Work function of sodium is $2.75eV$. What will be the KE of emitted electrons when a photon of energy $3.54eV$ is incident on the surface of sodium?

Hint: Remember Einstein’s photoelectric concept. Einstein theorized that when a photon hits the surface of a metal, the energy of the entire photon is transferred to the electron emitted from the metal’s surface.

Complete step by step solution:
The emission of a photoelectron is the result of the interaction between a single photon of the incident radiation and an electron in the metal.
As we know, the energy of the photon is transferred to the emitted electron, energy of the photon and emitted electron are the same.
From the Einstein’s photoelectric theory we know that, the energy of the electron,
$E = W + K.E$
Where, $W$ is the work function
$K.E$ is the kinetic energy.
From the above equation we get,
Kinetic energy of the emitted electron from the surface of sodium is given by,
Kinetic energy, \[K.E = E - W\] ………………………………. (1)

From the question it is known the values of the energy of the electron and the work function.
Energy of the electron =$3.54eV$
Work function of sodium = $2.75eV$.
Applying these values to the equation (1) we get,
Kinetic energy, \[K.E = 3.54eV - 2.75eV\]
\[ \Rightarrow K.E = 0.79eV\]

So the final answer is Kinetic energy of the emitted electron = \[0.79eV\].

Note: Under the right circumstances light can be used to push electrons, freeing them from the surface of a solid. This process is called the photoelectric effect. The velocity, thus, the kinetic energy and the energy of the liberated electron are dependent on each other. Einstein simply explained the photoelectric effect as follows: He assumed that the kinetic energy of the ejected electron was equal to the energy of the incident photon minus work function, which is defined as the amount of energy required to remove an electron from a material.