AI TutorTopicalsYearlysNotesDashboardPricing
Notes/Physics/Paper 4/Quantum Physics
CAIEA Level9702§22.1-22.4

Quantum Physics

Photon energy E = hf and the electronvolt, photon momentum p = E/c, the photoelectric effect and Einstein's equation hf = Φ + ½mv²max, wave-particle duality and the de Broglie wavelength λ = h/p, and discrete energy levels behind atomic line spectra.

200 min read 8 sub-topics
164
question parts
2021-2025 · 37 papers
9 marks
per paper
≈ 9% of the paper
1.9/3
avg difficulty
moderate
#5
most examined
of 16 topics by marks

Light already had a well-tested description by the end of the nineteenth century: a wave, spreading out, diffracting round obstacles, interfering to build bright and dark fringes exactly as Young's double-slit experiment demanded. Then two stubborn observations refused to fit. Shine light on a clean metal surface and electrons fly off instantly, with a maximum energy that depends on the light's colour, not its brightness — a wave theory predicts exactly the opposite. And fire a beam of electrons, solid little particles of matter, through a thin crystal, and they produce rings, the unmistakable signature of diffraction — something only a wave can do. Quantum physics is the resolution: electromagnetic radiation and matter each turn out to behave as both wave and particle, and which behaviour shows up depends on the experiment being done. This note builds that idea from the ground up — what a photon actually is and how much energy and momentum it carries, why the photoelectric effect makes sense only if light comes in discrete packets, the evidence and the equation (λ=h/p\lambda = h/p) for matter behaving as a wave too, and the discrete atomic energy levels that photons of exactly the right energy connect.

The bank places this squarely among the paper's major topics: 325 marks across 2021–2025 rank Quantum Physics 5th of the 16 A2 topics by marks, carried by 164 question parts across 37 paper sittings — close to 8.8 marks on every Paper 4. The mean difficulty of 1.95 sits almost exactly where Alternating Currents (1.95) and Magnetic Fields (2.10) sit: this is not a topic that demands hard algebra, but one with a cluster of small, similarly-shaped equations (E=hfE=hf, p=E/cp=E/c, λ=h/p\lambda = h/p, hf=Φ+12mvmax2hf = \Phi + \tfrac12mv_{max}^2) that are easy to blur together under exam pressure, and a data sheet that gives hh, cc and ee as constants but — unlike some other topics — does not hand candidates any of the equations connecting them. Every relation in this note has to be recalled, not looked up.

The route through is: §01 the photon itself — E=hfE=hf and the electronvolt as a unit sized for atomic-scale energies — §02 photon momentum, p=E/cp=E/c, and what it explains (radiation pressure, recoil) — §03 the photoelectric effect described qualitatively — threshold frequency, work function, and why intensity and frequency affect completely different things — §04 Einstein's photoelectric equation hf=Φ+12mvmax2hf=\Phi+\tfrac12mv_{max}^2 made quantitative, including the stopping-potential graph that measures hh itself — §05 wave-particle duality — the evidence, for both radiation and matter — §06 the de Broglie wavelength λ=h/p\lambda=h/p and electron diffraction made quantitative — §07 discrete energy levels and how they produce emission and absorption line spectra via hf=E1−E2hf=E_1-E_2 — and §08 [Legacy] band theory of conduction, kept for the pre-2021 papers still in the bank.

Before you start you should be able to

  • The wave properties of electromagnetic radiation from AS Waves and Superposition — frequency, wavelength, the wave equation c=fλc=f\lambda, and diffraction and interference as specifically wave phenomena — this note spends §03–§05 arguing that light is ALSO a particle, which only makes sense once the wave case is already solid

  • Kinetic energy Ek=12mv2E_k=\tfrac12mv^2 and work done by a p.d. accelerating a charge, W=qVW=qV, from AS Electricity — both appear repeatedly, in the photoelectric equation (§04) and in accelerating electrons before a de Broglie calculation (§06)

