Question

Above the threshold frequency, if the intensity of incident light falling on a photo sensitive material is increased, then the correct statement is:

• A. The number of emitted electrons increases
• B. The maximum kinetic energy of electrons increases
• C. The stopping potential increases
• D. The number of electrons emitted decreases
• E. The stopping potential decreases

Hint:

In the photoelectric effect, intensity controls the number of emitted electrons, while frequency controls the maximum kinetic energy and stopping potential.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This question deals with the photoelectric effect. We need to understand how changing the intensity and frequency of incident light affects the emission of photoelectrons.
- Frequency (\(f\)): The energy of individual photons is determined by their frequency, according to \(E = hf\). The maximum kinetic energy of an emitted electron depends on the photon's energy and the work function (\(\phi\)) of the material: \(K_{max} = hf - \phi\). The stopping potential (\(V_s\)) is directly related to \(K_{max}\) by \(eV_s = K_{max}\). - Intensity (\(I\)): The intensity of light is the power per unit area. For a monochromatic light source, this is proportional to the number of photons arriving per unit area per unit time.
Step 2: Key Formula or Approach:
The key principles of the photoelectric effect are:
1. Intensity determines the rate of electron emission. A higher intensity means more photons are hitting the surface per second. Assuming a one-to-one interaction (one photon ejects one electron), more incident photons will lead to more emitted electrons per second. This results in a higher photoelectric current.
2. Frequency determines the maximum kinetic energy of electrons. The energy of each individual photon depends only on its frequency. Therefore, the maximum kinetic energy of the photoelectrons and the stopping potential depend only on the frequency of the incident light, not its intensity.
Step 3: Detailed Explanation:
The question states that the intensity of the incident light is increased, while the frequency is kept constant (and above the threshold).
- Since the intensity is increased, the number of photons striking the photosensitive material per unit time increases.
- As a result, the number of electrons emitted per unit time will also increase. This means the photoelectric current increases. So, statement (A) is correct and statement (D) is incorrect.
- Since the frequency of the light is not changed, the energy of each individual photon (\(E = hf\)) remains the same.
- According to Einstein's photoelectric equation, \(K_{max} = hf - \phi\), the maximum kinetic energy of the emitted electrons depends only on the frequency and the work function, not on the intensity. Therefore, \(K_{max}\) remains unchanged. Statement (B) is incorrect.
- The stopping potential is a measure of the maximum kinetic energy (\(V_s = K_{max}/e\)). Since \(K_{max}\) does not change, the stopping potential also remains unchanged. Statements (C) and (E) are incorrect.
Step 4: Final Answer:
Increasing the intensity of incident light increases the number of emitted electrons. This corresponds to option (A).