Photoelectric Effect IV Curves

AI Thread Summary
In the discussion on photoelectric effect IV curves, the saturated current starts at the y-axis because it represents the maximum current when all emitted electrons are collected. The gradient at the negative x-axis indicates the rate of electron collection, which increases as the voltage approaches zero. This behavior reflects the relationship between photon energy and electron emission, as described by the equation hf = 1/2 mv^2 + e(Vs). The conversation emphasizes the distinction between the rate of emitting electrons and the rate of collecting them. Understanding these concepts is crucial for interpreting the IV characteristics of photoelectric materials.
Knightycloud
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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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Knightycloud said:

The Attempt at a Solution


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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?
 
Knightycloud said:
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.
 
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