Let K1 and K2 be the maximum kinetic energies of photo-electrons emitted when two monochromatic beams of wavelength $$\lambda$$1 and $$\lambda$$2, respectively are incident on a metallic surface. If $$\lambda$$1 = 3$$\lambda$$2 then :
The de Broglie wavelengths for an electron and a photon are $$\lambda$$e and $$\lambda$$p respectively. For the same kinetic energy of electron and photon, which of the following presents the correct relation between the de Broglie wavelengths of two ?
An $$\alpha$$ particle and a carbon 12 atom has same kinetic energy K. The ratio of their de-Broglie wavelengths $$({\lambda _\alpha }:{\lambda _{C12}})$$ is :
A metal surface is illuminated by a radiation of wavelength 4500 $$\mathop A\limits^o $$. The ejected photo-electron enters a constant magnetic field of 2 mT making an angle of 90$$^\circ$$ with the magnetic field. If it starts revolving in a circular path of radius 2 mm, the work function of the metal is approximately :