Worked example: Light of wavelength 400 nm falls on a metal with work function φ = 2.0 eV. Do electrons emerge, and with what maximum kinetic energy? Step 1: Photon energy E = hc/λ = (1240 eV·nm) / 400 nm = 3.10 eV. Step 2: Compare with φ: 3.10 eV > 2.0 eV, so yes, electrons are ejected. Step 3: Kmax = hν − φ = 3.10 − 2.0 = 1.10 eV. Step 4: Check the classical puzzle: doubling the light intensity doubles the number of photons (more electrons) but each photon still has 3.10 eV, so Kmax stays 1.10 eV — exactly what experiments show. Conclusion: frequency, not intensity, decides electron energy — the photon picture in one calculation.