IB Physics SL Topic 5 — The Atomic & Nuclear World Paper 1 & 2 E = hf = hc/λ ~7 min read

Photon Energy

Light doesn’t dribble out of an atom like water from a tap. It comes in tiny, indivisible parcels called photons — each one carrying a fixed amount of energy set entirely by its colour. Bluer light means fatter energy parcels; redder light means slimmer ones. Once you can put a number on a single photon, you can read an energy-level diagram like a map and work out the exact colour any electron jump produces.

📘 What you need to know

The Photon Model

Think of light not as a smooth wave pouring energy out, but as a stream of tiny bullets. Each bullet is a photon — a single, indivisible packet of electromagnetic energy. A photon can’t be split in half: an atom either hands over one whole photon or none at all. This “all-or-nothing” delivery is the heart of quantum physics, and it’s very different from the everyday picture of a wave that can carry any amount of energy you like.

The size of each packet depends only on the light’s frequency — effectively, its colour. High-frequency light (violet, ultraviolet) comes in big energy packets; low-frequency light (red, infrared) comes in small ones. The link is beautifully simple:

Photon energy from frequency E = hf

Here E is the photon’s energy in joules, f is its frequency in hertz, and h is Planck’s constant (6.63 × 10−34 J s) — a fixed number from the data booklet that sets the “exchange rate” between frequency and energy.

Writing it in terms of wavelength

Frequency and wavelength are tied together by the wave equation, c = fλ, where c is the speed of light. Rearranging gives f = c ÷ λ, and slotting that into E = hf gives the version you’ll reach for most often:

Photon energy from wavelength E = hc ÷ λ

Look closely at that form. Wavelength is on the bottom, so a longer wavelength gives a smaller energy. A red photon (long wavelength) is a low-energy parcel; a violet photon (short wavelength) is a high-energy one. That single fact explains almost every spectra question you’ll meet.

higher frequency
→ shorter wavelength →
more energy per photon

Energy Levels & Transitions

Electrons in an atom can’t sit just anywhere. They’re only allowed on a fixed set of energy levels, like rungs on a ladder with no space in between. The lowest rung is the ground state — the most stable place to be. Higher rungs are excited states. An electron can hop between rungs, but only by dealing in whole photons.

Because the rungs are fixed, the gaps are fixed, so the photon energies — and therefore the wavelengths — are fixed too. That’s why an atom’s light shows up as sharp, separate spectral lines rather than a smooth rainbow.

Photon energy = the energy gap ΔE = hf = E2E1

where E2 is the higher level and E1 the lower one. Combine this with E = hc ÷ λ and you can turn any energy gap straight into a wavelength:

Wavelength from the energy gap λ = hc ÷ (E2E1)
n = 4 n = 3 n = 2 n = 1 (ground state) energy 0 small gap red, long λ bigger gap blue, shorter λ largest gap UV, shortest λ
Bigger jumps release fatter photons. The large drop to the ground state gives a short-wavelength ultraviolet photon, while the small drop between upper levels gives a long-wavelength red one — the wiggles show relative wavelength.
Quick recap: a photon carries energy E = hf = hc ÷ λ; electrons absorb photons to jump up and emit them to drop down; and the photon energy always equals the exact gap between the two levels.

🧭 Turning an energy gap into a wavelength

  1. Find the two levels and take the gap: ΔE = E2E1 (the bigger minus the smaller, so it comes out positive)
  2. Convert eV to joules if needed — multiply by 1.60 × 10−19. This is the step people forget.
  3. Rearrange E = hc ÷ λ into λ = hc ÷ ΔE
  4. Substitute h = 6.63 × 10−34 and c = 3.0 × 108, then divide
  5. Check the size — visible light lives around 400–700 nm (10−7 m). If your answer is wildly off, suspect a missed eV–joule conversion.
WE 1

In a hydrogen atom, an electron drops from a level at −1.51 eV to a level at −3.40 eV, emitting a photon. Calculate (a) the photon’s energy in joules, and (b) its wavelength. State which region of the spectrum this belongs to.

Part (a) — photon energy ΔE = E2 − E1 = −1.51 − (−3.40) = 1.89 eV convert to joules: × 1.60 × 10−19 ΔE = 1.89 × 1.60 × 10−19 ΔE ≈ 3.0 × 10−19 J Part (b) — wavelength λ = hc ÷ ΔE λ = (6.63 × 10−34 × 3.0 × 108) ÷ (3.02 × 10−19) λ ≈ 6.6 × 10−7 m = 660 nm Around 660 nm → the red part of the visible spectrum. This is hydrogen’s famous red line.
WE 2

A red laser pointer emits light of wavelength 630 nm at a power of 5.0 mW. Calculate (a) the energy of a single photon, and (b) the number of photons the pointer emits in 15 s.

Part (a) — one photon’s energy E = hc ÷ λ, with λ = 630 nm = 6.30 × 10−7 m E = (6.63 × 10−34 × 3.0 × 108) ÷ (6.30 × 10−7) E ≈ 3.2 × 10−19 J Part (b) — number of photons in 15 s total energy = power × time = 5.0 × 10−3 × 15 = 0.075 J number = total energy ÷ energy per photon N = 0.075 ÷ (3.16 × 10−19) N ≈ 2.4 × 1017 photons A tiny laser fires hundreds of quadrillions of photons in seconds — which is why light feels perfectly smooth to us.

💡 Top tips

⚠ Common mistakes

That wraps up Atomic Structure — you can now go from Rutherford’s nucleus, through nuclear notation and spectra, to putting real numbers on each spectral line. Up next in Topic 5 we move from the atom to the nucleus itself: radioactive decay and how unstable nuclei shed energy.

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