Photoelectric Effect: Intensity vs Frequency Dispute
A metal surface has a work function of 2.3 eV. In trial 1, light of frequency 6.0 × 10¹⁴ Hz shines on the metal and no photoelectrons are emitted. In trial 2, the student doubles the intensity of the same 6.0 × 10¹⁴ Hz light and still sees no electrons. In trial 3, the student switches to light of frequency 1.2 × 10¹⁵ Hz at the original intensity, and electrons are emitted. The student concludes that 'the first light just wasn't bright enough - more intensity should eventually knock electrons loose.' Use h = 4.14 × 10⁻¹⁵ eV·s and treat the metal as ideal.
1. Compute the photon energy in trials 1–2 and in trial 3, compare each to the work function, and determine the maximum kinetic energy of emitted electrons in trial 3.
Read the worked answer
Trials 1–2: E = hf = (4.14×10⁻¹⁵)(6.0×10¹⁴) = 2.484 eV ≈ 2.5 eV. Work function φ = 2.3 eV. Since 2.5 eV > 2.3 eV, wait - this is above threshold, so electrons should be emitted. Recheck: 4.14 × 6.0 = 24.84; ×10⁻¹⁵ × 10¹⁴ = 10⁻¹ → 2.484 eV. That exceeds 2.3 eV, so the premise that no electrons appear at 6.0×10¹⁴ Hz is inconsistent with an ideal metal - flag this: either the stated work function or frequency is off, or the setup is non-ideal. With the given numbers, threshold frequency is f₀ = φ/h = 2.3/(4.14×10⁻¹⁵) ≈ 5.6×10¹⁴ Hz, which is below 6.0×10¹⁴ Hz.
Trial 3: E = (4.14×10⁻¹⁵)(1.2×10¹⁵) = 4.968 eV ≈ 5.0 eV. K_max = E − φ = 4.968 − 2.3 = 2.668 eV ≈ 2.7 eV.
Note the inconsistency in trials 1–2 and state that with the given values electrons would be expected there too; the qualitative conclusion (trial 3 gives electrons with K_max ≈ 2.7 eV) holds.
2. Regardless of the numerical inconsistency, evaluate the student's claim that 'more intensity would eventually knock electrons loose.' Explain what intensity and frequency each control in the photoelectric effect.
Read the worked answer
The claim is false in principle. Photoelectric emission depends on single-photon energy, E = hf, compared with the work function φ. If hf < φ, no single photon carries enough energy to free an electron, and increasing intensity only increases the number of photons - each still individually below threshold. So no amount of intensity produces emission below the threshold frequency.
Frequency controls whether electrons are emitted and their maximum kinetic energy: K_max = hf − φ. Intensity controls how many electrons are emitted per second (the photocurrent), assuming f is above threshold. Doubling intensity at fixed frequency doubles the photon arrival rate and thus the saturation current, but leaves K_max unchanged.
This is the key evidence for the photon model: a classical wave picture would predict that brighter light (more energy delivered) should eventually eject electrons regardless of frequency, which is not observed.