Photoelectric Effect: The Impact of Light Intensity on Photocell Response

Click For Summary
SUMMARY

The discussion centers on the photoelectric effect, specifically analyzing the impact of light intensity on photocell response when using blue and red light of equal intensity. Participants concluded that while both lights have the same power per area, red light, having a higher photon count, results in a higher current. The conversation highlighted confusion regarding the definition of intensity, with some interpreting it as the number of photons per second rather than power per area. Ultimately, the consensus is that the question posed was misleading due to its phrasing.

PREREQUISITES
  • Understanding of the photoelectric effect
  • Familiarity with the equation E = nhf
  • Knowledge of intensity as power per area
  • Basic principles of quantum efficiency
NEXT STEPS
  • Research the relationship between photon energy and frequency in the context of the photoelectric effect
  • Explore the effects of different materials on photocell response
  • Study the concept of quantum efficiency in detail
  • Investigate the implications of light intensity definitions in experimental physics
USEFUL FOR

Students studying physics, educators teaching the photoelectric effect, and researchers exploring light-matter interactions in photocells.

Biker
Messages
416
Reaction score
52

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?
 
  • Like
Likes   Reactions: physics123
Physics news on Phys.org
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.
 
  • Like
Likes   Reactions: Biker
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.
 

Similar threads

  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 35 ·
2
Replies
35
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 10 ·
Replies
10
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
3
Views
2K
  • · Replies 25 ·
Replies
25
Views
4K
  • · Replies 25 ·
Replies
25
Views
4K