Help: Photoelectric effect problem

In summary: KE=eV...? and if KE is zero... that makes the vmax zero too?No, the kinetic energy of the electrons is not zero when there is no voltage applied. KE=eV does not always mean that the kinetic energy is equal to the energy of the electron. In this case, the kinetic energy is still greater than zero because the electrons are moving.
  • #1
kas8
7
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In a photoelectric cell, the stopping voltage is 2.00V. If the voltage applied across this cell is zero, what is the maximum speed of the electrons from the photoelectric surface?

Confused.
 
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  • #2
First of all, tell us what you understand as the definition of "stopping voltage", and its significance in the photoelectric effect experiment.

Zz.
 
  • #3
well i know that its used to increase the intensity of the negative plate to repel the electrons... the stopping voltage is the point at which the current stops because of that increased negative intensity...
but i don't understand why there's a stopping voltage AND zero voltage applied... ??
 
  • #4
kas8 said:
well i know that its used to increase the intensity of the negative plate to repel the electrons... the stopping voltage is the point at which the current stops because of that increased negative intensity...
but i don't understand why there's a stopping voltage AND zero voltage applied... ??

Er... now it is my turn to not understand what you wrote. You do know that when you apply the stopping voltage, there IS an applied field that is being used to slow down and stop the photoelectrons from reaching the anode? The voltage whereby the most energetic electrons are just stopped (i.e. where the current first reaches zero) is the definition of the stopping potential.

Now, see whether this jives with your understanding of what the stopping potential is, and see if you can figure out that if it takes this much potential to stop that electron, what was its kinetic energy that it started out with?

Zz.
 
  • #5
ya i understand that part...
so, if the applied voltage is zero, doesn't that make the kinetic energy zero? since KE=eV...? and if KE is zero... that makes the vmax zero too?
 
  • #6
im trying to understand what the question is asking me for? the KE when the Vstop is 2.00? or when the voltage is zero?
 
  • #7
kas8 said:
In a photoelectric cell, the stopping voltage is 2.00V. If the voltage applied across this cell is zero, what is the maximum speed of the electrons from the photoelectric surface?

The fact that the stopping voltage is 2.00 V does not mean that there is an applied voltage at all. It means that if there was one, it should provide 2.00 V to stop the electrons before reaching the anode.
 
  • #8
so what does the question mean when it asks "if the applied voltage is zero...?"
 
  • #9
Wouldn't really like to tell this one directly.

Imagine your photocell which is in a circuit and tell me what objects are usually in a similar photocell circuit and why .
 
  • #10
so i should figure out what the ke is if the Vstop is 2.00V...then u use that information to solve for vmax?
 
  • #11
If you know the [itex] U_{stop} [/itex] you can figure out the [itex] E_{kin} [/itex] from that you know [itex] v_{max} [/itex].

However, I recommend you to understand/revise the theory, too.
 
  • #12
im still not sure what voltage being zero has to do with this question?
i came up with 2.65x10^7 m/s
theory in physics is NOT my forte! :S
appreciate the help, big time!
 
  • #13
The proper English isn't mine...

The incoming photons give their energy to the cathode. If these energy quantums are big enough ([itex] \phi ≤ [/itex] ) the cathode emit electrons which might (and usually) have kinetic energy. If you apply a voltage to this circuit you can speed up or slow down the emitted electrons.

When you slow them down there will be a voltage which will not let the electrons to reach the anode. So at the end (when the electrons almost reached the anode) the electrons' kinetic energy will be zero. This voltage is the stopping voltage.

So by knowing the stopping voltage you know the kinetic energy of the electrons which is constant if there isn't any applied voltage (and if λ is constant too).
 
  • #14
kas8 said:
ya i understand that part...
so, if the applied voltage is zero, doesn't that make the kinetic energy zero? since KE=eV...? and if KE is zero... that makes the vmax zero too?

OK, I think we have narrowed down a fundamental error in your understanding of the photoelectric effect.

Most of the photoelectrons that come out of the metal already have an initial kinetic energy!. So they are already moving!

This is why you apply a reverse voltage to stop them.

Zz.
 

Related to Help: Photoelectric effect problem

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 of a certain frequency or higher. This was first observed by Heinrich Hertz in 1887 and explained by Albert Einstein in 1905.

2. How does the photoelectric effect work?

When a photon of light with enough energy strikes a material, it can knock an electron out of its orbit. This process is called photoemission. The energy of the photon must be equal to or greater than the work function of the material, which is the minimum energy required to remove an electron. The remaining energy of the photon is transferred to the electron, giving it kinetic energy.

3. What is the equation for the photoelectric effect?

The equation for the photoelectric effect is E = hf - φ, where E is the energy of the photon, h is Planck's constant, f is the frequency of the light, and φ is the work function of the material.

4. What is the significance of the photoelectric effect?

The photoelectric effect is significant because it provided evidence for the particle nature of light and helped to develop the concept of photons. It also led to advancements in technology such as solar cells, photomultiplier tubes, and digital cameras.

5. What factors affect the photoelectric effect?

The photoelectric effect is affected by the intensity and frequency of the light, as well as the type of material and its work function. It is also influenced by the angle of incidence of the light and the surface area of the material exposed to the light.

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