Question

Which figure shows the correct variation of applied potential difference (\(V\)) with photoelectric current (\(I\)) at two different intensities of light (\(I_1<I_2\)) of same wavelengths:

• A.
• B.
• C.
• D.

Hint:

In the photoelectric effect, the stopping potential is independent of intensity and depends only on the frequency of the incident light. The saturation current increases with intensity.

Answer 1

The Correct Option is C

Step 1: Photoelectric Effect Fundamentals
The photoelectric effect describes the emission of electrons from a metal surface upon illumination with light of a specific minimum frequency (or maximum wavelength). This phenomenon established the quantum nature of light, conceptualized as discrete energy packets known as photons.

Key observations of the photoelectric effect include:

• Electron emission energy is dictated by incident light frequency \( u \), irrespective of its intensity.
• A threshold frequency \( u_0 \) exists; below this frequency, no electron emission occurs, irrespective of light intensity.

The photoelectric equation is: \[ K_{\text{max}} = hu - \phi \] Where:

• \( K_{\text{max}} \) represents the maximum kinetic energy of emitted electrons.
• \( h \) is Planck’s constant.
• \( u \) is the incident light frequency.
• \( \phi \) is the metal's work function.

 

To determine \( K_{\text{max}} \), a reverse (retarding) voltage, termed the stopping potential \( V_0 \), is applied, satisfying: \[ eV_0 = K_{\text{max}} = hu - \phi \] This stopping potential is the minimal voltage required to prevent the most energetic photoelectrons from reaching the anode. Crucially, it is dependent solely on light frequency and independent of light intensity.

Conversely, saturation current denotes the maximal current achieved when all emitted photoelectrons are collected. Increased light intensity (at constant frequency) corresponds to more photons striking the surface, leading to greater electron emission and thus a higher saturation current. Therefore:

• Saturation current increases with light intensity.
• Stopping potential remains constant for a fixed frequency, irrespective of intensity.

 

Step 2: Graph Interpretation
In typical photoelectric effect experiments:

• The x-axis typically denotes the wavelength \( \lambda \) or frequency \( u \) of incident light.
• The y-axis can represent either the stopping potential or the saturation current.

 

When comparing two light intensities \( I_1 \) and \( I_2 \) (\( I_2 > I_1 \)) at the same wavelength:

• The stopping potential will be identical for both intensities, as it is determined by photon energy (frequency).
• The saturation current will be higher for \( I_2 \) due to the increased number of emitted electrons resulting from more incident photons.

 

Conclusion: The accurate graph will depict an unchanging stopping potential between \( I_1 \) and \( I_2 \), while showing a greater saturation current for \( I_2 \) compared to \( I_1 \). Accordingly, the correct selection is (C).