Photoelectric Effect: The Impact of Light Intensity on Photocell Response

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Homework Help Overview

The discussion revolves around the photoelectric effect, specifically examining the impact of light intensity on the response of a photocell when exposed to blue and red light of equal intensity. Participants explore the implications of intensity as power per area versus the number of photons reaching the cell.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between light intensity and the number of photons, questioning whether the term "intensity" refers to power per area or photon count. There is a consideration of how this affects the photocell's current response to different light frequencies.

Discussion Status

There is an ongoing exploration of the question's phrasing and its implications. Some participants express confusion regarding the definition of intensity in the context of the problem, while others note the variability of photocell response based on material composition.

Contextual Notes

Participants reference textbook definitions and personal interpretations of intensity, indicating a potential discrepancy between educational materials and the problem statement. The discussion highlights the complexity of the photoelectric effect and the factors influencing photocell performance.

Biker
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Homework Statement


If blue and red light are used on a photoelectric cell and they both have the same intensity (Power per area). Choose the correct option from the following
UEqSI8P.png


Homework Equations


E = nhf
Intensity = power/area
Quantum effeciency = 100%

The Attempt at a Solution


if they have the same intensity as in same power per area then the number of photons reaching the cell differs. Red has more photons than blue light does. Thus red should have higher current.

a to d from left to right.
The correct answer was a, Was the question phrased wrong and they meant by intensity the number of photons reaching the plate per sec? Or did they neglect the small difference between the two currents?
 
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Biker said:
The correct answer was a, Was the question phrased wrong and they meant by intensity the number of photons reaching the plate per sec? Or did they neglect the small difference between the two currents?
In my textbook when it speaks about photoelectric effect, and it mentions "same intensity" I am supposed to think of intensity as the number of photons reaching the surface every second.
 
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Asmaa Mohammad said:
In my textbook when it speaks about photoelectric effect, and it mentions "same intensity" I am supposed to think of intensity as the number of photons reaching the surface every second.
Yea I do that too, But in this question they insisted that it is power per area. I guess this question is just weird.

Thanks for help
 
Biker said:
Yea I do that too, But in this question they insisted that it is power per area. I guess this question is just weird.

Thanks for help
You are welcome!
 
Asmaa Mohammad said:
In my textbook when it speaks about photoelectric effect, and it mentions "same intensity" I am supposed to think of intensity as the number of photons reaching the surface every second.
Blue photons have more energy per photon than red ones (=hf). So I would not go with that view.

Also, the question IS weird since photocell reponse to frequency varies greatly with the componds used to fabricate the photocell.
 

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