An electron in the ground state of the hydrogen atom has the orbital radius of $5.3 \times 10^{-11} \mathrm{~m}$ while that for the electron in third excited state is $8.48 \times 10^{-10} \mathrm{~m}$. The ratio of the de Broglie wavelengths of electron in the ground state to that in the excited state is

UV light of $$4.13 \mathrm{~eV}$$ is incident on a photosensitive metal surface having work function $$3.13 \mathrm{~eV}$$. The maximum kinetic energy of ejected photoelectrons will be:

A proton, an electron and an alpha particle have the same energies. Their de-Broglie wavelengths will be compared as :

A proton and an electron have the same de Broglie wavelength. If $$\mathrm{K}_{\mathrm{p}}$$ and $$\mathrm{K}_{\mathrm{e}}$$ be the kinetic energies of proton and electron respectively, then choose the correct relation :