Light of wavelength 4000 angstrom is incident on a metal plate whose function is 2 eV. The maximum kinetic energy of emitted photoelectron will be
A: 12.01 eV
B: 21.22 eV
C: 23.12 eV
D: 10.37 eV

Hint: This is an example of photoelectric effect where photoelectrons are emitted due to electromagnetic radiation. When light hits a metallic surface, it ejects electrons and they are released with an energy that depends on the energy of incident radiation and work function.

Formulas used:
Work function $\varphi =hf-KE$
Where $hf$the incident energy of the light and KE is the kinetic energy of the ejected photoelectron.

Complete step by step answer:
We are given that the wavelength of the incident light is $4000{{\alpha }^{0}}$ and the value of the work function is 2eV.
We know that work function $\varphi =hf-KE$
We have to find the kinetic energy of the emitted photoelectron.
Hence, $KE=hf-\varphi $
We know that $f=\dfrac{c}{\lambda }$
Hence, $KE=\dfrac{hc}{\lambda }-\varphi =\dfrac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{4000\times {{10}^{-10}}\times 1.6\times {{10}^{-19}}}-2$ ($1.6\times {{10}^{-19}}$ is used to get the value in terms of eV)

$\Rightarrow KE=10.37eV$
Hence, the value of the kinetic energy of the ejected photoelectron is $10.37eV$
Thus, option D is the correct answer among the given options.

Note:Photoelectric effect is extremely important in the field of physics as it helps to understand and study the quantum nature of light as well as electrons. It has also made an impact in the formation of the concept of the duality of the wave and the particle. The photoelectric effect is widely used in order to study and investigate the energy levels of electrons in matter.