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E.2

Quantum physics

Theme E · Nuclear and quantum physics · HL only

🎯What you need to be able to do

  • Describe the photoelectric effect and explain why the wave model of light cannot account for it.
  • Use Einstein’s photoelectric equation, work function, threshold frequency and stopping voltage.
  • Use photon energy and photon momentum.
  • Use the de Broglie relation, and cite electron diffraction as evidence for matter waves.

📚The physics

The photoelectric effect is the emission of electrons when light falls on a metal surface. Four observations matter, and it is the ones that contradict wave theory that carry the marks.

  • Below a threshold frequency no electrons are emitted at all, however intense the light or however long you wait. Wave theory says energy should accumulate until emission occurs, so any frequency should eventually work.
  • Emission is instantaneous above the threshold, even for very dim light. Wave theory predicts a measurable delay while energy builds up.
  • The maximum kinetic energy of the electrons depends on frequency, not intensity.
  • Intensity changes only the number of electrons emitted per second.

Einstein’s resolution was that light arrives as discrete quanta of energy \(hf\), and that one photon interacts with one electron. If the photon carries less than the work function \(\phi\) — the minimum energy needed to free an electron from that metal — nothing happens, no matter how many such photons arrive. If it carries more, the surplus becomes kinetic energy:

\[ E_{\max} = hf - \phi \]

Every one of the four observations follows from that single sentence, which is why it won a Nobel Prize and why the exam keeps asking for it.

Stopping voltage is the experimental handle. Apply a reverse potential until the photocurrent just falls to zero; then \( eV_s = E_{\max} \). Plotting \(V_s\) against \(f\) gives a straight line of gradient \(h/e\) and intercept \(-\phi/e\) — a genuinely elegant way to measure Planck’s constant on a school bench.

Photons carry momentum as well as energy: \( p = h/\lambda = E/c \). That a massless particle has momentum is strange but well tested; it is the basis of radiation pressure and solar sails.

Matter waves. De Broglie proposed the symmetric idea — if waves behave like particles, particles should behave like waves — with

\[ \lambda = \frac{h}{p} \]

The wavelength is inversely proportional to momentum, which is why the effect is invisible for everyday objects: a walking person has a de Broglie wavelength around \( 10^{-35} \) m, far too small to matter.

Electron diffraction is the evidence. Fire electrons at a thin graphite film and they produce concentric diffraction rings on a screen — a pattern only a wave can make. Increase the accelerating voltage and the rings shrink, exactly as \( \lambda = h/p \) predicts, because faster electrons have shorter wavelengths. This is also why an electron microscope resolves far finer detail than an optical one.

✏️Worked example

Light of wavelength 380 nm falls on a caesium surface of work function 2.1 eV. Take \( h = 6.63 \times 10^{-34} \) J s, \( c = 3.00 \times 10^{8} \) m s\(^{-1}\), \( e = 1.60 \times 10^{-19} \) C.

(a) Photon energy. \( E = hc/\lambda = (6.63 \times 10^{-34} \times 3.00 \times 10^{8})/(380 \times 10^{-9}) = 5.23 \times 10^{-19} \) J \( = 3.27 \) eV.

(b) Maximum kinetic energy of the emitted electrons. \( E_{\max} = 3.27 - 2.1 = 1.17 \) eV, or \( 1.9 \times 10^{-19} \) J.

(c) Stopping voltage. \( V_s = E_{\max}/e = 1.2 \) V.

(d) Threshold wavelength for caesium. At threshold \( hf = \phi \), so \( \lambda_0 = hc/\phi = (6.63 \times 10^{-34} \times 3.00 \times 10^{8})/(2.1 \times 1.60 \times 10^{-19}) = 5.9 \times 10^{-7} \) m, about 590 nm — yellow. Anything redder than yellow will not eject electrons from caesium at all, and turning up the lamp will not change that.

The trap. Mixing electronvolts and joules mid-calculation. Convert once, at the start, and stay in one system.

🔭See it happen

PhET, Photoelectric Effect. Set the light to red and turn the intensity to maximum — nothing happens. Then switch to dim ultraviolet and electrons fly off immediately. That single comparison demolishes the wave model more convincingly than any paragraph, which is exactly what the 1905 argument was.

📝Practise

Work through these, then reveal the answer. Each question targets a different objective from the list above.

1. Find the energy of a photon of wavelength 500 nm, in joules and in electronvolts.
\( E = \dfrac{hc}{\lambda} = \dfrac{6.63 \times 10^{-34} \times 3.00 \times 10^{8}}{500 \times 10^{-9}} = 3.98 \times 10^{-19} \) J. Dividing by \( 1.60 \times 10^{-19} \) gives 2.5 eV.
2. Light of photon energy 3.1 eV falls on a metal of work function 2.3 eV. Find the maximum kinetic energy of the emitted electrons and the stopping voltage.
\( E_{\max} = hf - \phi = 3.1 - 2.3 = 0.8 \) eV. Since \( eV_s = E_{\max} \) and the energy is already in electronvolts, \( V_s = 0.80 \) V.
3. Find the threshold frequency for a metal of work function 2.3 eV.
At threshold \( hf_0 = \phi \), so \( f_0 = \dfrac{2.3 \times 1.60 \times 10^{-19}}{6.63 \times 10^{-34}} = 5.6 \times 10^{14} \) Hz. Below this no electrons are emitted, however intense the light.
4. Find the de Broglie wavelength of an electron travelling at \( 1.0 \times 10^{6} \) m s\(^{-1}\). Take \( m = 9.11 \times 10^{-31} \) kg.
\( \lambda = \dfrac{h}{mv} = \dfrac{6.63 \times 10^{-34}}{9.11 \times 10^{-31} \times 1.0 \times 10^{6}} = 7.3 \times 10^{-10} \) m — comparable to atomic spacing, which is why electrons diffract off crystals.
5. Find the momentum of a photon of wavelength 500 nm.
\( p = \dfrac{h}{\lambda} = \dfrac{6.63 \times 10^{-34}}{500 \times 10^{-9}} = 1.3 \times 10^{-27} \) kg m s\(^{-1}\). A massless particle can still carry momentum, which is the basis of radiation pressure and solar sails.
6. Explain why increasing the intensity of the light does not increase the maximum kinetic energy of photoelectrons, and say what it does change.
One photon interacts with one electron, and the photon’s energy \(hf\) depends only on frequency. Raising the intensity delivers more photons per second, not more energetic ones, so each ejected electron still receives the same \( hf - \phi \). What changes is the number of electrons emitted per second, and therefore the photocurrent. Wave theory predicted the opposite, which is precisely why the photoelectric effect killed it.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

  • HyperPhysics — photoelectric effect and de Broglie wavelengths
  • The Physics Hypertextbook — quantum physics