Test a photoelectric threshold

Challenge an intensity-only explanation

Imagine two competing predictions. One says that sufficiently bright light ejects energetic electrons whatever its frequency. The other says each electron’s emission event depends on the energy of an absorbed photon, with intensity mainly changing how many opportunities occur. A useful comparison changes frequency and power separately.

The virtual surface is assigned work function φ=2 eV. This is a teaching value, not a claim about a particular metal preparation. At frequency 4×10¹⁴ Hz, photon energy is about 1.654 eV, below the removal cost. In the stipulated one-photon regime there is no emission even at the largest offered power. At 8×10¹⁴ Hz, photon energy is about 3.309 eV, leaving at most 1.309 eV kinetic energy.

Derive the threshold

Energy conservation for an electron escaping with maximum kinetic energy gives Kmax=hf−φ when hf exceeds φ. Set Kmax=0 to identify threshold frequency f₀=φ/h. Convert φ to joules before using h in joule seconds; the result is approximately 4.836×10¹⁴ Hz. A lower frequency cannot supply the required removal energy in one event.

The maximum matters: real emitted electrons can lose energy before leaving, so their kinetic energies need not all equal Kmax. At exact threshold, ideal limiting arguments predict zero excess kinetic energy; this model does not claim a measurable finite collected current there. It reports emission only strictly above threshold and with nonzero illumination.

Interpret stopping voltage

A retarding voltage opposes collection. In the ideal energy argument, an electron with charge magnitude e loses energy eV across the retarding potential. Thus eVstop=Kmax. A maximum kinetic energy of 1.309 eV corresponds numerically to 1.309 V stopping potential. The equal numerical values use different units and are not interchangeable physical quantities.

Limits of the comparison

The model omits surface contamination, contact potentials, electron-energy distributions and multiphoton processes at very high intensities. It also has no actual vacuum tube or detector. Use the calculation to distinguish predictions, not to plan a high-voltage or bright-beam experiment. Model failure outside the stated regime does not invalidate its successful bounded use.

Source and authorship

Reference for the physical relationship. Source checked 30 October 2026. The explanations, numerical cases, questions and diagrams here are original; reference material is not copied. Exact h,c,e follow NIST SI definitions.