Q.In photoelectric emission, a radiation whose frequency is 4 times the threshold frequency $\nu_0$ of a certain metal is incident on the metal. Then the maximum possible velocity of the emitted electron will be

Building directly on Planck's quantization idea, Einstein in 1905 proposed that light of frequency $\nu$ should itself be thought of as a stream of discrete energy quanta (photons), each carrying energy $h\nu$. Applying this to the photoelectric effect: when a single photon of energy $h\nu$ strikes a metal surface, it is completely absorbed by one single electron; part of this energy, equal to the metal's work function $\phi_0$, is spent freeing the electron from the surface, and the remainder becomes the ejected electron's kinetic energy. Conservation of energy for this one-photon-one-electron process gives Einstein's photoelectric equation, $h\nu=\phi_0+\tfrac12mv^2$; if the electron loses no energy internally on its way out, it emerges with the maximum possible kinetic energy, giving the most commonly used form $K_{max}=h\nu-\phi_0$. At the threshold frequency $\nu_0$, the photoelectron is emitted with essentially zero kinetic energy, so $h\nu_0=\phi_0$, meaning the threshold frequency is simply the photon frequency whose energy exactly equals the work function. …