When a certain metal surface is illuminated with light of frequency $$v$$, the stopping potential for photoelectric current is $$\mathrm{V}_0$$. When the same surface is illuminated by light of frequency $$\frac{v}{2}$$, the stopping potential is $$\frac{\mathrm{V}_0}{4}$$, the threshold frequency of photoelectric emission is

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