To determine the correct graph for the photo current \((i)\) vs stopping potential \((V_s)\), given the same frequency but different intensities \((I_1 > I_2)\), we need to apply the concepts of the photoelectric effect.
The photoelectric effect is described by the equation:
\(h\nu = eV_s + \phi\)
Where:
\(h\) is Planck's constant.
\(\nu\) is the frequency of the incident light.
\(e\) is the charge of an electron.
\(V_s\) is the stopping potential.
\(\phi\) is the work function of the material.
Key Concepts:
The stopping potential \((V_s)\) depends only on the frequency \((\nu)\) of the incident light and not on its intensity.
The intensity of the light affects the photo current \((i)\). Higher intensity means more photons hitting the material, producing more photoelectrons and hence a higher current.
Therefore, when the frequency is constant, but the intensity changes, the stopping potential remains the same; however, the photo current \((i)\) increases with increased intensity.
Let's evaluate the options:
Option 1: Incorrect as it shows variation of stopping potential with intensity, which is not correct.
Option 2: Correct. It shows higher photo current for higher intensity at the same stopping potential.
Option 3: Incorrect, as the shape of the graph does not change for increased intensity according to the photoelectric theory, except for the current level.
Option 4: Incorrect, as it wrongly shows a dependency of stopping potential based on intensity.
Conclusion: The correct graph is the second option, which displays the increase in photo current \((i)\) with increased intensity while keeping the stopping potential \((V_s)\) constant.
