Question

With respect to electromagnetic theory of light, the photoelectric effect is best explained by statement-
A. Light waves carry energy and when light is incident on the metallic surface, the energy absorbed by the metal may somehow concentrate on the individual electrons and reappear as their kinetic energy when ejected.
B. Particles of light (photons) collide energy when ejected the electrons take this energy and may eject.
C. When light waves fall on a metallic surface, the stability of atoms is disturbed and the electrons come out to make the system stable.
D. None of the above.

Answer 1

Energy can be converted from one form to the other. The ejected electron is associated with certain velocity.

Complete step by step answer:
In physics, electromagnetic radiation (EM radiation or EMR) refers to electromagnetic field waves (or their quanta, photons), propagating (radiating) across space, bearing electromagnetic radiant energy. Radio waves, microwaves, infrared, (visible) light, ultraviolet, X-rays, and gamma rays are included.
Classically, electromagnetic radiation consists of electromagnetic waves, which are electric and magnetic field oscillations that are coordinated. Electromagnetic waves propagate within a vacuum close to the speed.
When electromagnetic radiation, such as light, strikes an object, the photoelectric effect, resulting in the emission of electrons. In this way, electrons released are called photoelectrons. In condensed matter mechanics, and solid state and quantum chemistry, the phenomenon is studied to make inferences regarding the properties of atoms, gases, and solids.
In classical electromagnetic theory, the energy transfer from constant light waves to an electron is due to the photoelectric effect. Attempting to change the light intensity would affect the kinetic energy of the released electrons, and sufficiently dim light would prolong the emission by the time it would take the electrons to gain sufficient energy to escape the material. However, the experimental findings differ with all expectations. Instead, they prove that only when the light reaches a threshold frequency, electrons are expelled. It is now a simple implementation of an equation to find the kinetic energy of the emitted electron.
The equation which gives the estimation of kinetic energy of the expelled electron is:
$K{E_e} = hf - BE$
Where,
$h$ indicates Planck’s constant.
$f$ indicates frequency, and
$BE$ indicates binding energy.

So, the correct answer is “Option A”.

Note:
In this question, all the three options look similar in terms of photoelectric effect, but however, they have some differences. It is important to remember that energy can neither be destroyed nor be created, however it is converted from one form to the other. The sufficiently high intensity with which light beam hits the metal surface, ejects electrons from the surface. In meantime, it transfers its energy to the electron, which appears as kinetic energy.