  • Momentum p=mvp=mv and conservation of momentum from AS Dynamics — needed to compare a photon's momentum (§02) against an ordinary particle's, and to handle photon emission/absorption as a momentum-conserving event

  • Standard form, significant figures, and confident unit conversion (nm to m, eV to J) — every calculation in this note mixes very large and very small numbers, and a slip in the powers of ten is the single most common way marks are lost here

  • Reading a straight-line graph — gradient and both intercepts — from AS Practical Skills; §04's stopping-potential graph is a full worked example built entirely on this skill

By the end of this page you can

  • Understand that electromagnetic radiation has a particulate nature

  • Understand that a photon is a quantum of electromagnetic energy

  • Recall and use E=hfE = hf

  • Use the electronvolt (eV) as a unit of energy

  • Understand that a photon has momentum and that the momentum is given by p=E/cp = E/c

  • Understand that photoelectrons may be emitted from a metal surface when it is illuminated by electromagnetic radiation

  • Understand and use the terms threshold frequency and threshold wavelength

  • Explain photoelectric emission in terms of photon energy and work function energy

  • Recall and use hf=Φ+12mvmax2hf = \Phi + \tfrac12 mv_{max}^2

  • Explain why the maximum kinetic energy of photoelectrons is independent of intensity, whereas the photoelectric current is proportional to intensity

  • Understand that the photoelectric effect provides evidence for a particulate nature of electromagnetic radiation while phenomena such as interference and diffraction provide evidence for a wave nature

  • Describe and interpret qualitatively the evidence provided by electron diffraction for the wave nature of particles

  • Understand the de Broglie wavelength as the wavelength associated with a moving particle

  • Recall and use λ=h/p\lambda = h/p

  • Understand that there are discrete electron energy levels in isolated atoms (e.g. atomic hydrogen)

  • Understand the appearance and formation of emission and absorption line spectra

  • Recall and use hf=E1−E2hf = E_1 - E_2

01

Photon energy E = hf and the electronvolt

Syllabus requirement · §22.1

“

understand that electromagnetic radiation has a particulate nature; understand that a photon is a quantum of electromagnetic energy; recall and use E = hf; use the electronvolt (eV) as a unit of energy

”

Light meets its own graininess

Every description of light met so far — reflection, refraction, diffraction, interference, polarisation — treats it as a wave: a continuously varying disturbance that can carry any amount of energy, so long as the source is made brighter or dimmer. That picture is not wrong; Young's double-slit fringes and diffraction gratings are proof it works. But a handful of stubborn observations — how a metal surface releases electrons when light shines on it (§03–§05 of this note tell that story properly, with the actual evidence) — refused to fit a purely continuous wave at all. The resolution physics settled on is not that the wave picture was wrong, but that it was incomplete: electromagnetic radiation also behaves, in the right experiment, as a stream of individual particle-like packets. This section introduces the vocabulary for that packet and the single equation that says how much energy each one carries. The evidence for why this is necessary comes later — but the equation itself is needed everywhere from here on, including inside the evidence itself.

The photon

A photon is a single quantum of electromagnetic energy — one indivisible "packet" of energy carried by the radiation. "Quantum" here means exactly what it sounds like: a smallest indivisible amount. A beam of light does not deliver its energy in any size parcel you like; it delivers it in whole photons, and a detector sensitive enough (a single retinal cell in the dark, a photomultiplier tube) really does register light arriving one photon at a time, not as a smooth continuous trickle.

Every photon of a given frequency carries exactly the same energy as every other photon of that same frequency — that fixed amount is the entire content of the next equation.

E = hf — the energy of one photon

The frequency ff in this equation is nothing new — it is the same wave frequency already met in AS Waves: the number of complete oscillations the associated electromagnetic wave performs every second, in hertz. What is new is the claim that a single photon of that frequency carries a fixed, non-negotiable energy proportional to it:

E=hfE = hf

where hh is the Planck constant, h=6.63×10−34 J sh = 6.63\times10^{-34}\ \text{J s} — a fixed number of nature, given on the data sheet. Doubling the frequency doubles the energy of each photon; the light does not get "brighter" in the everyday sense, each individual photon simply carries more energy.

