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: Understanding the Photoelectric Effect
The photoelectric effect refers to the phenomenon where electrons are emitted from a metal surface when light of a certain minimum frequency (or maximum wavelength) shines on it. This effect led to the quantum understanding of light as being made up of discrete packets called photons.

The key observations from the photoelectric effect are:

• The energy of the emitted electrons depends on the frequency \( \nu \) of the incident light, not its intensity.
• There exists a minimum frequency (called the threshold frequency \( \nu_0 \)) below which no electrons are emitted, regardless of the light's intensity.

 

The photoelectric equation is given by: \[ K_{\text{max}} = h\nu - \phi \] Where:

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

 

To measure \( K_{\text{max}} \), we apply a reverse (retarding) voltage called the stopping potential \( V_0 \), such that: \[ eV_0 = K_{\text{max}} = h\nu - \phi \] This stopping potential is the minimum voltage needed to stop the most energetic photoelectrons from reaching the anode. Importantly, it depends only on the frequency of the light and is independent of its intensity.

On the other hand, the saturation current is the maximum current achieved when all emitted photoelectrons are collected. Since more light intensity means more photons hitting the surface (assuming frequency is constant), it leads to the emission of more electrons, thereby increasing the saturation current. Hence:

• Saturation current increases with intensity.
• Stopping potential remains the same for a fixed frequency, regardless of intensity.

 

Step 2: Interpreting the Graphs
In typical photoelectric effect experiments:

• The x-axis represents the wavelength \( \lambda \) (or frequency \( \nu \)) of the incident light.
• The y-axis may represent either the stopping potential or the saturation current.

 

If two different light intensities \( I_1 \) and \( I_2 \) (with \( I_2 > I_1 \)) are used but with the same wavelength:

• The stopping potential will be the same for both, because it's determined by the energy (frequency) of individual photons.
• The saturation current will be higher for \( I_2 \), as more photons result in more emitted electrons.

 

Conclusion: The correct graph is the one where the stopping potential remains unchanged between \( I_1 \) and \( I_2 \), but the saturation current for \( I_2 \) is greater than for \( I_1 \). Thus, the correct option is (C).