Photon Energy, Momentum & the Electronvolt
- Compute photon energy from frequency or wavelength
- Convert between joules and electronvolts fluently
- Compute photon momentum and explain why a massless particle has momentum
Energy comes in lumps
Light of frequency f delivers energy only in discrete amounts E = hf, one photon at a time. Since c = fλ, this can also be written E = hc/λ, which is usually the more convenient form because wavelength is what gets measured. Shorter wavelength means more energetic photons — which is why ultraviolet damages skin and radio waves do not, however intense the radio signal is.
Why the electronvolt exists
Atomic energies in joules are numbers like 3.2 × 10⁻¹⁹, which are awkward and invite exponent errors. The electronvolt is the energy an electron gains crossing one volt, 1.60 × 10⁻¹⁹ J, and it makes atomic quantities human-sized: visible photons are 2–3 eV, atomic ionization is around 10 eV, and nuclear energies are in MeV. Note the useful shortcut: E(eV) = 1240 / λ(nm), which turns a wavelength into an energy in one division.
Momentum without mass
A photon has zero mass and yet carries momentum p = E/c = h/λ. This is not a contradiction; it is a signal that p = mv is a low-speed approximation rather than the definition of momentum. The consequence is measurable: light exerts radiation pressure, which is what pushes a solar sail and what slightly perturbs spacecraft trajectories. Shorter wavelength means larger momentum, which is why gamma rays transfer momentum so effectively.
The photoelectric threshold
The work function φ is the minimum energy needed to liberate an electron from a metal surface, so KE_max = hf − φ. Three consequences the exam tests: there is a threshold frequency below which no electrons are emitted regardless of intensity; raising the intensity raises the number of electrons but not their maximum energy; and raising the frequency raises the maximum energy. All three defeat the wave picture and are why the photon picture was accepted.
Light of wavelength 400 nm strikes a metal with a work function of 2.0 eV. Find the photon energy in eV, the maximum kinetic energy of emitted electrons, and the threshold wavelength.
- 1.Photon energy: E = 1240/400 = 3.1 eV.
- 2.Maximum kinetic energy: KE = 3.1 − 2.0 = 1.1 eV.
- 3.Threshold is where KE = 0, so hf = φ = 2.0 eV.
- 4.Threshold wavelength: λ = 1240/2.0 = 620 nm.
Increasing the intensity does not increase the maximum kinetic energy of photoelectrons. It increases how many are emitted. Only a higher frequency gives each electron more energy — and that separation is the whole reason the photon model was needed.
A photon has a wavelength of 620 nm. Its energy is approximately:
The intensity of light striking a photoelectric surface is doubled at the same frequency. The maximum kinetic energy of emitted electrons:
A photon has zero mass but nonzero momentum. This tells us that:
Decide whether the question wants joules or electronvolts before computing, and state the unit at every step. Mixing the two is the dominant error in this unit, and it produces answers off by 19 orders of magnitude.
Answer the 3 checkpoints as you read.
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