The "photoelectrons" emitted as a result of the photoelectric effect have a certain kinetic energy, which can be measured. This kinetic energy (for each photoelectron) is independent of the intensity of the light, but depends linearly on the frequency; and if the frequency is too low (corresponding to a photon energy that is less than the work function of the material), no photoelectrons are emitted at all, unless a plurality of photons, whose energetic sum is greater than the energy of the photoelectrons, acts virtually simultaneously (multiphoton effect)  Assuming the frequency is high enough to cause the photoelectric effect, a rise in intensity of the light source causes more photoelectrons to be emitted with the same kinetic energy, rather than the same number of photoelectrons to be emitted with higher kinetic energy.
What does the kinetic energy of a photoelectron lose? (If the question is unanswerable, say "unanswerable")
unanswerable