E=hfE = hf

Photon energy — h = 6.63×10⁻34 J s (Planck constant, given on the data sheet), f in Hz.

From frequency to wavelength: E = hc/λ

Wavelength, not frequency, is very often the quantity actually given — from a diffraction grating, a spectrometer, or a stated colour. Getting from E=hfE=hf to a wavelength form is a short substitution, not a new law.

Start from the wave equation relating speed, frequency and wavelength (AS Waves), true for every electromagnetic wave in free space:

c=fλc = f\lambda

Rearrange for frequency:

f=cλf = \frac{c}{\lambda}

Substitute this expression for ff into E=hfE = hf:

E=h(cλ)=hcλE = h\left(\frac{c}{\lambda}\right) = \frac{hc}{\lambda}

Nothing here is a separate law to memorise on top of E=hfE=hf — it is the same relation, written with whichever of ff or λ\lambda a question happens to give.

E=hcλE = \frac{hc}{\lambda}

Photon energy from wavelength — reached by substituting f = c/λ into E = hf. c = 3.00×10⁸ m s⁻¹ (given).

A unit sized for the atom: the electronvolt

A photon energy computed from E=hfE=hf or E=hc/λE=hc/\lambda for visible light or X-rays comes out as a number like 10−1910^{-19} or 10−1510^{-15} joules — correct, but an awkward size to write and easy to lose a power of ten in. Physicists working at atomic and photon scales use a differently-sized unit instead: the electronvolt (eV), defined as the energy transferred to a single electron accelerated through a potential difference of exactly 1 volt.

Recall from AS Electricity that the work done moving a charge QQ through a potential difference VV is

W=QVW = QV

Set the charge equal to the charge on a single electron, Q=e=1.60×10−19 CQ = e = 1.60\times10^{-19}\ \text{C}:

W=eVW = eV

Let the potential difference be exactly V=1 VV = 1\ \text{V}, and evaluate:

W=(1.60×10−19)×1=1.60×10−19 JW = (1.60\times10^{-19})\times1 = 1.60\times10^{-19}\ \text{J} 1 eV=1.60×10−19 J1\ \text{eV} = 1.60\times10^{-19}\ \text{J}

That single number is the whole content of the unit: "an energy of 1 eV" just means "1.60×10−19 J1.60\times10^{-19}\ \text{J}", nothing more exotic. Any energy in electronvolts converts to joules by multiplying by 1.60×10−191.60\times10^{-19}, and any energy in joules converts to electronvolts by dividing by the same number — always as its own explicit step, since the eV is not an SI unit and cannot be substituted into a formula such as E=hfE=hf alongside quantities in metres, seconds and kilograms until it has been converted.

1 eV=1.60×10−19 J1\ \text{eV} = 1.60\times10^{-19}\ \text{J}

The electronvolt — energy gained by ONE electron accelerated through a p.d. of exactly 1 V. Not an SI unit; convert to J before combining with anything else.

Not on the data sheet

The data sheet gives the constants hh, cc and ee as numbers — it does not give the equations E=hfE=hf, E=hc/λE=hc/\lambda, or 1 eV=1.60×10−19 J1\ \text{eV}=1.60\times10^{-19}\ \text{J}, or any other quantum-physics relation in this note. Every equation connecting these constants has to be recalled from memory in the exam; candidates used to looking mechanics equations up on the front page can freeze exactly here, because this topic's own relations are never printed.

Invented demo — one photon of visible light, in J and in eV

Visible light near the middle of the spectrum has wavelength approximately 500 nm500\ \text{nm}. Calculate the energy of one photon of this light, (a) in joules, (b) in electronvolts.

Show full working
  1. 1

    Convert the wavelength to metres:

    λ=500 nm=5.00×10−7 m\lambda = 500\ \text{nm} = 5.00\times10^{-7}\ \text{m}

    The nm must become m before any SI formula can use it — writing this out as its own step is what prevents a stray factor of 10⁹ later.

