3
$\begingroup$

While I understand the effect of varying wavelength and frequencies on the photoelectric effect, I can't seem to turn my mind around that question... I suspect it has to do with quantas and the non continuous aspect of the electron's nature but I am really not sure...

Thanking you in advance.

$\endgroup$
  • $\begingroup$ Such models have been proposed and from a strictly atomic physics point of view it is not absolutely necessary to quantize the electromagnetic field to shoehorn it into non-relativistic quantum mechanics, but the procedure just doesn't buy you much. The problem is not to explain individual effects with semi-classical ad-hoc solutions, the problem is to explain all of nature with a self-consistent theory. The photoelectric effect is just one of many radiation-matter interactions and the real theory that explains them all is quantum electrodynamics. $\endgroup$ – CuriousOne Apr 30 '16 at 19:56
  • $\begingroup$ For a well known paper on the topic see e.g. "The photoelectric effect without photons" by Lamb, W. E., Jr. Scully, M. O.. Despite the qualifications of the authors, I don't think that this line of reasoning is anywhere in the mainstream of physics. $\endgroup$ – CuriousOne Apr 30 '16 at 20:05
  • $\begingroup$ Is this a duplicate of physics.stackexchange.com/questions/45851/…? $\endgroup$ – CuriousOne Apr 30 '16 at 20:08
  • $\begingroup$ The Lamb and Scully article gets a lot of bad press... I don't know, I haven't looked at it closely. However, Mandel and Wolf in their book on Quantum Coherence does provide a photon-free analysis of the photoelectric effect which seems perfectly good to me. $\endgroup$ – garyp Apr 30 '16 at 20:23
  • $\begingroup$ @garyp: The Lamb/Scully paper is a well known example of these kinds of attempts and I think everybody should have heard about it for that reason alone. One can do a lot of good atomic physics with classical fields and that's indication enough that not every light-matter interaction needs the full QED treatment. We know what the real precision tests for QED are and they were, of course, not available in 1905, when the photoeffect was front and center. My take is that Einstein guessed correctly, but at the end of the day it's not a watertight argument... nor did it have to be. $\endgroup$ – CuriousOne Apr 30 '16 at 20:33
0
$\begingroup$

The photoelectric effect is explained by a cool experiment which exactly displays why the wave model doesn't apply to the idea of photons.

Here's a diagram of the experiment: image

The wave model insinuates that the strength or amplitude of a light wave is proportional to its brightness, suggesting that a bright light should be strong enough to generate a large current.

EDIT: Physically, this means that the energy of the electrons increases as the intensity of the lamp increases, which implies that the speed of the electrons travelling in the tube from the cathode to the anode is also higher.

However, the experiment shows that, as long as the lamp is on (intensity $> 0\%$), the energy of the electrons remain the same regardless of the intensity of the lamp (energy = $x$ Joules whether the intensity is 50 or 100%). Only the number of electrons emitted from the cathode increases, causing an increase in the current of the circuit. This implies that there's a relation between current and intensity (which is directly proportional).

The experiment performed with varying frequencies of light shows that the energy of the electrons increases with higher frequencies (a linearly proportional relationship exists). This relationship is depicted by $ E_e = hf - \phi =\frac{hc}{\lambda} - \phi$, where $h$ is Planck's constant and $\phi$ is the work function of the metal (the minimum amount of energy required by electrons to escape the surface of the cathode).

EDIT: Einstein was inspired by Planck's black body radiation results and used the idea of obtaining energy from the electromagnetic field in discrete portions, indicating to him of the existence of quanta called photons.

These experimental results strongly indicate that the wave model for electromagnetic waves is not compatible with the photoelectric effect, as no energy variation in the electrons is observed with a change in intensity.

EDIT: Here's a link to an interactive Java animation which allows you to "play" around with this experiment beautifully.

