Photoelectric Effect IV Curves

[Knightycloud]

Homework Statement

In photoelectric effect, why the saturated current starts at the y-axis (0,y)? and what is the reason for that gradient at the negative x-axis (Shown as a red line)?

Homework Equations

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The Attempt at a Solution

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[epenguin]

This part was missing

The Attempt at a Solution

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[Knightycloud]

2. Homework Equations
hf = 1/2 mv^2 + e(Vs)

3. The Attempt at a Solution
Because the rate of emitting electrons is increasing from zero to that value.

Am I right?

 

[epenguin]

2. Homework Equations
hf = 1/2 mv^2 + e(Vs)

3. The Attempt at a Solution
Because the rate of emitting electrons is increasing from zero to that value.

Am I right?

Well, it's more the rate of collecting electrons I'd say, but that's only a description of what's observed and what the graph says, not an explanation of why it's like that.

 

[DeeNos]

The reason for the saturated current starting at the y-axis (0,y) in the IV curves for the photoelectric effect is due to the threshold frequency of the incident light. The threshold frequency is the minimum frequency required for a photon to have enough energy to eject an electron from the metal surface. Below this frequency, the electrons do not have enough energy to overcome the binding energy of the metal and therefore no current is produced. As the frequency of the incident light increases, the energy of the photons also increases, and more electrons are able to be ejected, resulting in an increase in current.

The gradient at the negative x-axis (shown as a red line) is due to the stopping potential, which is the minimum potential required to stop the current from flowing. As the potential difference increases, the electrons are able to overcome the opposing electric field and reach the anode, resulting in an increase in current. However, once the stopping potential is reached, any further increase in potential difference does not result in an increase in current as all the electrons have already been ejected. This results in a plateau in the IV curve, creating the sharp gradient at the negative x-axis.