Which of the following best describes the threshold frequency ν_0 of cesium in the context of the photoelectric effect?

• A. It is the frequency at which cesium emits visible light.
• B. It is the frequency at which cesium atoms ionize in solution.
• C. It is the minimum frequency of incident light required to eject electrons from cesium.
• D. It is the maximum frequency at which cesium can absorb photons.

Answer: (C)

Verified step by step guidance

1. Understand the concept of threshold frequency (\( \nu_0 \)) in the photoelectric effect: it is the minimum frequency of incident light needed to provide enough energy to eject electrons from the surface of a metal.
2. Recall the photoelectric equation: the energy of the incident photon is given by \( E = h \nu \), where \( h \) is Planck's constant and \( \nu \) is the frequency of the light.
3. Recognize that the work function \( \phi \) of the metal (cesium in this case) is the minimum energy required to remove an electron from the metal surface, so \( \phi = h \nu_0 \).
4. Therefore, the threshold frequency \( \nu_0 \) corresponds to the frequency where the photon energy exactly equals the work function, meaning electrons just begin to be ejected with zero kinetic energy.
5. Conclude that the correct description of \( \nu_0 \) is the minimum frequency of incident light required to eject electrons from cesium, not related to emission of visible light, ionization in solution, or maximum absorption frequency.