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See according to Einstein equation KE(Max)= Energy of photons - work function of surface, but since work function is not given i have assumed it to be 0. And energy of photons is equal to hc/lamda and this energy is equal to KE(Max) or the stopping potential so using these equations and i am...

Yes i tried but i am unable to solve it.

When a cm thick surface is illuminated with light of wavelength lambda,the stopping potential is V. When the same surface is illuminated by light of wavelength 2lambda , the stopping potential is V/3.The threshold wavelength for the surface is ? Homework EquationsThe Attempt at a Solution