The work function of cesium metal is $2.14\,eV$ . When light of frequency $6 \times {10^{14}}\,Hz$ is incident on the metal surface, photoemission of electrons occurs. find the (i) energy of incident photons (ii) maximum kinetic energy of photoelectrons.

Hint: To find the energy of the incident photons, simple use the formula $E = h\upsilon $ since the frequency of the incident light rays is given in the question. To find the maximum kinetic energy of photoelectrons, find the difference between the energy of incident light rays and the work function of the given cesium metal.

Complete step by step answer:
We will approach the solution exactly as explained in the hint section of the solution to the asked question. First, we will find the energy of the incident photons and then, to find the maximum kinetic energy of photoelectrons, all we need to do is to find the difference of the energy of the incident light rays or photons and the work function of the metal surface.
First, let’s find out the energy of the incident photons or light rays:
The frequency of the incident light rays is given us to be:
$\upsilon = 6 \times {10^{14}}\,Hz$
To find the energy of the incident photons, we can apply the formula of energy of light rays:
$E = h\upsilon $
Where, $E$ is the energy of the photons
$h$ is the Planck’s constant with a value of $6.626 \times {10^{ - 34}}\,Js$
$\upsilon $ is the frequency of the photons
In the question, the frequency of the photons is given to be $\upsilon = 6 \times {10^{14}}\,Hz$
Substituting the value of frequency and Planck’s constant, we get:
$\implies$ $E = 6.626 \times {10^{ - 34}}\, \times 6 \times {10^{14}}\,J$
Upon solving, we get:
$\implies$ $E = 39.78 \times {10^{ - 20}}\,J$
But we need the energy to be in electron-volts and not in joules, so:
$
  E = \dfrac{{39.78 \times {{10}^{ - 20}}}}{{1.6 \times {{10}^{ - 19}}}}eV \\
  \implies E \approx 2.5\,eV \\
$
Hence, the energy of the incident photons is $E = 2.5\,eV$
Now, to find the maximum kinetic energy of the photoelectrons, we can easily do:
$K{E_{\max }} = E - {\phi _o}$
Where, $K{E_{\max }}$ is the maximum kinetic energy of the photoelectrons
${\phi _o}$ is the work function of the given metal
And, $E$ is the energy of the incident light rays or photons
The question has already given us the value of work function of cesium metal plate as: ${\phi _o} = 2.1\,eV$
We have already found out the value of the energy of incident light rays or photons as: $E = 2.5\,eV$
Substituting both, we get:
$
  \implies K{E_{\max }} = 2.5 - 2.1\,eV \\
 \implies K{E_{\max }} = 0.4\,eV \\
$
Note: The electrons acquire kinetic energy if the incident photon energy is greater than the work function of the metal, else if the incident energy is just equal to the work function of metal, electrons come out with zero speed or negligible energy.