Understanding the Photoelectric Effect and Current-Voltage Characteristics:
In the photoelectric effect, the photoelectric current \( I \) depends on the intensity of the incident light and the applied potential difference \( V \).
The key points to note are:
- For a given frequency (or wavelength), the stopping potential \( V_0 \) remains constant, as it depends only on the frequency of the incident light and not on its intensity.
- The saturation current (maximum current) is proportional to the intensity of the incident light. Therefore, higher intensity light (e.g., \( I_2 \)) results in a higher saturation current than lower intensity light (e.g., \( I_1 \)).
Selecting the Correct Graph:
Since the wavelength (or frequency) of the light is the same for both intensities, the stopping potential \( V_0 \) will be the same for both \( I_1 \) and \( I_2 \). However, since \( I_2 > I_1 \), the saturation current for \( I_2 \) will be greater than that for \( I_1 \).
Option (3) correctly shows:
- The stopping potential \( V_0 \) is the same for both intensities.
- The saturation current for \( I_2 \) is greater than for \( I_1 \), consistent with the higher intensity of \( I_2 \).
Conclusion:
Therefore, the correct graph is Option (3), as it accurately represents the photoelectric current variation with applied potential for two different light intensities of the same wavelength.
Given below are two statements: one is labelled as Assertion (A) and the other one is labelled as Reason (R).
Assertion (A): Emission of electrons in the photoelectric effect can be suppressed by applying a sufficiently negative electron potential to the photoemissive substance.
Reason (R): A negative electric potential, which stops the emission of electrons from the surface of a photoemissive substance, varies linearly with the frequency of incident radiation.
In light of the above statements, choose the most appropriate answer from the options given below: