An electron and proton are separated by a large distance. The electron starts approaching the proton with energy 3 eV. The proton captures the electron and forms a hydrogen atom in second excited state. The resulting photon is incident on a photosensitive metal of threshold wavelength 4000$$\mathop A\limits^o $$. What is the maximum kinetic energy of the emitted photoelectron?

An electron moving with speed v and a photon moving with speed c, have same D-Broglie wavelength. The ratio of kinetic energy of electron to that of photon is :

When radiation of wavelength $$\lambda$$ is incident on a metallic surface, the stopping potential of ejected photoelectrons is 4.8 V. If the same surface is illuminated by radiation of double the previous wavelength, then the stopping potential becomes 1.6 V. The threshold wavelength of the metal is :

What should be the order of arrangement of de-Broglie wavelength of electron ($$\lambda$$_e), an $$\alpha$$-particle ($$\lambda$$_a) and proton ($$\lambda$$_p) given that all have the same kinetic energy?