From a metallic surface photoelectric emission is observed for frequencies $$v_1$$ and $$v_2\left(v_1 > v_2\right)$$ of the incident light. The maximum values of the kinetic energy of the photoelectrons emitted in the two cases are in the ratio $$1: \mathrm{x}$$. Hence the threshold frequency of the metallic surface is

If the kinetic energy of a free electron doubles, it's de Broglie wavelength ($$\lambda$$) changes by a factor

A light of wavelength '$$\lambda$$' and intensity '$$\mathrm{I}$$' falls on photosensitive material. If '$$\mathrm{N}$$' photo electrons are emitted, each with kinetic energy 'E', then

In a photoelectric experiment, a graph of maximum kinetic energy $$(\mathrm{KE}_{\text {max }})$$ against the frequency of incident radiation (v) is plotted. If $$\mathrm{A}$$ and $$\mathrm{B}$$ are the intercepts on the $$\mathrm{X}$$ and $$\mathrm{Y}$$ axis respectively then the Planck's constant is given by