In a photoelectric experiment, increasing the intensity of incident light:

In the photoelectric effect we illuminate a metal surface with light and observe that electrons are emitted. According to Einstein’s explanation, each electron absorbs one whole photon, gains the photon’s energy and then uses part of that energy to escape the metal. The precise energy balance is given by the photoelectric equation

$$h\nu = \phi + K_{\text{max}},$$

where $$h$$ is Planck’s constant, $$\nu$$ (Greek letter “nu”) is the frequency of the incident photon, $$\phi$$ is the work function of the metal (the minimum energy required to pull an electron out of the surface) and $$K_{\text{max}}$$ is the maximum kinetic energy of the emitted electrons.

From this formula we immediately see that $$K_{\text{max}}$$ depends only on $$\nu$$ and not on the intensity of the light. If we keep the frequency fixed and shine brighter light, we are merely sending more photons of exactly the same energy toward the metal. Each individual photon still carries the same energy $$h\nu$$, so every emitted electron still receives the same maximum kinetic energy $$K_{\text{max}}$$.

Intensity, in simple words, tells us how much energy comes per unit area per unit time. For monochromatic (single-frequency) light, intensity is proportional to the number of photons arriving per unit area per unit time. Mathematically,

$$I \propto N_{\text{photons}},$$

where $$I$$ is the intensity and $$N_{\text{photons}}$$ is the photon arrival rate. Thus increasing intensity, with frequency held constant, only increases $$N_{\text{photons}}$$; it does not alter $$\nu$$, and therefore it does not alter $$K_{\text{max}}$$.

Now we examine each option in the light of these facts:

A. “Increases the number of photons incident and also increases the K.E. of the ejected electrons.” This is partly correct (the number of photons does increase) but the statement about kinetic energy is wrong. So Option A is rejected.

B. “Increases the frequency of photons incident and increases the K.E. of the ejected electrons.” Intensity has nothing to do with frequency, so this option is wrong on both counts. Option B is rejected.

C. “Increases the number of photons incident and the K.E. of the ejected electrons remains unchanged.” This matches exactly what we deduced from the photoelectric equation and the definition of intensity. Option C is correct.

D. “Increases the frequency of photons incident and the K.E. of the ejected electrons remains unchanged.” Again, intensity does not affect frequency, so Option D is also wrong.

Hence, the correct answer is Option C.