The electric field of light wave is given as $$$\overrightarrow E = {10^{ - 3}}\cos \left( {{{2\pi x} \over {5 \times {{10}^{ - 7}}}} - 2\pi \times 6 \times {{10}^{14}}t} \right)\mathop x\limits^ \wedge {{\rm N} \over C}$$$ This light falls on a metal plate of work function 2eV. The stopping potential of the photoelectrons is :
Given, E (in eV) = 12375/$$\lambda $$(inÅ)

A nucleus A, with a finite de-broglie wavelength $$\lambda $$_A, undergoes spontaneous fission into two nuclei B and C of equal mass. B flies in the same direction as that of A, while C flies in the opposite direction with a velocity equal to half of that of B. The de-Broglie wavelengths $$\lambda $$_B and $$\lambda $$_C of B and C are respectively :

Two particles move at right angle to each other. Their de-Broglie wavelengths are $$\lambda _1$$ and $$\lambda _2$$ respectively. The particles suffer perfectly inelastic collision. The de-Broglie wavelength $$\lambda _2$$ of the final particle, is given by :

In a Frank-Hertz experiment, an electron of energy 5.6 eV passes through mercury vapour and emerges with an energy 0.7 eV. The minimum wavelength of photons emitted by mercury atoms is close to :