Question

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

• A. increases the number of photons incident and also increases the K.E. of the ejected electrons.
• B. increases the number of photons incident and the K.E. of the ejected electrons remains unchanged.
• C. increases the frequency of photons incident and increases the K.E. of the ejected electrons.
• D. increases the frequency of photons incident and the K.E. of the ejected electrons remains unchanged.

Hint:

Remember the two key relationships in the photoelectric effect:
- \textbf{Intensity} \(\leftrightarrow\) \textbf{Number of Photons} \(\leftrightarrow\) \textbf{Photoelectric Current}.
- \textbf{Frequency} \(\leftrightarrow\) \textbf{Energy of Photons} \(\leftrightarrow\) \textbf{Kinetic Energy of Electrons} (and Stopping Potential).
Intensity does not affect K.E., and frequency does not affect the number of photons (for a fixed intensity).

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
This question asks about the effect of changing the intensity of incident light on the photoelectric effect, specifically on the number of photons and the kinetic energy (K.E.) of the photoelectrons.
Step 2: Key Formula or Approach:
The key concepts are based on Einstein's explanation of the photoelectric effect.
1. Intensity and Photons: The intensity of light is defined as energy per unit area per unit time. In the quantum picture, this corresponds to the number of photons striking the surface per unit area per unit time. Therefore, increasing intensity means increasing the number of photons.
2. Einstein's Photoelectric Equation: The maximum kinetic energy of an ejected electron is given by \(K.E._{max} = h\nu - \phi_0\), where \(h\) is Planck's constant, \(\nu\) is the frequency of the incident light, and \(\phi_0\) is the work function of the metal surface.
Step 3: Detailed Explanation:
From the principles above:
- Increasing the intensity of light means that more photons are hitting the metal surface each second. Assuming a one-to-one interaction (one photon ejects one electron), an increase in the number of incident photons will lead to an increase in the number of ejected photoelectrons, which results in a larger photoelectric current.
- The photoelectric equation, \(K.E._{max} = h\nu - \phi_0\), shows that the maximum kinetic energy of the ejected electrons depends only on the frequency (\(\nu\)) of the incident light and the properties of the metal (work function \(\phi_0\)). It does not depend on the intensity of the light.
Therefore, when we increase the intensity of incident light (while keeping the frequency constant), the number of incident photons increases, but the maximum kinetic energy of the ejected electrons remains unchanged.
Step 4: Final Answer:
The correct statement is that increasing the intensity of incident light increases the number of photons incident and the K.E. of the ejected electrons remains unchanged.