IB Physics HLTopic 5 — Atomic & NuclearPaper 1 & 2E = hf~16 min read
Photon Energy
The spectra told us light comes in specific colours; the photon model tells us why. Light isn’t a smooth continuous stream — it arrives in tiny indivisible packets called photons, each carrying one fixed dose of energy. And that energy depends on just one thing: the frequency. High frequency means a high-energy photon; low frequency means a low-energy one. This single equation, E = hf, connects the colour of a spectral line to the exact energy-level jump that produced it.
📘 What you need to know
A photon is a massless “packet” or quantum of electromagnetic energy
Photon energy is discrete — each photon delivers its energy all in one go
Photon energy: E = hf, or using the wave equation, E = hc / λ
Higher frequency → higher energy; longer wavelength → lower energy
Electrons sit in discrete energy levels; the lowest is the ground state
Excitation: an electron absorbs a photon and moves up a level
De-excitation: an electron emits a photon and moves down a level
Ionisation energy is the minimum energy to remove an electron from the ground state
The photon energy equals the gap between levels: ΔE = hf = E2 − E1
The photon model
Photons are the fundamental particles of all electromagnetic radiation. The crucial idea is that a photon’s energy is quantised — it carries a specific amount and transfers it all at once, unlike a classical wave which delivers energy continuously.
Photon energyE = hf = hc / λE = energy (J) • h = Planck’s constant (6.63 × 10−34 J s) • f = frequency (Hz) • c = 3.0 × 108 m s−1 • λ = wavelength (m)
Reading the equation tells you the trends: energy rises with frequency, and falls with wavelength (since f = c/λ). So a violet photon carries more energy than a red one; an X-ray photon far more than a radio-wave photon.
Two forms, one equation. Use E = hf when the question gives you a frequency, and E = hc/λ when it gives you a wavelength. They’re identical — just substitute f = c/λ. Both h and c are in your data booklet, but memorising them saves precious time in Paper 1.
Atomic energy levels
Electrons in an atom can only occupy certain energy states called energy levels. They naturally settle into the lowest available level — the ground state — because that’s the most stable arrangement. To move an electron between levels, energy must be absorbed or emitted as a photon.
Excitation: an electron moves up to a higher level by absorbing a photon — the atom is now “excited”
De-excitation: an electron moves down by emitting a photon
Ionisation: an electron is removed entirely; the atom becomes an ion
Ionisation energy = the minimum energy to remove an electron from the ground state
Energy levels are discrete, negative, and get closer together near the top. An electron absorbs a photon to climb (blue) and emits one to fall (red). The photon’s energy equals the gap between the two levels.
Why are the energies negative? Because a bound electron has less energy than a free one. We set the “free” electron (fully removed) at 0 eV, so anything still trapped in the atom sits below that — hence the minus signs. The ground state is the most negative (−13.6 eV for hydrogen), meaning it’s the most tightly bound. To ionise from the ground state you must supply the full 13.6 eV to reach 0.
Linking photons to energy levels
Here’s where it all connects. Each spectral line is a photon, and each photon comes from a jump between two energy levels. The photon’s energy is exactly the difference between those levels.
Photon energy = energy-level gapΔE = hf = E2 − E1rearranged for wavelength: λ = hc / (E2 − E1)
Energy gap E2 − E1
= photon energy
E = hf
λ = hc/E
Wavelength of the line
A bigger energy gap means a higher-energy photon, and therefore a shorter wavelength. A small gap gives a low-energy, long-wavelength photon. That’s why big jumps down to the ground state produce ultraviolet lines, while small jumps between high levels give infrared.
WE 1
A photon has a frequency of 6.0 × 1014 Hz. (a) Calculate its energy in joules. (b) Convert this to electronvolts. (h = 6.63 × 10−34 J s, 1 eV = 1.60 × 10−19 J)
(a) energy from E = hfE = hf = (6.63 × 10⁻³⁴)(6.0 × 10¹⁴)E = 4.0 × 10⁻¹⁹ J(b) convert to eVE = (3.98 × 10⁻¹⁹) / (1.60 × 10⁻¹⁹)E = 2.5 eV6.0 × 10¹⁴ Hz is orange-ish visible light, and ~2.5 eV is a typical visible-photon energy. To convert J → eV, divide by 1.60 × 10⁻¹⁹; to go the other way, multiply.
