Question

A Photon of energy $3.4\,eV$ is incident on a metal having work function $2\,eV$. The maximum kinetic energy of photoelectron is equal to
A. $1.4\,eV$
B. $1.17\,eV$
C. $5.4\,eV$
D. $6.8\,eV$

Answer 1

We know that in photoelectric effect the energy of the photon is converted into the work function and the rest of the energy is given as the kinetic energy of the electrons. Hence if we subtract the work function from the energy of the photon, we can get the value of maximum kinetic energy.

Complete step by step answer:
We know that in photoelectric effect the energy of the photon is converted into the work function and the rest of the energy is given as the kinetic energy of the electrons. Hence if we subtract the work function from the energy of the photon, we can get the value of maximum kinetic energy.
Step by step solution:
Photoelectric effect is the emission of electrons from the surface of a metal when energy in the form of photons is given to the metal. The electrons will be emitted only if the incoming radiation has the minimum energy to break the electron from the metal surface. This energy is called the threshold energy.
We know that the energy of the photon is give an as
$E = h\upsilon$
Where $h$ is the Planck’s constant and $\upsilon$ is the frequency of the incident radiation.
So, for a fixed frequency the energy carried by each photon is fixed.
This energy is converted into the energy required to break the electron from the surface which is known as the work function and the remaining energy is converted as a kinetic energy of the electron.
We know that the photoelectric equation is given as
$hv = \dfrac{1}{2}m{v^2} + \phi$
Where $hv$ is the energy of a photon, $\dfrac{1}{2}m{v^2}$ is the kinetic energy of an emitted electron and $\phi$ is the work function.
From this we can get the kinetic as
$\Rightarrow hv - \phi = \dfrac{1}{2}m{v^2}$ -------------(1)
Since energy of the incident photon and the value of the work function is given, we can subtract the work function from the photon energy to get the maximum value of kinetic energy.
Let us substitute the given values in equation 1. Then we get
$\Rightarrow 3.4\,eV - 2\,eV = \dfrac{1}{2}m{v^2}$
$\Rightarrow \dfrac{1}{2}m{v^2} = 1.4\,eV$
This is the value of maximum kinetic energy of the emitted photoelectrons.

Hence, the correct answer is option A.

Note:
All the electrons emitted from the surface will have this maximum kinetic energy. But the electrons that are away from the surface will have a kinetic energy less than this maximum value because some energy is used for bringing those electrons to the surface. For the affixed frequency of the incident photon, the maximum of kinetic energy is fixed. If we use a radiation of higher frequency then the kinetic energy can be increased.