+}$ emission to neon-22.

(a) Write the nuclear equation for this decay, including the emitted particle(s).

(b) Describe, at the quark level, what happens to a nucleon inside the sodium-22 nucleus during this decay, naming the exchange particle involved.

(c) Explain why $β^{+}$ decay can occur for a proton bound in this nucleus, even though an isolated free proton cannot undergo $β^{+}$ decay.

Show worked solution

(a):

_{11}^{22} Na \to_{10}^{22} Ne + e^{+} + ν_{e}

(b) An up quark within a proton in the nucleus converts to a down quark, emitting a virtual $W^{+}$ boson: $u \to d + W^{+}$ ; the $W^{+}$ then decays, $W^{+} \to e^{+} + ν_{e}$ — overall, a proton within the nucleus converts to a neutron.

(c) A free proton has less rest energy than a free neutron, so an isolated proton undergoing $β^{+}$ decay would violate conservation of energy; inside sodium-22, the binding energy difference between the parent and daughter nuclides supplies the extra energy needed, making the decay energetically possible.

Mark scheme · 5 marks

Writes the correct nuclear equation $_{11}^{22} Na \to_{10}^{22} Ne + e^{+} + ν_{e}$ with correct mass and atomic numbers 1 mark
States an up quark converts to a down quark within a proton in the nucleus 1 mark
Names the exchange particle as a (virtual) $W^{+}$ boson 1 mark
States the $W^{+}$ decays to a positron and an electron neutrino 1 mark
Explains that the nucleus supplies the extra energy needed, since a free proton's rest energy is less than a free neutron's 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

3.2.2

Electromagnetic radiation and quantum phenomena

The photoelectric effect 3.2.2.1

Definitions

Work function, $ϕ$ : the minimum energy needed to free an electron from a metal surface.
Threshold frequency, $f_{0}$ : the minimum light frequency that can eject a photoelectron, however intense the light, $f_{0} = ϕ / h$ .

Key results

Photon energy: $E = h f = \frac{h c}{λ}$ .
Einstein's photoelectric equation: $h f = ϕ + E_{k (m a x)}$ , i.e. $E_{k (m a x)} = h f - ϕ$ .

Notes

The photoelectric effect is the clearest evidence that light's quantisation is real rather than a convenient bookkeeping trick.
Below $f_{0}$ : no electrons are emitted, no matter how intense the light. Above $f_{0}$ : electrons are emitted instantly, and their maximum kinetic energy rises linearly with frequency rather than with intensity.
Intensity controls the number of photons arriving per second (and so electrons emitted per second), but each photon-electron interaction is one-to-one — a single photon either has enough energy to free an electron or it does not.
Photoelectric emission follows photon absorption with no measurable delay for energy to 'accumulate' — direct evidence that a single photon transfers its whole energy to a single electron in one event, not that energy builds up gradually as the classical wave picture would suggest.

Frequency, intensity and stopping potential 3.2.2.1

Definitions

Stopping potential, $V_{s}$ : the magnitude of the retarding potential difference that just stops even the fastest photoelectrons reaching the collector.

Key results

$e V_{s} = E_{k (m a x)}$ — stopping potential gives a direct experimental route to $E_{k (m a x)}$ without needing to measure electron speed.
The gradient of a graph of $E_{k (m a x)}$ against $f$ gives the Planck constant $h$ directly, and the (negative) intercept gives $- ϕ$ .

Notes

Increasing frequency above threshold increases $E_{k (m a x)}$ (and so $V_{s}$ ) directly, following Einstein's equation.
Increasing intensity at fixed frequency (above threshold) increases the RATE of photoelectron emission — more photons arriving per second, so more electrons freed per second — but leaves $E_{k (m a x)}$ and $V_{s}$ completely unchanged, since each photon still only carries $h f$ , however many photons arrive.
This intensity-independence of $E_{k (m a x)}$ is one of the clearest pieces of evidence against the classical wave picture, in which a more intense wave should be able to deliver more energy to a single electron.

Electron collisions, excitation and ionisation 3.2.2.2

Definitions

Excitation: a bound electron gains energy from a collision and moves to a higher (still bound) energy level, without leaving the atom.
Ionisation: a bound electron gains enough energy to leave the atom entirely, becoming free and leaving behind a positive ion.

