Photoelectric Effect IV Curves

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Discussion Overview

The discussion revolves around the characteristics of IV curves in the context of the photoelectric effect, specifically addressing why the saturated current begins at the y-axis and the implications of the gradient at the negative x-axis.

Discussion Character

  • Homework-related
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant questions why the saturated current starts at the y-axis (0,y) and seeks an explanation for the gradient at the negative x-axis.
  • Another participant reiterates the equation hf = 1/2 mv^2 + e(Vs) in their attempt to address the question.
  • A participant suggests that the increase in the rate of emitting electrons corresponds to the observed behavior of the current, but they express uncertainty about whether this is a correct interpretation.
  • Another participant agrees with the notion of increasing rates but emphasizes that it may be more accurate to describe it as the rate of collecting electrons, noting that this is an observational description rather than a definitive explanation.

Areas of Agreement / Disagreement

Participants express differing views on the interpretation of the current behavior in the IV curve, with no consensus reached on the underlying reasons for the observed characteristics.

Contextual Notes

There are limitations in the explanations provided, particularly regarding the assumptions made about the relationship between electron emission and collection rates, as well as the lack of detailed mathematical reasoning.

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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Attachments

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    Radiation IV Graph.JPG
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This part was missing
Knightycloud said:

The Attempt at a Solution


-
 
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.
 
Last edited:

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