Light of wavelength 400 nm illuminates a metal with work function 2.20 eV. Calculate photon energy, maximum photoelectron kinetic energy, stopping potential and maximum electron speed. Predict what changes when wavelength becomes 300 nm at the same photon rate, and outline how stopping potential should be measured without confusing it with the voltage that reduces current merely to a small nonzero value. Use h = 6.626 × 10⁻³⁴ J s and electron mass 9.11 × 10⁻³¹ kg.

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At 400 nm: photon energy = 3.10 eV, K_max = 0.900 eV, stopping potential = 0.900 V, and v_max = 5.63 × 10⁵ m s⁻¹. At 300 nm, K_max and stopping potential rise to about 1.93 eV and 1.93 V; saturation current is unchanged for equal photon rate and efficiency. Photon energy is hc/λ = (6.626 × 10⁻³⁴)(2.998 × 10⁸)/(400 × 10⁻⁹) = 4.966 × 10⁻¹⁹ J = 3.10 eV. Einstein's equation gives K_max = hf−Φ = 3.10−2.20 = 0.900 eV, so stopping potential is K_max/e = 0.900 V. Convert kinetic energy to joules and use ½mv²: v = √[2(0.900)(1.602 × 10⁻¹⁹)/(9.11 × 10⁻³¹)] = 5.63 × 10⁵ m s⁻¹. At 300 nm photon energy is 4.13 eV, so K_max ≈ 1.93 eV and stopping potential ≈ 1.93 V. Equal photon rate and unchanged quantum efficiency give the same saturation electron rate, not greater current merely because each photon is more energetic. Reverse the collector voltage and extrapolate the photocurrent curve to zero current, correcting dark current and contact potential. Calling an arbitrary small-current voltage the stopping potential introduces detector-sensitivity bias.

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