Write Einstein’s photoelectric equation.

Hint To answer this question, we need to consider the photoelectric effect and we need to define it properly. Then, after writing the required equation which is given as ${K_{\max }} = h\nu - {\varphi _0}$, we need to explain each of the terms and also the significance of this equation.

Complete step by step answer
Einstein's photoelectric equation basically represents the photoelectric effect. When the light is incident on the surface of a metal, then the electrons present at the outermost layer of the metal receive the energy from the photons of the incident light, and hence they become photoelectrons. This equation is written as
${K_{\max }} = h\nu - {\varphi _0}$ (i)
Here, ${K_{\max }}$ is the maximum kinetic energy of the photoelectrons emitted, $\nu $ is the frequency of the incident light, and ${\varphi _0}$ is the work function of the metal used.
This equation basically tells us that if the incident light has a minimum frequency so that it has energy greater than or equal to the work function of the metal, then the electrons are liberated from the metal surface. The part of the energy of the incident radiation, which is equal to the work function of the metal, is utilized by the electrons in getting free from the metallic bond. The rest of the part, if any, appears as the kinetic energy of the photoelectrons.
It must be noted that all the photoelectrons do not have the maximum kinetic energy. Only the electrons present at the outermost layer of the metal possess this maximum value.
Now, the maximum kinetic energy of the photoelectrons is written in the form of the stopping potential as
${K_{\max }} = e{V_0}$ (ii)
From (i) and (ii) we get
$e{V_0} = h\nu - {\varphi _0}$
Dividing by $e$ we get
${V_0} = \dfrac{h}{e}\nu - \dfrac{{{\varphi _0}}}{e}$
So, from the above equation we can say that the stopping potential is directly proportional to the frequency of the incident radiation. This is an important conclusion of the photoelectric equation.

Note
The photoelectric effect, which is discussed in this question, has many applications in different fields. The examples include the Photostats, which are used to generate the copies of a printed paper. Also, it is used in the photovoltaic cells, which are used in the solar panels for converting the solar energy of the sun into the electricity.