Explain Einstein's photoelectric effect and derive Einstein's photoelectric equation.

(N/A) In $1905$,Einstein proposed a new theory to explain the photoelectric effect.
According to this theory,the emission and absorption of energy (called a photon) take place in discrete units. These units are called quanta of energy of radiation.
Each quantum (photon) has energy $E = h\nu$,where $\nu$ is the frequency of radiation. An electron on the surface will absorb this energy $h\nu$.
If the energy absorbed by an electron on the surface is greater than the minimum energy required (work function $\phi_{0}$),then the electron will be emitted with maximum kinetic energy $K_{\max}$.
Let the energy of the incident radiation be $h\nu$,the maximum kinetic energy of the electron be $K_{\max}$,and the work function of the metal be $\phi_{0}$. Then,according to the law of conservation of energy:
$h\nu = K_{\max} + \phi_{0}$
Therefore,$K_{\max} = h\nu - \phi_{0} \quad \dots (1)$
More tightly bound electrons will emerge with kinetic energy less than the maximum value. With an increase in the intensity of light,the number of electrons emitted per second increases. However,the maximum kinetic energy of the emitted electron is determined solely by the energy of the photon.
In Einstein's equation,the maximum kinetic energy is $K_{\max} = e V_{0}$,where $V_{0}$ is the stopping potential. Substituting this into equation $(1)$:
$e V_{0} = h\nu - \phi_{0} \quad \dots (2)$
Rearranging this gives $V_{0} = \left(\frac{h}{e}\right)\nu - \frac{\phi_{0}}{e}$.
The graph of stopping potential $V_{0}$ versus frequency $\nu$ is a straight line. The slope of the $V_{0}-\nu$ graph is $\frac{h}{e}$,which is a universal constant and does not depend on the type of material used.