A metal surface is illuminated with two different frequencies. With slightly below the threshold frequency, no electrons are emitted even at very high intensity. With slightly above threshold, electrons are emitted immediately, and increasing intensity increases the number emitted per second. Which statement best accounts for the lack of emission at ?
- The intensity at is too low to heat the metal enough
- Photons at have insufficient energy to overcome the work function (correct answer)
- Electrons at need more time to accumulate wave energy
- The emitted electrons at are too slow to be detected
Explanation: The photoelectric effect. Each photon carries energy E = hf, and an electron can only be ejected if a single photon's energy exceeds the metal's work function. At frequency f₁ below threshold, each photon has insufficient energy to overcome the work function, making emission impossible regardless of how many photons strike the surface. At f₂ above threshold, each photon has enough energy to eject an electron immediately upon absorption. Choice C reflects the classical wave misconception that electrons could gradually accumulate energy from multiple photons or from the wave over time. The fundamental principle is that photoelectric emission is an all-or-nothing process determined by individual photon energy, not cumulative effects.