  2. 2

    (a) Find the frequency, using c=fλc=f\lambda:

    f=cλ=3.00×1085.00×10−7=6.00×1014 Hzf = \frac{c}{\lambda} = \frac{3.00\times10^{8}}{5.00\times10^{-7}} = 6.00\times10^{14}\ \text{Hz}

    Going via f and E = hf here, rather than jumping straight to E = hc/λ, keeps every substitution visible — both routes give the same answer.

  3. 3

    Apply E=hfE=hf:

    E=(6.63×10−34)×(6.00×1014)=3.98×10−19 JE = (6.63\times10^{-34})\times(6.00\times10^{14}) = 3.98\times10^{-19}\ \text{J}

    A direct substitution of the value of h given on the data sheet and the frequency just found.

  4. 4

    (b) Convert this energy to electronvolts:

    EeV=3.98×10−191.60×10−19=2.49 eVE_{eV} = \frac{3.98\times10^{-19}}{1.60\times10^{-19}} = 2.49\ \text{eV}

    Dividing by the eV-to-J conversion factor, as its own final step — this is the size of number the electronvolt unit exists to produce.

Answer

E ≈ 3.98×10⁻19 J = 2.49 eV.

This same photon returns in §02 to find its momentum — keep both the 3.98×10⁻19 J and the 5.00×10⁻7 m results on hand.

radio> 10⁻¹ mmicrowave10⁻¹–10⁻³infrared10⁻³–7×10⁻⁷visible400–700 nmultra-violet4×10⁻⁷–10⁻⁸X-rays10⁻⁸–10⁻¹¹gamma< 10⁻¹¹ mwavelength increasesfrequency increasesvisible light: 400 nm – 700 nmviolet ≈ 400 nm … red ≈ 700 nm — recall this rangeband widths are schematic, not to scale — all EM waves travel at c = 3.00 × 10⁸ m s⁻¹ in free space

The same spectrum met in AS Waves, now read through E = hc/λ: every step from radio towards gamma is a step to SHORTER wavelength, and therefore — inversely — to LARGER photon energy. A single gamma-ray photon can carry many billions of times the energy of a single radio photon, even though both travel at the same speed c.

A linked real question — definition, wavelength, and photon rate

9702/42 F/M 2025 Q86 marks

A laser emits red light of a single wavelength. The light is produced when electrons move from a higher energy level to a lower energy level. The difference in energy between the two levels is 1.96 eV1.96\ \text{eV}.

(a) State what is meant by a photon. [2]

(b)(i) Calculate the wavelength of the light. [3]

(b)(ii) The power of the beam emitted by the laser is 1.0×10−2 W1.0\times10^{-2}\ \text{W}. Calculate the number of photons emitted per unit time by the laser. [1]

Show full working
  1. 1

    (a) A photon is a quantum of energy of electromagnetic radiation.

    The mark scheme awards one mark for 'quantum of energy' and a separate second mark for 'of electromagnetic radiation' — both halves of the definition are needed, not just the word 'packet'.

  2. 2

    (b)(i) Convert the given energy to joules:

    E=1.96×(1.60×10−19)=3.136×10−19 JE = 1.96\times(1.60\times10^{-19}) = 3.136\times10^{-19}\ \text{J}

    The energy was given in eV, but E = hc/λ needs SI joules — converting is always the first move, before any other substitution.

  3. 3

    Rearrange E=hc/λE = hc/\lambda for λ\lambda:

    λ=hcE\lambda = \frac{hc}{E}

    Naming the rearrangement as its own step, before any numbers go in, keeps the algebra separate from the arithmetic that follows.

  4. 4

    Substitute:

    λ=(6.63×10−34)×(3.00×108)3.136×10−19\lambda = \frac{(6.63\times10^{-34})\times(3.00\times10^{8})}{3.136\times10^{-19}}

    Every value plugged in directly — h and c from the data sheet, E from the previous step — with nothing simplified yet.

