0
$\begingroup$

I'm an engineering graduate revisiting my text on semiconductor physics, and I've hit a snag.

My book claims that demonstrations of the photoelectric effect clearly show that "...light energy is contained in discrete units rather than a continuous distribution of energies."

Now as I read this with only the vaguest understanding that classical mechanics discusses "state transition" in detail.

Is this a good model to have developed from the claims that I am reviewing?

Basically the book is saying this cannot be the case: Continuous case

Because this is observed in all experiments: Discrete case

Please help, I'm not sure how to understand the structure being described here.

$\endgroup$
0

2 Answers 2

1
$\begingroup$

I'm a bit confused as to your question. Are you asking what those diagrams mean?

The energy oh a photon is proportional to its frequency. So increasing the frequency of the light is another way of saying you are increasing the energy of the photons.

The first diagram says that if you increase the energy of the photons continuously, you get a continuously increasing output current. In other words, no minimum step size. They are continuously proportional to each other.

The second diagram says they aren't. It says that to observe a flow of charge, the energy of the photons must exceed some threshold dependent on the sample material. In other words, there is a minimum photon energy required to produce eject electrons and produce current.

But I personally find the diagrams a bit lacking in that they do not seem to be comparing like-with-like since the first describes an charge response to a range of energies whereas the second diagram only describes the zero response and minimal non-zero response.

I also don't see how the second diagram necessarily implies that photon energies are discretized rather than electrons transition levels being discrete. It seems to me you only need either photons or electron transition states to be discrete to get a discrete response.

I thought that the concept for photons having discrete energies came from the math Planck had to do to working with blackbody radiation to solve the ultraviolet catastrope. Because integrating the photon energy levels across the blackbody radiation gave you infinities if it was continuous.

$\endgroup$
6
  • $\begingroup$ Those are my diagrams to demonstrate the best understanding I can muster. I found an enthusiastic high school professor which at least highlighted to me that my threshold assertion is correct, but that the drift rate/ current of the returned electrons would correlate solely with frequency $\endgroup$
    – user306661
    Commented Jul 10, 2021 at 1:41
  • $\begingroup$ @meltyness Oh, you made those up? They aren't from your notes? $\endgroup$
    – DKNguyen
    Commented Jul 10, 2021 at 1:42
  • $\begingroup$ Yes, those are my interpretations of the descriptions I'm reading. It's possible it's a suspension of rigor thing here. I'm sure I didn't read into this carefully the first time through. $\endgroup$
    – user306661
    Commented Jul 10, 2021 at 1:43
  • $\begingroup$ @meltyness Now you have me wondering how we know electron transition levels are discrete. The stuff about Planck solving the ultraviolet catastrope makes sense and stands on its own about discrete photon energies. But it seems you need another way (i.e. not photons) to prove electron transition levels are discrete $\endgroup$
    – DKNguyen
    Commented Jul 10, 2021 at 1:46
  • $\begingroup$ I'm sure discrete shells are a result of spectra $\endgroup$
    – user306661
    Commented Jul 10, 2021 at 1:49
0
$\begingroup$

Yes, a two level system state you (me) described is indeed quantized.

The distinction between:

  • no electrons,
  • and electrons with E sub m / K sub max

is exactly quantized. And is exactly strangeness the early modern physicists identified.

...and if you look at the history, Planck described the extrinsic behavior of systems with highly mobile down to immobile temperatures, wheras Einstein appears to have invented the transistor to demonstrate the same effects, simplifying the model.

$\endgroup$
1
  • $\begingroup$ It seems that this was break through thinking, but now it seems obvious because of all the other such effects I'm already aware of. Maybe if I had been steeped in bad hypotheses this would be more bewildering, but I think I got this burned into my brain now. $\endgroup$
    – user306661
    Commented Jul 10, 2021 at 11:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.