Q.Light of wavelength $400\ \text{nm}$ is incident on a metal surface of work function $2.0\ \text{eV}$. Calculate the maximum kinetic energy of the emitted photoelectrons (in eV) and the stopping potential for this surface.

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.

Plotting $K_{max}$ against $\nu$ gives a straight line whose slope is $h$ and whose $y$-intercept (extended back to zero frequency) is $-\phi_0$ -- exactly the graph R. A. Millikan produced experimentally for several metals, finding a common slope $h=6.626\times10^{-34}$ Js across every metal tested, giving the first precise experimental measurement of Planck's constant from the photoelectric effect itself. Einstein's equation resolves every one of the photoelectric effect's experimental laws at once: more photons per second (higher intensity) means proportionally more electrons ejected per second, since each absorbed photon liberates exactly one electron; $K_{max}=h\nu-\phi_0$ depends only on frequency, never on intensity, since intensity does not appear in the formula at all; a definite threshold frequency $\nu_0=\phi_0/h$ must exist below which no single photon carries enough energy to free even one electron, however many such photons arrive; and because each photon is absorbed and its energy transferred in one complete, instantaneous event rather than being slowly accumulated, there is no measurable time lag between illumination and emission.