PhysicsGrade 12–13

Photoelectric effect

Red light can be as bright as you like, it frees not a single electron. Short-wave light does it at once. A puzzle for waves, not for photons.

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What is it about?

Light falling on a metal can release electrons: the photoelectric effect. Below a threshold frequency no electrons are released, however bright the light. More intensity gives more electrons, but not faster ones. Einstein explained this with photons of energy EPh=h⋅fE_\text{Ph} = h \cdot f. One photon gives its energy to one electron, which needs the work function WAW_\text{A} to leave the metal:

Ekin,max=h⋅f−WA.E_\text{kin,max} = h \cdot f - W_\text{A}.

With a retarding voltage UU between anode and cathode the current vanishes at e⋅UG=Ekin,maxe \cdot U_\text{G} = E_\text{kin,max}. Plotting e⋅UGe \cdot U_\text{G} against ff gives Einstein’s line with slope h≈4.14⋅10−15 eV⋅sh \approx 4.14 \cdot 10^{-15}\,\text{eV}\cdot\text{s} and threshold frequency fG=WAhf_\text{G} = \frac{W_\text{A}}{h}.

Work functions (rounded textbook values): caesium 1.94 eV1.94\,\text{eV}, potassium 2.25 eV2.25\,\text{eV}, sodium 2.28 eV2.28\,\text{eV}, lithium 2.46 eV2.46\,\text{eV}, zinc 4.34 eV4.34\,\text{eV}, copper 4.48 eV4.48\,\text{eV}.

Try it

  1. Watch the electrons at U=0U = 0, then increase the retarding voltage. Load example
  2. Press “Adjust the voltage”. No current
  3. Double the intensity from 50 %50\,\% to 100 %100\,\%: the current doubles, UGU_\text{G} stays. 50 % · 100 %
  4. Try red light. Red light
  5. Record an automatic series and read off the slope.

Tasks

Task 1: Violet light on caesium

λ=400 nm\lambda = 400\,\text{nm}, WA=1.94 eVW_\text{A} = 1.94\,\text{eV}. Find EPhE_\text{Ph}, Ekin,maxE_\text{kin,max} and UGU_\text{G}. Load task

Show solution

EPh=1240 eV⋅nm400 nm≈3.10 eVE_\text{Ph} = \frac{1240\,\text{eV}\cdot\text{nm}}{400\,\text{nm}} \approx 3.10\,\text{eV}, Ekin,max=1.16 eVE_\text{kin,max} = 1.16\,\text{eV}, UG=1.16 VU_\text{G} = 1.16\,\text{V}. Show in the simulation

Task 2: Threshold frequency

Find fGf_\text{G} for caesium. Does red light (650 nm650\,\text{nm}) release electrons?

Show solution

fG=1.94 eV4.136⋅10−15 eV⋅s≈4.69⋅1014 Hzf_\text{G} = \frac{1.94\,\text{eV}}{4.136 \cdot 10^{-15}\,\text{eV}\cdot\text{s}} \approx 4.69 \cdot 10^{14}\,\text{Hz} (λG≈639 nm\lambda_\text{G} \approx 639\,\text{nm}). At 650 nm650\,\text{nm}, EPh≈1.91 eV<WAE_\text{Ph} \approx 1.91\,\text{eV} < W_\text{A}: no electrons. Show in the simulation

Task 3: Zinc

Why does 400 nm400\,\text{nm} light release no electrons from zinc (WA=4.34 eVW_\text{A} = 4.34\,\text{eV})?

Show solution

3.10 eV<4.34 eV3.10\,\text{eV} < 4.34\,\text{eV}; zinc needs λ<286 nm\lambda < 286\,\text{nm} (UV). Show in the simulation

Ideas for the classroom

  • Collect the three surprising observations first, then compare with the wave model.
  • Common stumbling blocks:
    • “Brighter light makes faster electrons.” – More light means more photons, not more energetic ones.
    • “Ekin=h⋅fE_\text{kin} = h \cdot f.” – The work function is missing.