Photoelectric effect refers to the emission of electrons from certain metal surfaces when irradiated with high frequency light (em radiation). (UV rays falling on cesium metal is a good example).
Number of electrons emitted is proportional to the intensity of light. KE of electron emitted depends on the frequency of incident light, it does not depend on intensity of light. The minimum frequency of light required to initiate photoelectric effect is called threshold frequency \( \left(v_{0}\right) \). The minimum energy required to eject one electron from a metal surface is called its work function and is given by \( \phi=h \mathrm{v}_{0} \). Maximum \( \mathrm{KE} \) of emitted electron when light of frequency \( v \) falls on the metal surface is given by \( \mathrm{KE}_{\max }=h v-\phi \) ( \( \phi \) is the work function) Negative potential can be applied to the electrons to stop them first outside the metal surface.
A particular metal can emit electrons when green light falls on it. In which of the following case, the maximum KE of emitted electrons cannot be more, compared to this case?
(A) Using blue light with the same metal
(B) Using a metal of less work function with the same light (green)
(C) Using red light with the first metal
(D) Using UV light with the second metal
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