Calculating Electron Emission and Reflection in the Photoelectric Effect

In summary, the conversation discusses how to calculate the number of electrons released by a light source, the percentage of photons absorbed, and the behavior of photons that do not eject electrons. The steps include finding the power of the light, using the voltage from a graph, and understanding how energy is conserved.
  • #1
Zeynaz
29
0
Homework Statement
The intensity of the light that falls onto the cathode is 6.0 W/m2. The area of the cathode of the photocell is 3.5 cm^2

b) Calculate the percentage of the photons incident on the cathode that release an electron.

c) what happens to the energy of the photons that do not release an electron
Relevant Equations
E=hf
V=IR
P=IV
I=n(q-electron)
241969


The full questions is in the picture. I already solved a) and found 5.6E14 electrons per second

For b) i first found the power of the light but just multiplying the intensity with the area: (6.0 W/m2)(3.5E-4 m^2) = 0.0021 W
Then I tried to use the voltage from the graph but i am not sure which value i should use because i don't know the voltage that corresponds to the intensity given.

So could you show me a way that can help me work it out?

After finding the number of electrons that the light releases i would calculate the percetage by dividing the value in question a) by the value i find and multiply it with 100.
(The correct answer should be 13% but i don't know how to get there)

For c) i said that because the energy of the photons are not enough to eject the electrons on the plate are a reflected back. (But I am not sure if that's a correct statement)
 
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  • #2
Zeynaz said:
i am not sure which value i should use because i don't know the voltage that corresponds to the intensity given.

I don't think they're related. The intensity is a property of the incoming light, and it's the same light source with the same intensity no matter what you're doing with the photocell, or even if you don't have a photocell, you just replace it with a rock.

That's just telling you how much energy is incident from the outside light source.

A photon releases an electron, by being absorbed. Since you know how many electrons are released (each second) then you know how many photons were absorbed each second. The question is asking what percentage that is, of the photons which were incident in that same second.
 
  • #3
Zeynaz said:
For b) i first found the power of the light but just multiplying the intensity with the area: (6.0 W/m2)(3.5E-4 m^2) = 0.0021 W
Then I tried to use the voltage from the graph but i am not sure which value i should use because i don't know the voltage that corresponds to the intensity given.

So could you show me a way that can help me work it out?
After multiplying with area of plate you found the power transferred to the plate. Now power is energy transferred per second. So of course you need to find the energy of each photon first.

For part c, the photons which don't eject electrons can be absorbed by the material of the plate to generate heat, or they can be reflected, scattered etc. Energy is conserved.
 
  • #4
Thanks! Found the correct answer
 

Related to Calculating Electron Emission and Reflection in the Photoelectric Effect

1. What is the photoelectric effect?

The photoelectric effect is the phenomenon where electrons are emitted from a material when it is exposed to light. This was first observed by Heinrich Hertz in 1887.

2. How does the photoelectric effect work?

The photoelectric effect occurs when photons from light strike the surface of a material, causing electrons to be emitted. The energy of the photons must be greater than the work function of the material in order for electrons to be emitted.

3. What is the significance of the photoelectric effect?

The photoelectric effect played a crucial role in the development of quantum mechanics and the understanding of the particle nature of light. It also has many practical applications, such as in solar cells and photodetectors.

4. What is the work function in the photoelectric effect?

The work function is the minimum amount of energy required to remove an electron from the surface of a material. It is different for each material and is dependent on factors such as the type of material and its surface conditions.

5. How does the intensity of light affect the photoelectric effect?

The intensity of light does not affect the photoelectric effect, as long as the energy of the photons is greater than the work function of the material. However, increasing the intensity of light will result in more electrons being emitted, but their kinetic energy will remain the same.

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