Relationship between stopping potential and the intensity of light

[tnecniv]

Homework Statement

The higher the intensity of the light we will get a higher stopping potential.
Why is that so? is there any theory and formula to justifiy this behavior?

How is wavelength affecting the intensity of light?

Homework Equations

not sure ,and would like to know.

The Attempt at a Solution

tried several transmission density and it proved that the higher intensity in light the higher stopping potential% Transmission => Stopping V
100% =>0.48
66 =>0.45
45 =>0.44Is there a formula or theory behind this behavior?

 

[Jhero]

Yes, there is a formula that explains the relationship between intensity of light and stopping potential. This is known as the photoelectric effect equation, which is given by:

KE = hν - Φ

where KE is the kinetic energy of the emitted electron, h is Planck's constant, ν is the frequency of the incident light, and Φ is the work function of the material.

From this equation, we can see that the kinetic energy of the emitted electron is directly proportional to the frequency of the incident light. Therefore, as the intensity of light increases, the frequency of the incident light also increases, resulting in a higher stopping potential.

The wavelength of light also affects the intensity of light because intensity is defined as the amount of energy per unit area per unit time. This means that as the wavelength of light decreases, the frequency and energy of the light increases, resulting in a higher intensity. Therefore, both intensity and wavelength play a role in determining the stopping potential.