0
$\begingroup$

I'm fairly noobish over here, but delving into the details of the major experiments. The particle nature of a photon has me stumped though. So hopefully there is already an explanation to this I have missed that someone can link too.

I can't figure out why light is considered to have a particle nature: We use the photoelectric effect to kick off a tiny amount of energy (at some point the amplitude isn't enough to excite an electron when we hit infrared) and a photo multiplier tube to detect that tiny amount of energy on the other end. We are measuring, though, that an electron has been excited and produced a tiny current that is then multiplied. Why do we then say that that electron was excited by a particle? As opposed to a peak of a tiny wave from one or more electron(s) being excited on the emission side? How do we know it comes from a point and ends at at point; assuming that this is the fundamental essence of 'particle' nature?

| cite | improve this question | |
$\endgroup$
  • 1
    $\begingroup$ It looks as if you're basically asking what photons are, in which case see my answer to Do photons truly exist in a physical sense or are they just a useful concept like $i=\sqrt{-1}$. $\endgroup$ – John Rennie Mar 27 '18 at 7:37
  • $\begingroup$ Great explanation! And pretty much the answer I was expecting at the end :P.. thanks.. why is no one bothered by the notion of a 'particle' nature to light if there is no real evidence? $\endgroup$ – T.Donaldson Mar 27 '18 at 7:50
  • $\begingroup$ There is abundant experimental evidence that photons exist. The point of my answer is that they are stranger objects than most people imagine. $\endgroup$ – John Rennie Mar 27 '18 at 7:53
  • $\begingroup$ Would you be so kind as to point me in the direction of some of that evidence? :) $\endgroup$ – T.Donaldson Mar 27 '18 at 7:59
  • $\begingroup$ The statement, "at some point the amplitude isn't enough to excite an electron when we hit infrared" is filled with misconceptions. Straighten out those misconceptions, and you'll have a much clearer understanding of what the word "photon" refers to. 1) the amplitude of the light is nearly irrelevant to the photoelectric effect: only the frequency or wavelength of the light matters. This fact shows that frequency (wavelength) relates directly to the amount of energy available to kick out an electron. $\endgroup$ – S. McGrew Mar 27 '18 at 8:22
0
$\begingroup$

Let's say that the "ripple" does not end at a point, i.e. it is continuous. A continuous stream of electromagnetic waves that carry some energy. Using this hypothesis, one could easily predict the following :

1. We let this "continuous ripple" fall on the surface for some time and let the surface (rather the electrons) absorb some energy. If we wait for some more time, the surface will absorb some more energy and after a certain period of time, the electrons will gain enough energy to leave the surface. So, given any small amount of energy in the "continuous ripple", electrons will always be ejected.

2. Since absorption of energy takes time, there should be a time gap between the falling of the "continuous ripple" and the ejection of an electron.

But... Unfortunately, that is exactly what does not happen! Pretty sad.

So, what happens?

1. The ejection of electrons occurs above a certain energy of the "ripple". Given a "ripple" with less energy than the minimum value, even if you wait for a very very long amount of time, not a single electron will be ejected.

Strike 1 to "continuous ripple"

2. There is no detectable time lag between the falling of the ripple and the ejection of electrons.

Strike 2

3. The minimum energy that we talked about before depends on the frequency of the "ripple".

Strike 3 and out

Innings changes to "start and end point ripple".

Does "start and end point ripple" answer our concerns? Is it experimentally consistent? Turns out it is. These ripples are defined to have an energy proportional to its frequency (an idea that Einstein took from Max Planck after the latter conceived of this idea regarding the energy of electromagnetic oscillators for his theory of blackbody radiation).

1. As per definition, "start and end point ripples" are discontinuous, and thus the electrons cannot continuously absorb energy. They absorb energy in the small values that these ripples bring along. If it has an energy lesser than that energy required to eject the electron, then nothing happens. If it is just equal or more, then it is ejected. Thus, we can explain the minimum energy of the ripple that is required.

2. Since these "start and end point ripples" are extremely small in their width, they are basically (very very crudely) like a billiard ball, and when a billiard ball hits another billiard ball, there is almost no time lag between the hit and the movement of the second billiard ball. The same with this scenario we are pursuing. The no-time-lag is also consistent.

3. Finally, by definition again, energy is directly proportional to the frequency, and thus the explanation of the frequency dependence of the minimum energy is also done with.

Getting this experimentally consistent model is what turned the favor towards the "start and end point ripple" view of this phenomenon.

In literature, we actually talk of the "continuous ripples" as waves and the "start and end point ripples" as wave packets. Here the wave packet is called a photon because it is an excitation of the electromagnetic field that we deal in QFT.

In general, all "so-called particles" are excitations in their corresponding fields. The time when Einstein developed this theory, there was nothing to guide him, no Schrodinger wave equation, no QED, QFT nothing. So he basically dealt with a "billiard ball like particle", but we now know that particles are just excitations in their respective fields.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks for that.. it seems the issue here is my notion of a particle being something other than a discrete ripple :).. once again, difference in definition is the culprit! $\endgroup$ – T.Donaldson Mar 27 '18 at 8:14
  • $\begingroup$ Hehe, happens with all. :) $\endgroup$ – Yuzuriha Inori Mar 27 '18 at 8:20

Not the answer you're looking for? Browse other questions tagged or ask your own question.