  5. 5

    Evaluate:

    λ=6.34×10−7 m≈6.3×10−7 m\lambda = 6.34\times10^{-7}\ \text{m} \approx 6.3\times10^{-7}\ \text{m}

    6.3×10⁻7 m (630 nm) sits comfortably in the red part of the visible spectrum, matching the question's own description of the laser as 'red light' — a good consistency check on the arithmetic.

  6. 6

    (b)(ii) Name the relation between power and photon rate:

    Nt=PE\frac{N}{t} = \frac{P}{E}

    Power is energy delivered per second; since every photon here carries the same fixed energy E, dividing the total energy-per-second by the energy of ONE photon gives the number of photons arriving per second — the 'discrete lumps' idea from earlier in this section, put to work.

  7. 7

    Substitute and evaluate:

    Nt=1.0×10−23.136×10−19=3.2×1016 s−1\frac{N}{t} = \frac{1.0\times10^{-2}}{3.136\times10^{-19}} = 3.2\times10^{16}\ \text{s}^{-1}

    A genuinely enormous number — a reminder of just how small a single photon's energy is, and part of why the electronvolt exists as a more human-sized alternative.

Answer

(a) A photon is a quantum of energy of electromagnetic radiation. (b)(i) λ = 6.3×10⁻7 m. (b)(ii) N/t = 3.2×10¹⁶ s⁻¹.

Whenever a question gives an energy in eV and later needs a rate (photons per second, or — see §02 — a force), convert to joules FIRST and carry that joule value through every later part. Converting twice, or forgetting the first conversion, is the most common way marks bleed out of a linked multi-part question like this one.

A real 'show that' — minimum X-ray wavelength from an accelerating voltage

9702/42 M/J 2025 Q6(c)(i)3 marks

Many electrons are accelerated from rest from plate X to plate Y in Fig. 6.1, through a potential difference of 58 kV58\ \text{kV}. When the electrons hit plate Y, the absorption of their kinetic energies results in the emission of electromagnetic waves.

Show that the minimum wavelength of these electromagnetic waves is 21 pm21\ \text{pm}.

Fig. 6.1 from the question paper: two parallel plates X and Y in a vacuum, separated by 0.041 m, with an electron at rest at the centre of plate X. A p.d. of 58 kV accelerates the electron from X towards Y; later in the same question, many such electrons are fired at plate Y and their kinetic energy is converted into electromagnetic radiation on impact.

Fig. 6.1 from the question paper: two parallel plates X and Y in a vacuum, separated by 0.041 m, with an electron at rest at the centre of plate X. A p.d. of 58 kV accelerates the electron from X towards Y; later in the same question, many such electrons are fired at plate Y and their kinetic energy is converted into electromagnetic radiation on impact.

Show full working
  1. 1

    Identify what "minimum wavelength" means here: the SHORTEST wavelength corresponds to the HIGHEST-energy photon that can be produced, and the highest-energy photon possible is made when a SINGLE electron gives up ALL of its kinetic energy — the energy eVeV it gained from the accelerating p.d. — to ONE photon in a single event.

    This is the physical reasoning a 'show that' question hides behind the equation — without stating it, it is not obvious why the accelerating p.d. alone fixes a minimum wavelength.

  2. 2

    Equate that maximum photon energy to the electron's kinetic energy, eVeV, and to the photon-energy formula:

    eV=hcλeV = \frac{hc}{\lambda}

    eV here is the standard work-energy relation for an accelerated charge from AS Electricity; the right-hand side is this section's own E = hc/λ. Setting them equal is the one physical idea in the whole question.

  3. 3

    Rearrange for λ\lambda:

    λ=hceV\lambda = \frac{hc}{eV}

    Isolating λ before any numbers go in, exactly as in the worked example above.

