How Does Light Intensity and Frequency Affect Photoelectron Emission?

AI Thread Summary
Low-intensity light fails to release photoelectrons if its frequency is below the metal's threshold frequency, and increasing brightness will not change this. When low-intensity light emits photoelectrons, increasing its brightness raises the number of emitted photoelectrons due to increased intensity. If low-intensity light does not emit photoelectrons, gradually increasing the frequency will eventually lead to photoelectron emission once the threshold frequency is surpassed. The discussion clarifies that brightness relates to the number of photons, while frequency affects photon energy. Understanding these distinctions is crucial for accurately addressing the homework questions.
chef99
Messages
75
Reaction score
4

Homework Statement



Use Max Planck’s quantum theory to explain the following behavior of photoelectrons.

a) Low-intensity light does not release any photoelectrons. What will happen if the light is made brighter? Explain your reasoning

b) Low-intensity light releases photoelectrons. What will happen if the light is made brighter? Explain your reasoning.


c) Low-intensity light does not release any photoelectrons. What will happen if the frequency of the light is gradually increased? Explain your reasoning.2. Homework Equations
n/a

The Attempt at a Solution

a)[/B]
For a metal to emit photoelectrons, the frequency of the light must be above the threshold frequency of the metal. Since this light doesn’t emit any photoelectrons, its frequency must be below the metal’s threshold frequency. Therefore, making the light brighter will still not produce any photoelectrons from the metal.
b)
Since this low-intensity light emits photoelectrons, its frequency is above the threshold frequency. If the brightness is increased, the intensity is also increased, meaning the number of photoelectrons emitted will increase.c)
If the frequency is increased past the threshold frequency, then the light will begin to emit photoelectrons.I feel I might be mixing something up here, as my answers to b and c seem very similar. I don't know if they are supposed to. Any help here would be great.
 
Physics news on Phys.org
chef99 said:
b) Low-intensity light releases photoelectrons. What will happen if the light is made brighter? Explain your reasoning.
<snip>
Since this low-intensity light emits photoelectrons, its frequency is above the threshold frequency. If the brightness is increased, the intensity is also increased, meaning the number of photoelectrons emitted will increase.

Are you thinking that making the light brighter means increasing the number of photons or the energy of the photons?
 
Fewmet said:
Are you thinking that making the light brighter means increasing the number of photons or the energy of the photons?

Yes, but that is what I am confused about. Initially, I thought the brightness doesn't have an effect on frequency, which I know does have an effect on the energy of photons. Everything I have been able to find has said that increasing the brightness means increasing the intensity, which I believe would also increase the number of photons emitted. However, this is just what I've been able to research, as I initially didn't know the answer.
From what I understand, increasing frequency increasing the energy of the photons, and increasing the intensity increases the number of photons. Is that correct?
 
I think the textbook question is a little ambiguous, but I might be overlooking a convention I have not noticed before.

The text I usually work from discusses this in the context of Einstein's experiments, and that description specifies that he increased the intensity at a constant frequency. Unless your text's discussion indicates otherwise, the way the question is structured leaves me confident the intent is that the light is made brighter while keeping the frequency constant.

Does that resolve your question?
 
Fewmet said:
I think the textbook question is a little ambiguous, but I might be overlooking a convention I have not noticed before.

The text I usually work from discusses this in the context of Einstein's experiments, and that description specifies that he increased the intensity at a constant frequency. Unless your text's discussion indicates otherwise, the way the question is structured leaves me confident the intent is that the light is made brighter while keeping the frequency constant.

Does that resolve your question?

Yes, thank you for your help.
 
chef99 said:
I feel I might be mixing something up here, as my answers to b and c seem very similar. I don't know if they are supposed to. Any help here would be great.

No, you're not mixing things up. However, you can be a bit more explicit in your description for (c). You could say that as the frequency increases, at first there is no photoelectrons emitted until it reaches the threshold frequency. Beyond this value, photoelectrons are detected as frequency keeps on increasing.

You do not have to worry about intensity, etc. for this question since it doesn't say what happens to the intensity. All it cares about, and all you should care about, is to say what you can say with the given information. So you do not have to describe if you detect more or less photoelectrons. The only thing you can say is if you detect or do not detect photoelectrons.

Zz.
 
  • Like
Likes chef99
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top