In photoelectric effect, the work function of a metal is 3.5eV. The emitted electrons can be stopped by applying a potential of 1.2 V. Find energy of incident photons.

Hint: This question is based on the photoelectron effect which we will discuss first and then the stopping potential in the photoelectrons, then we will use Einstein's equation of photoelectric effect to get the final answer.

Formula used:  According to Einstein’s law on photoelectric effect

$E = {\omega _0} + K$

Where E is the energy of the ejected electron

${\omega _0}$ is the work function 

K is the maximum kinetic energy of the photon

Given, The work function 

${\omega _0} = 3.5eV$

The stopping potential of the electron    \[{V_0} =   1.2V\]

As we know the maximum kinetic energy of the emitted electron is given by 

$K =   e{V_0} =   1.2eV$

 According to Einstein’s law on photoelectric effect

$E = {\omega _0} + K$

Substituting the value of the work function and maximum kinetic energy, we get

Additional Information: According to the wave theory, energy is distributed uniformly across the wavefront and depends on the intensity of the beam only. That implies that electrons' kinetic energy increases with the intensity of light. But the kinetic energy was independent of the intensity of light. Because, according to the wave theory, energy is dependent on intensity, the low-intensity light should emit electrons after some time so that the electrons can acquire enough energy to emit. However, the emission of electrons was spontaneous no matter how small the light intensity.