  4. 4

    Substitute the given accelerating voltage, V=58 kV=58×103 VV = 58\ \text{kV} = 58\times10^{3}\ \text{V}, and the constants from the data sheet:

    λ=(6.63×10−34)×(3.00×108)(1.60×10−19)×(58×103)\lambda = \frac{(6.63\times10^{-34})\times(3.00\times10^{8})}{(1.60\times10^{-19})\times(58\times10^{3})}

    The kV → V conversion is written out explicitly as part of the substitution — a silent ×1000 error here is exactly the kind of slip a 'show that' question is designed to catch.

  5. 5

    Evaluate numerator and denominator separately, then divide:

    numerator=1.989×10−25 J m,denominator=9.28×10−15 J\text{numerator} = 1.989\times10^{-25}\ \text{J m}, \qquad \text{denominator} = 9.28\times10^{-15}\ \text{J} λ=1.989×10−259.28×10−15=2.14×10−11 m\lambda = \frac{1.989\times10^{-25}}{9.28\times10^{-15}} = 2.14\times10^{-11}\ \text{m}

    Splitting the fraction into its top and bottom before dividing keeps each power-of-ten calculation small and checkable, rather than risking an error buried inside one long division.

  6. 6

    Convert to picometres to compare with the given answer:

    λ=2.14×10−11 m=21.4 pm≈21 pm, as required.\lambda = 2.14\times10^{-11}\ \text{m} = 21.4\ \text{pm} \approx 21\ \text{pm, as required.}

    1 pm = 10⁻¹² m, so shifting the decimal point three places converts m to pm — carrying the extra figure (21.4) before rounding is what makes a 'show that' answer trustworthy rather than a suspiciously exact match.

Answer

λ = hc/(eV) = 21 pm (2 s.f.), as required.

'Minimum wavelength' in an X-ray / accelerated-electron context always means 'maximum photon energy', which always means 'one electron's entire kinetic energy converted into one photon' — recognising that physical statement is the real content of these questions; the algebra afterwards is routine.

Common mistakes

  • Adding or comparing an energy given in eV directly with one already in J, e.g. writing "total energy = 1.96+3.14×10−191.96 + 3.14\times10^{-19}".

    Convert the eV value to joules FIRST (multiply by 1.60×10−191.60\times10^{-19}), and only then combine it with any quantity already in SI units.

    eV and J are different-sized units for the same physical quantity — treating the raw numbers as directly comparable is exactly like adding a distance in miles to one in metres without converting.

  • Trying to recall E=hfE=hf by looking for it on the data sheet, or misremembering it as E=h/fE=h/f or E=f/hE=f/h.

    E=hfE=hf — and every other quantum-physics equation in this note — must be memorised. The data sheet supplies only the constants hh, cc, ee, never the relations between them.

    This is a genuine, recurring trap: candidates used to looking equations up on the front page can freeze exactly here, because this topic's own relations are never printed.

  • Substituting a wavelength given in nm directly into E=hc/λE=hc/\lambda without converting to metres first, e.g. using λ=500\lambda=500 instead of 5.00×10−75.00\times10^{-7}.

    Convert nm to m as an explicit first step (1 nm=10−9 m1\ \text{nm}=10^{-9}\ \text{m}) — write it out before it goes anywhere near the formula.

    This produces an energy that is out by a factor of 10⁹ — a wildly wrong but still numerically 'plausible-looking' answer that is easy to miss without a sanity check.

Your turn

A routine E = hc/λ calculation with an eV conversion, a real accelerating-voltage question, and an invented comparison across the EM spectrum.

  1. 14 marks

    A sodium street lamp emits yellow light of wavelength 589 nm589\ \text{nm}. Calculate the energy of one photon of this light, giving your answer in both joules and electronvolts.

    Stuck? Show hint

    Convert the wavelength to metres first, then use E=hc/λE=hc/\lambda before converting the result to eV.

    Show solution
    1. 1

      Convert the wavelength:

      λ=589 nm=5.89×10−7 m\lambda = 589\ \text{nm} = 5.89\times10^{-7}\ \text{m}

      Written out before the formula, exactly as the mistakes box above warns.

