Question

How does one explain the emission of electrons from a photosensitive surface with the help of Einstein's photoelectric equation?

Answer 1

Hint: Use the concept of photoelectric effect. Photoelectric emission is the emission of electrons when electromagnetic radiation such as light hits a material. According to wave theory, ejection of electrons can take place at any frequency of light waves but electron emission occurred only for frequencies greater than threshold frequencies.

Complete step by step answer:
We know when a photon of sufficient energy is taken with $E = h\nu $ strikes a metal surface; some of the energy is used to liberate the electron. This energy is called the work function. It is denoted by ${\phi _o}$. The surplus energy propels the electron, giving it kinetic energy. If there is no loss in energy in any other form, the kinetic energy is known as maximum kinetic energy $K{E_{\max }}$ .
Thus we can say that,
$h\nu = {\phi _o} + K{E_{\max }}$
If we go on decreasing the value of $\nu $, the value of $K{E_{\max }}$ decreases. There will be a point when $K{E_{\max }} = 0$. Thus,
${\phi _o} = h{\nu _o}$
Where ${\nu _o}$ is the minimum frequency of light for the ejection of the light, known as threshold frequency. Putting this value in the above equation we get that,
$h\nu = h{\nu _o} + K{E_{\max }}$
$ \Rightarrow h\nu - h{\nu _o} = K{E_{\max }}$
$ \Rightarrow h(\nu - {\nu _o}) = K{E_{\max }}$
$ \Rightarrow h(\nu - {\nu _o}) = K{E_{\max }} = \dfrac{1}{2}m{v_{\max }}^2$
This is Einstein’s equation of photoelectric effect.

Note: The Einstein’s equation of photoelectric effect explains the dual nature of light and justifies the presence of light as matter. Remember the equations correctly. Also know the meaning of the terms involved. The photocurrent depends on the intensity of light incident on the metal surface. The intensity increases and so does the photocurrent.