Question

{"given":"Photon energy $= 6 \\text{ eV}$, Maximum kinetic energy of photoelectrons $KE_{\\max} = 4 \\text{ eV}$","key_observation":"The stopping potential $V_s$ is the minimum retarding potential needed to stop the most energetic photoelectrons. It satisfies $eV_s = KE_{\\max}$, so $V_s = KE_{\\max}/e$.","option_analysis":[{"label":"(A)","text":"2 V","verdict":"incorrect","explanation":"This would correspond to $KE_{\\max} = 2 \\text{ eV}$, but the maximum kinetic energy given is $4 \\text{ eV}$."},{"label":"(B)","text":"4 V","verdict":"correct","explanation":"Using $eV_s = KE_{\\max} = 4 \\text{ eV}$, the stopping potential is $V_s = 4 \\text{ V}$."},{"label":"(C)","text":"6 V","verdict":"incorrect","explanation":"This equals the incident photon energy, not the maximum kinetic energy of the photoelectrons."},{"label":"(D)","text":"10 V","verdict":"incorrect","explanation":"This is the sum of photon energy and kinetic energy, which has no physical significance here."}],"answer":"(B)","formula_steps":[]}