    2. 2

      Apply E=hc/λE=hc/\lambda:

      E=(6.63×10−34)×(3.00×108)5.89×10−7E = \frac{(6.63\times10^{-34})\times(3.00\times10^{8})}{5.89\times10^{-7}}

      Substituting h and c straight from the data sheet, and the converted λ from the previous step.

    3. 3

      Evaluate:

      E=3.38×10−19 JE = 3.38\times10^{-19}\ \text{J}

      A result the same order of magnitude as the §01 demo photon (500 nm), as expected for two nearby visible wavelengths.

    4. 4

      Convert to eV:

      E=3.38×10−191.60×10−19=2.11 eVE = \frac{3.38\times10^{-19}}{1.60\times10^{-19}} = 2.11\ \text{eV}

      Dividing by the eV-to-J conversion factor, as its own final step.

    Answer

    E = 3.38×10⁻19 J = 2.11 eV.

  2. 29702/41 M/J 2024 Q8(c)(i)1 mark

    Fig. 8.1 shows a tube in which X-rays are produced at a metal target. Electrons are accelerated from a heated filament to the target by a constant high voltage applied across terminals X and Y.

    For an accelerating voltage of 32 kV32\ \text{kV}, determine the maximum energy, in MeV, of an X-ray photon produced at the target.

    Fig. 8.1: an X-ray tube. A heated filament emits electrons, which are accelerated across a vacuum by the high voltage between terminals X and Y towards an angled metal target, the whole assembly enclosed in a glass tube.

    Fig. 8.1: an X-ray tube. A heated filament emits electrons, which are accelerated across a vacuum by the high voltage between terminals X and Y towards an angled metal target, the whole assembly enclosed in a glass tube.

    Stuck? Show hint

    The maximum photon energy equals the full kinetic energy gained by one electron, E=eVE=eV — then convert from J to MeV.

    Show solution
    1. 1

      Apply E=eVE=eV for the maximum-energy electron (all of the accelerating p.d.'s energy given to a single electron):

      E=(1.60×10−19)×(32×103)E = (1.60\times10^{-19})\times(32\times10^{3})

      The same reasoning as the worked X-ray example above — the MAXIMUM photon energy corresponds to an electron losing its ENTIRE kinetic energy in one collision.

    2. 2

      Evaluate in joules:

      E=5.12×10−15 JE = 5.12\times10^{-15}\ \text{J}

      Kept in joules for now, since the conversion factor to MeV is itself defined via joules.

    3. 3

      Convert to MeV, using 1 MeV=1.60×10−13 J1\ \text{MeV}=1.60\times10^{-13}\ \text{J}:

      E=5.12×10−151.60×10−13=0.032 MeVE = \frac{5.12\times10^{-15}}{1.60\times10^{-13}} = 0.032\ \text{MeV}

      MeV is simply a larger multiple of the eV — one million electronvolts — used because X-ray and nuclear energies are awkwardly large numbers of eV, just as eV itself is a more convenient size than joules.

    Answer

    E = 0.032 MeV.

  3. 34 marks

    A gamma-ray photon emitted by a cobalt-60 source has energy 1.33 MeV1.33\ \text{MeV}. A radio-station transmitter emits photons of frequency 1.00×108 Hz1.00\times10^{8}\ \text{Hz} (100 MHz100\ \text{MHz}).

    Calculate the energy, in joules, of one photon of each type, and state the ratio of the gamma-ray photon energy to the radio photon energy.

    Stuck? Show hint

    Convert the gamma energy from MeV to J directly; find the radio photon's energy from its frequency using E=hfE=hf.

    Show solution
    1. 1

      Gamma photon — convert MeV to J:

      Eγ=1.33×(1.60×10−13)=2.13×10−13 JE_\gamma = 1.33\times(1.60\times10^{-13}) = 2.13\times10^{-13}\ \text{J}

      1 MeV = 1.60×10⁻13 J — the same conversion factor used in the exercise above, just applied to a different starting unit.

    2. 2

      Radio photon — apply E=hfE=hf directly, since the frequency is already given:

      Er=(6.63×10−34)×(1.00×108)=6.63×10−26 JE_r = (6.63\times10^{-34})\times(1.00\times10^{8}) = 6.63\times10^{-26}\ \text{J}

      No wavelength conversion needed here — the frequency was given directly, so E = hf is the more direct route than E = hc/λ.

