G he work functions of gold, aluminum, and cesium are 5.1 eV, 4.1 eV, and 2.1 eV, respectively. If light of a particular frequency causes photoelectrons to be emitted when the light is incident on an aluminum surface, explain if we know whether this means that photoelectrons are emitted from a gold surface or a cesium surface when the light is incident on those surfaces.

Explanation:

Work function of aluminium  is  4.1 eV .

Therefor light of minimum energy of 4.1 eV will be required to cause the emission of electrons.

If we use this frequency of light or energy of light  to effect ejection of photoelectrons from cesium , it will be easily   ejected because energy requirement for this metal is lower ( work function is lower 2.1 eV ). than the energy of photon falling on it.

If we use this frequency of light or energy of light  to effect ejection of photoelectrons from aluminium  , it will be not  ejected because energy requirement for this metal is high ( work function is higher 5.1 eV ) . Even if we increase the intensity of light , electrons will not be ejected because energy of each photon depends upon the frequency .

i think its d if wrong im sorry