Key results

Electronvolt, $1 eV$ : the energy gained by an electron accelerated through a potential difference of $1 V$ , $1 eV \approx 1.602 \times 1 0^{- 19} J$ .

Notes

An incident electron can only excite a bound electron to an energy level that actually exists in that atom — unlike a photon, though, an incident electron does not need to supply exactly the right amount of energy, since any energy not absorbed by the bound electron is simply kept by the scattered incident electron.
Ionisation needs at least the ionisation energy; any additional energy the incident electron carries beyond that is shared between the two resulting free electrons, not lost.
A fluorescent tube uses excitation and ionisation together: accelerated electrons collide with mercury atoms, exciting (and ionising, to sustain the electrical discharge) them; the excited mercury atoms then emit ultraviolet photons as they fall back down; that UV light then excites a phosphor coating on the tube's inner surface, which re-emits the absorbed energy as visible light — a two-step energy-conversion chain, not direct visible emission from the mercury itself.

Energy levels and line spectra 3.2.2.3

Definitions

Energy level: one of a set of discrete, allowed energy values a bound electron in an atom may have — conventionally negative, with $0 eV$ representing an unbound (ionised) electron.
Line spectrum: a spectrum consisting of specific, discrete wavelengths only, rather than a continuous range — direct evidence for discrete atomic energy levels.

Key results

Photon emission on a downward transition: $h f = E_{upper} - E_{lower} = Δ E$ .

Notes

A larger energy gap between levels corresponds to a shorter emitted wavelength, since $E \propto 1/ λ$ .
Excitation by photon absorption is the reverse process, and is far more restrictive than excitation by electron collision: the photon's energy $h f$ must match an allowed energy gap exactly, or it is not absorbed by that transition at all.
Worked example: hydrogen's $n = 3$ to $n = 2$ transition, $Δ E \approx (- 1.51) - (- 3.40) = 1.89 eV \approx 3.03 \times 1 0^{- 19} J$ , giving $λ = \frac{h c}{Δ E} \approx 656 nm$ — one of the visible hydrogen Balmer lines.
Because each element has its own unique set of energy levels, its line spectrum acts as a unique identifying fingerprint — the basis of identifying elements in distant stars from their absorption and emission lines.

Wave–particle duality 3.2.2.4

Key results

De Broglie wavelength: $λ = \frac{h}{p}$ , with $p \approx m v$ for $v ≪ c$ .
Electrons accelerated from rest through potential difference $V$ (non-relativistic): $λ \approx \frac{h}{2 m _{e} e V}$ .

Notes

Light shows particle-like behaviour (the photoelectric effect); matter shows wave-like behaviour (electron diffraction through a thin polycrystalline foil is the standard evidence) — wave-particle duality runs in both directions.
A faster (higher-momentum) electron has a shorter de Broglie wavelength, which produces a diffraction pattern with smaller-radius rings for the same foil and screen geometry — the opposite of what higher speed might naively suggest, but consistent with $λ \propto 1/ p$ .
The diffraction pattern is a genuinely wave property (it depends on wavelength, which depends on momentum), but each individual electron is still detected as a single, localised event on the screen — the pattern only emerges statistically once many electrons have arrived.

Worked examples

Worked example 3.2.2 · 5 marks

In a photoelectric experiment, the stopping potential $V_{s}$ is measured for light of several frequencies incident on a caesium surface.

A graph of $V_{s}$ against $f$ is a straight line of gradient:

4.14 \times 1 0^{- 15} V s

and $y$ -intercept $- 1.90 V$ .

(a) Use the gradient to find a value for the Planck constant, and state the percentage difference from the accepted value:

6.63 \times 1 0^{- 34} J s

(b) Use the intercept to find the work function of caesium, in eV.

Show worked solution

Rearranging $e V_{s} = h f - ϕ$ gives:

V_{s} = \frac{h}{e} f - \frac{ϕ}{e}

so gradient $= h / e$ . (a):

h = gradient \times e = 4.14 \times 1 0^{- 15} \times 1.60 \times 1 0^{- 19} \approx 6.62 \times 1 0^{- 34} J s

a difference of:

\frac{6.63 - 6.62}{6.63} \times 100% \approx 0.09%

from the accepted value.

(b) Intercept:

= - ϕ / e = - 1.90 V

so $ϕ \approx 1.90 eV$ .