    3. 3

      Form the ratio:

      EγEr=2.13×10−136.63×10−26≈3.2×1012\frac{E_\gamma}{E_r} = \frac{2.13\times10^{-13}}{6.63\times10^{-26}} \approx 3.2\times10^{12}

      Dividing the two energies found above makes the enormous span of photon energies across the EM spectrum concrete: a single gamma photon here carries over a trillion times the energy of a single radio photon.

    Answer

    Eγ = 2.13×10⁻13 J; Er = 6.63×10⁻26 J; ratio ≈ 3.2×10¹².

Practise photon energy and electronvolt questionsReal past-paper questions · Photon energy E = hf and the electronvolt

The rest of this note

Checking your access…

Can you do all of these?

  • E = hf is NOT given on the data sheet — only h, c and e are provided as constants; the equation itself must be recalled every time

  • Convert eV to J with 1 eV = 1.60×10⁻19 J as its OWN step before mixing with any other SI energy — never combine eV and J directly

  • p = E/c is a separate relation for a MASSLESS photon — never derive it from p = mv, and never use it for an electron or any particle with mass

  • Threshold frequency f0 (or threshold wavelength λ0) is defined by Φ = hf0 = hc/λ0 — the photon energy that leaves an electron with EXACTLY zero kinetic energy, the boundary case

  • Below the threshold frequency, NO photoelectrons are emitted no matter how intense the light — intensity cannot compensate for too little energy per photon

  • Intensity changes the RATE of photon arrival, so it changes the photoelectric CURRENT (electrons per second) — it does NOT change KEmax, which depends only on frequency

  • Frequency changes the ENERGY per photon, so it changes KEmax — a brighter but lower-frequency source below f0 still emits nothing

  • vmax in hf = Φ + ½mv²max belongs to the FASTEST photoelectrons only — those emitted from the very surface, losing no energy escaping; slower ones lost energy on the way out

  • eVS = ½mv²max converts the equation into VS = (h/e)f − Φ/e — a straight line in VS against f, with gradient h/e, y-intercept −Φ/e, and x-intercept exactly f0

  • The photoelectric effect is evidence for the PARTICLE nature of light; diffraction and interference are evidence for its WAVE nature — never cite one experiment as evidence for the other

  • Electron diffraction (rings from a thin crystal/graphite film) is the standard qualitative evidence for the WAVE nature of matter — know the observation (rings, not spots or a single bright patch) as well as the conclusion

  • λ = h/p is inversely proportional: DOUBLING momentum HALVES wavelength — check any 'what happens to λ if...' answer against this direction before writing it down

  • Accelerating a charge through p.d. V is a chain: eV = ½mv² → v → p = mv → λ = h/p — each arrow is its own step, never compressed into one line

  • A specific transition's photon energy is the MAGNITUDE |E1 − E2|; energy levels themselves are conventionally negative (bound states), but the emitted/absorbed photon energy is always quoted positive

  • An emission spectrum shows BRIGHT lines (photons actually reaching the detector); an absorption spectrum shows DARK lines on an otherwise continuous background, at the SAME wavelengths as the emission lines from those same transitions

  • With n discrete energy levels, count ALL pairwise transitions if asked how many spectral lines are possible — not just transitions to/from the ground state

  • [Legacy, pre-2021] Conductors have overlapping or partially-filled energy bands, so electrons move freely with almost no energy cost

  • [Legacy, pre-2021] Insulators have a full valence band separated from an empty conduction band by a LARGE forbidden gap; semiconductors have the same structure with a SMALL gap, so a few electrons cross it thermally

Now do the questions
164 real Paper 4 parts from 2021-2025, sorted by difficulty, with mark schemes
Previous
Alternating Currents
Next
Nuclear Physics