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 $$=3 \times 10^8 \mathrm{~m} / \mathrm{s}]$$
In photoelectric experiment keeping the frequency of incident radiation and accelerating potential fixed, if the intensity of incident light is increased,
de-Broglie wavelength associated with an electron accelerated through a potential difference '$$\mathrm{V}$$' is '$$\lambda$$'. When the accelerating potential is increased to '$$4 \mathrm{~V}$$', de-Broglie wavelength.