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 :

A proton and an electron are associated with same de-Broglie wavelength. The ratio of their kinetic energies is:

(Assume h = 6.63 $$\times 10^{-34} \mathrm{~J} \mathrm{~s}, \mathrm{~m}_{\mathrm{e}}=9.0 \times 10^{-31} \mathrm{~kg}$$ and $$\mathrm{m}_{\mathrm{p}}=1836$$ times $$\mathrm{m}_{\mathrm{e}}$$ )

When UV light of wavelength $$300 \mathrm{~nm}$$ is incident on the metal surface having work function $$2.13 \mathrm{~eV}$$, electron emission takes place. The stopping potential is :

(Given hc $$=1240 \mathrm{~eV} \mathrm{~nm}$$ )