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Question:

The maximum velocity of the photoelectrons emitted from the surface is v when light of frequency n falls on a metal surface. If the incident frequency is increased to 3n, the maximum velocity of the ejected photoelectrons will be:

Equal to √3v

Equal to 2v

More than √3v

Less than √3v

Solution:

E=12mV20+ϕohn=12mV2+ϕo.. (1)h×3n=12mv21+ϕo (2)From equations 1 and 2.3×12mv2+3ϕo=12mv21+ϕov21=3v2+4ϕomv21>3v2v1>√3v