A photon emitted during the de-excitation of electron from a state $$\mathrm{n}$$ to the second excited state in a hydrogen atom, irradiates a metallic electrode of work function $$0.5 \mathrm{~eV}$$, in a photocell, with a stopping voltage of $$0.47 \mathrm{~V}$$. Obtain the value of quantum number of the state '$$n$$'.

A particle of mass $$2 \mathrm{mg}$$ has the same wavelength as a neutron moving with a velocity of $$3 \times 10^5 \mathrm{~ms}^{-1}$$. The velocity of the particle is (mass of neutron is $$1.67 \times 10^{-27} \mathrm{Kg}$$)

The velocity of an electron so that its momentum is equal to that of a photon of wavelength $$660 \mathrm{~nm}$$ is

$$\mathrm{K}_1$$ and $$\mathrm{K}_2$$ are maximum kinetic energies of photoelectrons emitted when lights of wavelength $$\lambda_1$$ and $$\lambda_2$$ respectively are incident on a metallic surface. If $$\lambda_1=3 \lambda_2$$, then