The light intensity incident on a metallic surface with a work function of 3 eV produces photoelectrons with a maximum kinetic energy of 2 eV. The frequency of the light is doubled. Determine the maximum kinetic energy (in eV).

Did the proton move into a region of higher potential or lower potential? did the proton move into a region of higher potential or lower potential? because the proton is a negative charge and it accelerates as it travels, it must be moving from a region of higher potential to a region of lower potential.because the proton is a negative charge and it accelerates as it travels, it must be moving from a region of lower potential to a region of higher potential.because the proton is a positive charge and it slows down as it travels, it must be moving from a region of higher potential to a region of lower potential.because the proton is a positive charge and it slows down as it travels, it must be moving from a region of lower potential to a region of higher potential.request answerpart bwhat was the potential difference that stopped the proton? express your answer with the appropriate units.î”v î” v = nothingnothingrequest answerpart cwhat was the initial kinetic energy of the proton, in electron volts? express your answer in electron volts.ki k i = nothing ev request answerprovide feedback