Photoelectric effect calculation

In summary, the conversation discusses the photoelectric effect and calculating the frequency of monochromatic light by using the work function and maximum kinetic energy of an electron. The metal surface becomes charged to a steady positive potential of 1.0V relative to its surroundings and the equation hf=work function + max kinetic energy of an electron is used to find the frequency. The final answer is 9.69 x 10^14Hz.
  • #1
*now*
97
71
Photoelectric effect
1. Homework Statement

A metal surface is illuminated with monochromatic light and it becomes charged to a steady positive potential of 1.0V relative to its surroundings. The work function energy of the metal surface is 3.0eV, the electron charge is 1.6 x 10-19 C. Calculate the frequency of the light.

Homework Equations


hf=work function + max kinetic energy of an electron. E=hf

The Attempt at a Solution


wavespeed/wavelength = frequency. work function =3.0 eV electron charge = 1.6x10-19
 
Physics news on Phys.org
  • #2
Hi "now",

You haven't made an attempt at solution. What are your thoughts about how to approach the problem? What are the implications of "charged to a steady positive potential" if the metal surface continues to be illuminated? That is, what does it mean for electrons that absorb the light photons?
 
  • #3
Hi, I'm sorry, I'm having a lot of trouble understanding what is asked. We had our first lesson about the photoelectric effect today. Does 'charged to a steady positive potential' mean it isn't changing?
 
  • #4
*now* said:
Hi, I'm sorry, I'm having a lot of trouble understanding what is asked. We had our first lesson about the photoelectric effect today. Does 'charged to a steady positive potential' mean it isn't changing?
Yes. It rises to some value as electrons are "kicked" off of the surface by the photoelectric effect. Why do you think it reaches a steady value?
 
  • #5
gneill said:
Yes. It rises to some value as electrons are "kicked" off of the surface by the photoelectric effect. Why do you think it reaches a steady value?
It's monochromatic light so the value stays the same, or does it increase? Do you need to consider the speed of light?
 
  • #6
*now* said:
It's monochromatic light so the value stays the same, or does it increase? Do you need to consider the speed of light?
I'm not sure what you're thinking here. Perhaps you should try describing in your own words what happens, starting with the plate in an uncharged state and the light source is first directed at the metal surface. While you do this, think about the potentials and energies involved in the photoelectric effect.
 
  • #7
I think I needed to use the equation frequency is wave speed divided by wavelength thus 3.8x10^8 over 4.0 x10-7^-7 equalling 7.5 x10^14 Hz. This times planks constant is 4.95x10^-19 which is the answer for the amount of energy. My work has been marked but I want to check my thinking with you, does this seem right?
 
  • #8
*now* said:
I think I needed to use the equation frequency is wave speed divided by wavelength thus 3.8x10^8 over 4.0 x10-7^-7 equalling 7.5 x10^14 Hz. This times planks constant is 4.95x10^-19 which is the answer for the amount of energy. My work has been marked but I want to check my thinking with you, does this seem right?
You probably don't need the f = c/λ formula with the values that were given to you in the problem. But even so, where does your value of 4.0 x 10-7 come from? I don't see it as a given value in the problem statement.

I would look at the energy required for an electron to escape the plate when the plate's potential reaches its equilibrium state (1.0 V with respect to its surroundings).
 
  • #9
Sorry wrote wrong answer, had to get the energy by multiplying q and v to get 1.6x10^-19. Then I used the equation hf= work function + KEmax which goes to 4.8 x10^-19 (3eV) + 1.6 x 10^-19 which gives 6.4 x 10^-19. To find frequency we divide this by planks constant which gives 9.69 x 10^14Hz.
 

Related to Photoelectric effect calculation

1. What is the photoelectric effect?

The photoelectric effect is a phenomenon in which electrons are emitted from a material when it is exposed to electromagnetic radiation, such as light.

2. How is the photoelectric effect calculated?

The photoelectric effect is calculated using the equation E = hf - ΦE0, where E is the energy of the emitted electron, h is Planck's constant, f is the frequency of the incident light, and ΦE0 is the work function of the material.

3. 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 a characteristic property of the material and varies depending on the type of material.

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

The intensity of light does not directly affect the photoelectric effect. Instead, it is the frequency of the incident light that determines the energy of the emitted electrons. However, increasing the intensity of light can increase the number of photons that hit the material, which can result in more electrons being emitted.

5. What is the threshold frequency in the photoelectric effect?

The threshold frequency is the minimum frequency of light required to cause the photoelectric effect. Below this frequency, no electrons will be emitted regardless of the intensity of the light. Above this frequency, the number of emitted electrons will increase with increasing light intensity.

Similar threads

  • Introductory Physics Homework Help
2
Replies
35
Views
1K
  • Introductory Physics Homework Help
Replies
25
Views
2K
Replies
5
Views
686
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
19
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
3K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
824
Back
Top