ϕ = 1.90 \times 1.60 \times 1 0^{- 19} = 3.04 \times 1 0^{- 19} J

so:

f_{0} = \frac{3.04 \times 1 0 ^{- 19}}{6.62 \times 1 0 ^{- 34}} \approx 4.59 \times 1 0^{14} Hz

(equivalently,

- intercept / gradient

Mark scheme · 5 marks

Identifies gradient $= h / e$ from rearranging Einstein's photoelectric equation 1 mark
Calculates $h \approx 6.62 \times 1 0^{- 34} J s$ 1 mark
States the percentage difference from the accepted value as $\approx 0.09%$ 1 mark
Uses the intercept to find $ϕ \approx 1.90 eV$ 1 mark
Calculates the threshold frequency as $\approx 4.59 \times 1 0^{14} Hz$ 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.2.2 · 5 marks

$n$	Energy / eV
1	$- 13.6$
2	$- 3.40$
3	$- 1.51$
4	$- 0.850$

The table shows some energy levels of the hydrogen atom.

An electron in a hydrogen atom falls from the $n = 4$ level to the $n = 2$ level.

Calculate the energy and wavelength of the photon emitted, and state whether this transition corresponds to emission or absorption.

Show worked solution

Energy released:

Δ E = E_{4} - E_{2} = - 0.850 - (- 3.40) = 2.55 eV

which is:

2.55 \times 1.60 \times 1 0^{- 19} \approx 4.08 \times 1 0^{- 19} J

Wavelength:

λ = \frac{h c}{Δ E} = \frac{6.63 \times 1 0 ^{- 34} \times 3.00 \times 1 0 ^{8}}{4.08 \times 1 0 ^{- 19}} \approx 4.88 \times 1 0^{- 7} m = 488 nm

which lies in the visible spectrum (one of the hydrogen Balmer lines).

Since the electron falls from a higher to a lower energy level, this is a downward transition and so corresponds to emission, not absorption.

Mark scheme · 5 marks

Calculates the energy released as $Δ E = 2.55 eV$ 1 mark
Converts this to $\approx 4.08 \times 1 0^{- 19} J$ 1 mark
Calculates the wavelength as $\approx 4.88 \times 1 0^{- 7} m$ ( $488 nm$ ) 1 mark
States this wavelength lies within the visible spectrum 1 mark
States the transition is downward ( $n = 4$ to $n = 2$ ) and so corresponds to emission 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.2.2 · 5 marks

In an electron diffraction tube, electrons are accelerated from rest through a potential difference of $5.0 kV$ before diffracting through a thin polycrystalline graphite foil, producing rings on a fluorescent screen.

(a) Calculate the de Broglie wavelength of the electrons.

(b) The accelerating voltage is then increased.

State and explain what happens to the radius of the diffraction rings.

Without recalculating, state and explain whether its de Broglie wavelength would be larger or smaller than the electron's. (:

m_{e} = 9.11 \times 1 0^{- 31} kg

e = 1.60 \times 1 0^{- 19} C

h = 6.63 \times 1 0^{- 34} J s

)

Show worked solution

(a):

λ = \frac{h}{2 m _{e} e V} = \frac{6.63 \times 1 0 ^{- 34}}{2 \times 9.11 \times 1 0 ^{- 31} \times 1.60 \times 1 0 ^{- 19} \times 5000} \approx 1.74 \times 1 0^{- 11} m

(b) Increasing $V$ increases the electrons' kinetic energy and so their momentum; since $λ = h / p$ , a larger momentum gives a shorter wavelength, which produces a diffraction pattern with smaller-radius rings.

(c) For the same accelerating voltage, both particles gain the same kinetic energy $e V$ , but $p = 2 m E_{k}$ , so the much more massive proton has a much larger momentum than the electron; since $λ = h / p$ , the proton's de Broglie wavelength is correspondingly smaller than the electron's.

Mark scheme · 5 marks

Calculates the electron's de Broglie wavelength as $\approx 1.74 \times 1 0^{- 11} m$ 1 mark
States increasing $V$ increases the electrons' kinetic energy and momentum 1 mark
States this decreases the de Broglie wavelength, since $λ \propto 1/ p$ 1 mark
Deduces the diffraction ring radii decrease as $V$ increases 1 mark
States the proton's de Broglie wavelength would be smaller than the electron's, since equal kinetic energy $e V$ gives the far more massive proton a larger momentum ( $p = 2 m E_{k}$ ) and hence a shorter wavelength 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.