Question

The incident photon involved in the photoelectric effect experiment (\[\nu > {\nu _0}\])
A. Completely disappears
B. Comes out with increased frequency
C. Comes out with decreased frequency
D. Comes out with a change frequency

Answer 1

Hint: The energy of a photon of light is exactly proportional to its frequency. When the frequency of light is increased, the energy of the photon rises as well.

Formula used:
\[E = h\nu \]
where E is the energy of the photon of light with frequency \[\nu \] and $h$ is called the Plank’s constant.

Complete step by step solution:
Light is an electromagnetic wave which has an electric field and magnetic field component vibrating mutually perpendicular to the propagation of light. Due to the presence of an electric field and magnetic field component in the electromagnetic wave, it contains energy in the form of an electric field and a magnetic field.

The basic unit of energy of the electromagnetic wave is called a photon. The energy content of a photon is directly proportional to the frequency of the electromagnetic wave.
\[E = h\nu \]
When the photon is incident on the surface of the metal, then there are two cases of photoelectric effect shown by the metal surface,

Case-1: If the energy of the photon incident is less than the work function of the metal then the electron from the atomic orbital of the metal is not able to overcome the attractive energy which holds the electron in the orbit. In this case, the incident photon comes out with the same energy with which it was incident, i.e. there is no change in frequency of the photon.

Case-2: If the energy of the photon incident is larger than the work function of the metal then the electron from the atomic orbital of the metal is able to overcome the attractive energy which holds the electron in the orbit. In this case the incident photon transfers all of the energy to the electron. Some part of the energy transferred is used to balance the work function and the rest of the energy is converted to the kinetic energy of the ejected electron. And hence, the photon disappears. This happens when the frequency of the photon is greater than the threshold frequency of the metal.

As it is given that the frequency of the incident photon is larger than the threshold frequency, hence the photon disappears completely.

Therefore, the correct option is A.

Note: During the collision between the photon and the electron, energy is transferred from the photon to the electron. The greatest kinetic energy of the expelled electron is equal to the difference between the photon's energy and the metal's work function.