Question

In a photoelectric experiment, keeping the frequency of incident radiation and accelerating potential fixed, if the intensity of incident light is increased, then the

• A. photoelectric current decreases
• B. kinetic energy of emitted photoelectrons decreases
• C. photoelectric current increases
• D. kinetic energy of emitted photoelectrons increases

Hint:

Always separate the effects of frequency and intensity in your mind: Frequency controls energy (how fast the individual electrons fly out), whereas Intensity controls quantity (how many electrons fly out per second). Since current is a measure of quantity per second, it tracks intensity directly.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The problem describes a photoelectric effect experiment where two specific parameters are held perfectly constant: the frequency of the incident light waves ($f$) and the internal accelerating potential ($V$).
We need to determine the physical consequences on the system when the intensity of the incident light beam is increased.

Step 2: Key Formula or Approach:
1. Light intensity is defined as the total energy falling per unit area per unit time. In quantum mechanics, this is directly proportional to the flux of incoming photons:
$$\text{Intensity} \propto \text{Number of photons incident per second}$$ 2. According to Einstein's photoelectric theory, the maximum kinetic energy ($K.E._{max}$) of the ejected photoelectrons depends strictly on the frequency of the light and the material's work function ($\phi$):
$$K.E._{max} = hf - \phi$$

Step 3: Detailed Explanation:
When the intensity of the light beam increases at a fixed frequency, the energy of each individual photon ($hf$) remains completely unchanged. As a result, the maximum kinetic energy of the ejected electrons does not change.
However, a higher intensity means a larger number of photons strike the metal surface per second.
Since the photon-electron collision mechanism is a one-to-one interaction, a greater number of incident photons causes a proportionally higher number of photoelectrons to be knocked out of the surface per second.
An increase in the rate of emitted charge carriers directly translates to a higher electric current flowing through the circuit ($I = \frac{\Delta q}{\Delta t}$). Therefore, the photoelectric current increases.

Step 4: Final Answer:
The photoelectric current increases, which perfectly corresponds to option (C).