Photoelectric effect (2 Viewers)

[howcanibesmarter]

Is it correct in explaining that as frequency increases, stopping voltage also increases through the following method. Kmax=eVstopping, and Kmax=hf-phi, therefore hf-phi=eVstopping. Since h,phi are all constants, if we increase frequency, Vstopping must increase? The question was asking me to explain why stopping voltage increases as frequency increases.

Another thing, if we increase frequency, will photocurrent increase? I know it increases with intensity, but I'm not too sure with frequency.

 

[carrotsss]

Yes to the first question and no to the second.

Current is effectively a measure of how many electrons are travelling through a space per second. If you increase the light intensity, there are more photons to collide with electrons, so more electrons are being released, and so the number of electrons travelling through a space per second increases as well.

Meanwhile, if you increase frequency, you’re just increasing the energy of each photon, so each electron will have more excess energy after ionisation, which is converted into kinetic energy, but more kinetic energy in each electron just means that they travel faster, it doesn’t actually increase the number of electrons being released.

 

[howcanibesmarter]

Wait then how would you explain this graph? If increasing frequency does not increase photocurrent...

They've made the assumption that intensity is constant, and so I managed to come up with an answer for why it decreases at some point. Here's what I thought.

Since keeping intensity is essentially power/area, if you increase frequency then by E=hf, the energy of a single photon increases. Thus, for the same intensity, as energy of every photons increase, there will be less number of photons. Hence, photocurrent decreases. But idk how to explain the first part lol

 

[carrotsss]

Since keeping intensity is essentially power/area, if you increase frequency then by E=hf, the energy of a single photon increases. Thus, for the same intensity, as energy of every photons increase, there will be less number of photons. Hence, photocurrent decreases.

Iirc intensity of light (at least within the scope of the physics syllabus) isn’t power/area, it’s just the number of photons being released. For lasers the number of photons stays the same no matter the frequency, it’s not like black bodies.

Do you know where that graph is from? It kinda contradicts every single hsc textbook so idk if it’s beyond the syllabus scope or something..

 

[howcanibesmarter]

Iirc intensity of light (at least within the scope of the physics syllabus) isn’t power/area, it’s just the number of photons being released. For lasers the number of photons stays the same no matter the frequency, it’s not like black bodies.

Do you know where that graph is from? It kinda contradicts every single hsc textbook so idk if it’s beyond the syllabus scope or something..

Graph was a question from my friends tutoring lol. Hm maybe I'll just disregard that then. But I'm pretty sure intensity is power/area tho...

 

[wizzkids]

That graph is reasonably consistent with what HSC students know (or should know) about the photoelectric effect.
Also take into account that these are experimental data points, and the smooth line drawn through the points is not necessarily correct.
I have annotated the graph with red lines to highlight the important features.
It shows that there is a cut-off frequency below which there is no photocurrent.
The photocurrent rises rapidly after the energy of the incident photons exceeds the cut-off frequency, and then it plateaus at a constant value.
I wouldn't get too hung up about the slight drop off as frequency increases.

 

[howcanibesmarter]

That graph is reasonably consistent with what HSC students know (or should know) about the photoelectric effect.
Also take into account that these are experimental data points, and the smooth line drawn through the points is not necessarily correct.
I have annotated the graph with red lines to highlight the important features.
It shows that there is a cut-off frequency below which there is no photocurrent.
The photocurrent rises rapidly after the energy of the incident photons exceeds the cut-off frequency, and then it plateaus at a constant value.
I wouldn't get too hung up about the slight drop off as frequency increases.

View attachment 38843

could you perhaps explain why photocurrent increases with increase to frequency? Carrots said it doenst increase so im kinda a bit confused as to who to listen to

 

[wizzkids]

The photocurrent increases suddenly. The photocurrent is a step-wise function of frequency.

 

[howcanibesmarter]

Whats a step-wise function? Like what if a hsc question was like explain why frequency increases with photocurrent...surely that isn't what u're supposed to give right?

@carrotsss (u both hv contradiciting things so thats why i tagged u)

 

[wizzkids]

A step function in mathematics is any function whose graph consists of a series of discontinuous steps, like steps on a ladder (Google it).
In the example we are considering, there is a range of frequencies for which the photocurrent is essentially zero, then a range of frequencies for which the photocurrent is a large positive number. In the transition region, there is a sharp rise. It is too complicated to explain why all the photoelectrons are not released at the same value of "f" but essentially it is due to the range of potential energy values of the conduction band electrons in the metal target. It is the conduction band electrons that are going to be ejected by the incident photons (the valence shell electrons are too strongly bound to be ejected).

 

[howcanibesmarter]

more confused than ever ‍
@carrotsss pls help

 

[carrotsss]

more confused than ever ‍
@carrotsss pls help

don’t worry too much about it unless you’re interested in it, unless I’m heavily mistaken that level of stuff isn’t in the syllabus

 

[howcanibesmarter]

don’t worry too much about it unless you’re interested in it, unless I’m heavily mistaken that level of stuff isn’t in the syllabus

so as much as the hsc is concerned, frequency has no effect on photocurrent?

 

[carrotsss]

so as much as the hsc is concerned, frequency has no effect on photocurrent?

Yep, other than the threshold