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

A photon has wavelength $$3 \mathrm{~nm}$$, then its momentum and energy respectively will be $$[\mathrm{h}=6.63 \times 10^{-34} \mathrm{Js}, \mathrm{c}=$$ velocity of light $$=3 \times 10^8 \mathrm{~m} / \mathrm{s}]$$