Photoelectric Effect Classical Breakdown

In summary: Actually I suppose it would be better to solve for ##\tau## directly, and then use the given work function to find the minimum time.
  • #1
baseballfan_ny
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23
Homework Statement
In the photoelectric effect, photoelectrons sometimes leave the surface at es-
sentially the instant that the light is turned on. This is in contrast to the classical picture, in
which it would take a certain amount of time for enough energy to have been accumulated
at the surface of the metal. Suppose that a very weak beam of light of wavelength ##\lambda## and power W were to land on a piece of metal of work function ##\phi##, starting at time t = 0. The first photoelectron is observed at a time t = ##\tau## . How small must ##\tau## be if the measurement is to be incompatible with the classical picture of light?
Relevant Equations
##KE_{max} = E_{incoming} - \phi##
##E_{incoming} = \frac {hc} {\lambda}##
##P = \frac {Work} {t} ##
So I'm kind of confused about how to interpret the question and the idea of there being a small enough ##\tau## for the classical picture to break down.

I started with the max KE eqn: ##KE_{max} = E_{incoming} - \phi##

I suppose ##E_{incoming}## is the power (W) times time and that is also equal to the Planck energy...
## E_{incoming} =W*\tau = \frac {hc} {\lambda}##

And then I just solved for ##\tau##.

##\tau = \frac {hc} {W\lambda}##.

Is this even close? I feel like I'm really missing something here.
 
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  • #2
You did seem to obtain an answer relevant to the variables given when equating:
## E_{incoming} =W*\tau## and ## E_{incoming} =\frac {hc} {\lambda}##

However, I fail to see where your use of the max KE eqn came in, as you said that it was your starting point. This might be due to a gap in my own knowledge, but I was wondering if you used it at all here?
 
  • #3
sethshoneman said:
You did seem to obtain an answer relevant to the variables given when equating:
## E_{incoming} =W*\tau## and ## E_{incoming} =\frac {hc} {\lambda}##

However, I fail to see where your use of the max KE eqn came in, as you said that it was your starting point. This might be due to a gap in my own knowledge, but I was wondering if you used it at all here?

Thanks for the reply. You're right I started with the KE equation and then totally forgot about it, so I probably shouldn't have written it in the first place. Is that equation relevant here? I'm not sure I'm using the right tools to find how small ##\tau## should be for classical breakdown.
 
  • #4
I would wait for a more experienced responder than me, honestly. All I can say is that if those are the relevant equations (given to you, or taught) that does seem to be a reasonable method. I apologize but the only advice I have for now is to possibly see if solving for KE would be relevant, and if so substituting another value of
##E_{incoming}## into the equation, and utilizing the given work function.

I wish I could be more precise, but my own knowledge is very limited. Good luck!
 
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  • #5
sethshoneman said:
I would wait for a more experienced responder than me, honestly. All I can say is that if those are the relevant equations (given to you, or taught) that does seem to be a reasonable method. I apologize but the only advice I have for now is to possibly see if solving for KE would be relevant, and if so substituting another value of
##E_{incoming}## into the equation, and utilizing the given work function.

I wish I could be more precise, but my own knowledge is very limited. Good luck!

No worries appreciate your response.

Edit: Still working on this problem so if anyone has any feedback please do let me know :) thanks.
 
  • #6
Think classically. The EM wave delivers energy continuously at a constant rate of energy units per time unit. If all that energy were absorbed by a single electron, how long must the electron "wait" until it has collected enough energy to leave the surface?

Similar problem: You are stuck in Nowheresville completely broke. You get a job flipping burgers to make enough money for a $125.00 bus ticket home. Your pay is $0.25 per minute. What is the minimum time that you must work to make enough money to leave Nowheresville?
 
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  • #7
kuruman said:
Think classically. The EM wave delivers energy continuously at a constant rate of energy units per time unit. If all that energy were absorbed by a single electron, how long must the electron "wait" until it has collected enough energy to leave the surface?

Similar problem: You are stuck in Nowheresville completely broke. You get a job flipping burgers to make enough money for a $125.00 bus ticket home. Your pay is $0.25 per minute. What is the minimum time that you must work to make enough money to leave Nowheresville?

Thanks for the response. I think this is starting to make sense now.

Starting with the analogous question you gave, the minimum time should be the money required divided by the rate of income, which gives 500 minutes.

I suppose in a similar manner, the minimum time in the classical picture would just be the minimum energy required divided by the rate of energy; any ##\tau## smaller than that would be incompatible with the classical picture of light.

So I suppose the minimum energy required would be just enough to eject an electron with 0 KE, that would be ##\frac {hc} {\lambda} = \phi##?

So then the minimum time in the classical picture is just this minimum energy divided by the power, ##\frac {hc} {\lambda*W} = \frac {\phi} {W}##

So it would just be ##\tau \lt \frac {hc} {\lambda*W} = \frac {\phi} {W}## for the classical picture to break down?

Edit: Actually I suppose it would be better to say ##\tau \lt \frac {\phi} {W}## rather than ##\tau \lt \frac {hc} {\lambda*W}## since we just need the minimum energy to eject an electron and we don't know whether or not ##\lambda## exceeds our threshold.
 
  • #8
baseballfan_ny said:
Edit: Actually I suppose it would be better to say ##\tau \lt \frac {\phi} {W}## rather than ##\tau \lt \frac {hc} {\lambda*W}## since we just need the minimum energy to eject an electron and we don't know whether or not exceeds our threshold.
I think you got it. The quantity ## \frac {hc} {\lambda*W}## could be greater than ##\frac{\phi}{W}##. Why work for more hours to buy a plane ticket when a bus ticket is enough to get you out of Nowheresville?
 
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  • #9
Awesome, thank you so much!
 
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Related to Photoelectric Effect Classical Breakdown

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 and was later explained by Albert Einstein in 1905.

2. What is the classical breakdown of the photoelectric effect?

The classical breakdown of the photoelectric effect refers to the classical theory of light, which states that light is a wave and energy is transferred continuously. According to this theory, increasing the intensity of light should increase the number of electrons emitted, but this is not observed in the photoelectric effect.

3. Why does the classical breakdown of the photoelectric effect occur?

The classical breakdown occurs because the classical theory of light does not take into account the particle nature of light. In reality, light is made up of discrete packets of energy called photons, and the photoelectric effect can only be explained by considering the particle nature of light.

4. How does the photoelectric effect support the particle nature of light?

The photoelectric effect supports the particle nature of light because it shows that the energy of light is transferred in discrete packets (photons). The energy of a photon is directly proportional to the frequency of the light, and this is observed in the photoelectric effect where increasing the frequency of light increases the energy of the emitted electrons.

5. What are the practical applications of the photoelectric effect?

The photoelectric effect has many practical applications, including solar cells, photodiodes, and photoelectric sensors. It is also the basis for the operation of devices such as photocopiers and digital cameras. Additionally, the photoelectric effect has been used in experiments to determine fundamental constants, such as Planck's constant.

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