The work function of a metal is $\mathrm{X} \mathrm{eV} .$ When light of energy $2 \mathrm{X}$ is made incident on it then the maximum kinetic energy of emitted photoelectron will be
(A) $2.5 \mathrm{X} \mathrm{eV}$
(B) $2 \mathrm{X} \mathrm{eV}$
(C) $\mathrm{X} \mathrm{eV}$
(D) $3 \mathrm{X} \mathrm{eV}$

Hint: We know that the photoelectric effect works like this. If you shine light of high enough energy on to a metal, electrons will be emitted from the metal. Light below a certain threshold frequency, no matter how intense, will not cause any electrons to be emitted. Electrons can gain energy by interacting with photons. Study of the photoelectric effect led to important steps in understanding the quantum nature of light and electrons and influenced the formation of the concept of wave-particle duality. The photoelectric effect is also widely used to investigate electron energy levels in matter.

Complete step by step answer
We know that in physics, the kinetic energy (KE) of an object is the energy that it possesses due to its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes. Kinetic energy, a form of energy that an object or a particle has by reason of its motion. If work, which transfers energy, is done on an object by applying a net force, the object speeds up and thereby gains kinetic energy.
Each particle of light, called a photon, collides with an electron and uses some of its energy to dislodge the electron. The rest of the photon's energy transfers to the free negative charge, called a photoelectron. Understanding how this works revolutionized modern physics.

The maximum kinetic energy of emitted photoelectron is given as $\mathrm{E}_{\max }=\mathrm{h} v-\phi$
where, $v$ is the frequency of incident light and $\phi$ is photoelectric work function of metal.
$\mathrm{E}_{\max }=2 \mathrm{X}-\mathrm{X}$
$\mathrm{E}_{\max }=\mathrm{X} \mathrm{eV}$

So, the answer is option (C).

Note: We can conclude that the rest of the photon's energy transfers to the free negative charge, called a photoelectron. Understanding how this works revolutionized modern physics. Applications of the photoelectric effect brought us electric eye door openers, light meters used in photography, solar panels and photostatic copying. The photoelectric effect happens when light strikes a metal surface causing the emission of electrons from it (photoelectrons). On the other hand, if we increase the frequency the number of packets remains the same (emitting fewer electrons perhaps) but the energy carried by each of them increases. The most important conclusion of photoelectric effect is that it shows that light radiation has particle nature. That is light radiation is not continuous but discrete and photons are the carriers of light.