WE 2
Light of wavelength 490 nm is completely absorbed by a surface. The light has a power of 3.6 mW. Calculate the number of photons hitting the surface in 30 s. (h = 6.63 × 10−34 J s, c = 3.0 × 108 m s−1)
Step 1 — energy of one photonE = hc/λ = (6.63 × 10⁻³⁴)(3.0 × 10⁸) / (490 × 10⁻⁹)E = 4.06 × 10⁻¹⁹ JStep 2 — photons per second = power / energyn/s = P/E = (3.6 × 10⁻³) / (4.06 × 10⁻¹⁹)n/s = 8.87 × 10¹⁵ per secondStep 3 — total in 30 sN = (8.87 × 10¹⁵) × 30N = 2.7 × 10¹⁷ photonsThe chain is: power is energy per second, so dividing by the energy of one photon gives photons per second. Then multiply by the time. Note how many photons even a few milliwatts delivers — light is extraordinarily “grainy” only on the tiniest scale.
WE 3
In hydrogen, the n = 4 level is at −0.85 eV and the n = 2 level is at −3.40 eV. An electron drops from n = 4 to n = 2. (a) Calculate the energy of the emitted photon in joules. (b) Calculate the wavelength of the emitted light. (c) State the region of the spectrum this belongs to.
(a) Step 1 — the energy gapΔE = E₄ − E₂ = (−0.85) − (−3.40) = 2.55 eVin joules: ΔE = 2.55 × (1.60 × 10⁻¹⁹)ΔE = 4.08 × 10⁻¹⁹ J(b) Step 2 — wavelength from λ = hc/ΔEλ = (6.63 × 10⁻³⁴)(3.0 × 10⁸) / (4.08 × 10⁻¹⁹)λ = 4.9 × 10⁻⁷ m = 490 nm(c) region of the spectrumvisible light (blue-green)490 nm is one of hydrogen’s famous visible (Balmer) lines — the same blue-green line you’d see in its emission spectrum. Take the gap in eV, convert to joules, then use λ = hc/ΔE. The subtraction of two negatives is where marks are most often lost: (−0.85) − (−3.40) = +2.55.
⚡ Working a photon-energy question
Given frequency?E = hf. Given wavelength?E = hc/λ.
Convert units: nm → m, mW → W, and eV ↔ J (× or ÷ 1.60 × 10−19).
Number of photons?N = power / energy-per-photon × time.
Bigger gap = higher energy = shorter wavelength.
💡 Top tips
Use E = hf for frequency, E = hc/λ for wavelength — same equation.
Convert nm → m and eV → J before substituting.
Energy-level gap: ΔE = E2 − E1; take care with the negative level energies.
Bigger gap → shorter wavelength (higher-energy photon).
Photons per second = power ÷ photon energy.
h and c are in the data booklet, but memorise them for speed.
⚠ Common mistakes
Leaving wavelength in nm in E = hc/λ — convert to metres
Botching the double negative: (−0.85) − (−3.40) = +2.55, not −2.55
Forgetting to convert eV to J before using E = hf
Thinking longer wavelength means higher energy — it’s the opposite
Confusing frequency and wavelength in the two forms of the equation
Forgetting to multiply by time when finding a total number of photons
Quick recap: A photon is a quantum of EM energy, and its energy is E = hf = hc/λ — higher frequency (shorter wavelength) means more energy. Electrons occupy discrete energy levels, sitting lowest in the ground state; they absorb a photon to rise and emit one to fall, and ionisation energy frees an electron from the ground state. Each spectral line is a photon whose energy equals the gap between two levels: ΔE = hf = E2 − E1, so a bigger gap gives a shorter wavelength.
You can now turn any wavelength into a photon energy and match it to an energy-level jump. The natural next step is to look more closely at the experiment that measured the nucleus itself — using the energy of an alpha particle to work out how close it gets to a nucleus, and from that, the nuclear radius. That’s the Rutherford Scattering & Nuclear Radius page coming up in this topic.
Photon calculations tripping you up?
Book a free meeting and we’ll drill E = hf, unit conversions, energy-level gaps, and photon-counting until they’re automatic.