$\endgroup$
  • 2
    $\begingroup$ The photoelectric effect predates quantum mechanics and a proper understanding of atomic physics. It assigns the quantization to the electromagnetic field, but in fact one can reproduce the effect theoretically by having a classical electromagnetic field interact with quantized matter. We usually don't discuss this at this level because Einstein guessed correctly and the rest is science history, but if one wanted to be anal about it, his guess is not a truly firm argument for the quantization of the electromagnetic field. $\endgroup$ – CuriousOne Apr 30 '16 at 20:20
  • $\begingroup$ I believe that it's not a truly firm argument. However, I can't seem to find any decent material that adequately explains the quantized matter and classical electromagnetic field interactions. I guess I'll have to figure it out myself! A question, though. Does the quantization pertain to matter waves? $\endgroup$ – GodotMisogi Apr 30 '16 at 20:29
  • $\begingroup$ You can't find decent material for things that are just not so. Matter is quantized and radiation is quantized and both are actually just two expressions of one and the same thing. When we teach this topic, we tend to go along with the historical development because bypassing it and clobbering a beginner with the apparatus of quantum field theory promises absolutely no hope of understanding for 99.9% of students. $\endgroup$ – CuriousOne Apr 30 '16 at 21:22
  • 1
    $\begingroup$ @Arjit Seth, I find that your answer to the OP's question consists of a straw-man argument. No, the wave model of light doesn't insinuate that the energy of an electromagnetic wave varies by its intensity. That was merely a guess by Maxwell, and he was wrong. $\endgroup$ – David Reishi May 3 '16 at 14:02
  • $\begingroup$ @Arjit Seth, it's like saying that the biological model of the human body insinuates that blood-letting is valid. $\endgroup$ – David Reishi May 3 '16 at 18:02
1
$\begingroup$

Let us first describe a relevant experiment: You have a photomultiplier tube, hooked to a loudspeaker for convenience. If you shine on the detector with light you hear noise, which is louder if the light source is brighter. But if you only take a very feeble light, you'll notice a peculiar thing: The loudspeaker does not make noise anymore but produces distinct clicks (in apparently random intervals)! Also, all those clicks are equally loud and long (except perhaps if two are too close toghether and one cannot separate them), i.e. in every way identical. Getting curious you might experiment with different colors and notice that only blue light makes clicks, red does not etc. The intensity of the light source only influences the amount of clicks.

The above can obviously not be explained with the behaviour of ordinary waves, waves are inherently very continuous.

This chapter describes the mentioned things in more detail on an intuitive level and is certainly worth reading.

$\endgroup$
  • $\begingroup$ It's not quite this simple. The quantization of the photomultiplier tube's matter already provides a reason for the "clicks" (which, by the way, are not equally loud by any measure) and one can, with some strain, reproduce the behavior with a semi-classical approach using a continuous electromagnetic field and quantized matter. Where all of this really falls apart is detailed here en.wikipedia.org/wiki/Precision_tests_of_QED. $\endgroup$ – CuriousOne Apr 30 '16 at 20:16
  • $\begingroup$ Thank you for this input @CuriousOne, I had the impression however that OP wanted a rather simple explanation and this would confuse him. Couldn't you add your comments as an answer? I think they are very interesting and relevant. Why do you think the clicks are bot equally loud? I assumed each photon frees one initial electron and the starting energy is completely irrelevant in comparison to the strong field. $\endgroup$ – caconyrn May 1 '16 at 11:02
0
$\begingroup$

With a classical wave model for light and a classical mechanics model for matter, the energy which is absorbed by one particle will be a function of the wave amplitude $E$. If we reduce the wave amplitude the absorbed energy $W$ will smoothly go to zero. In mathematical terms: $$ W(E\rightarrow0,\omega) \rightarrow 0 $$ If an electron requires a minimum amount of energy to escape a material, a minimum wave amplitude is required to do that.

However this is not obsorved in the photoelectric effect. There is no minimum light intensity i.e. wave amplitude required to extract electrons from the cathode.

It turns out that both a quantum description of matter or radiation can explain this behaviour. Former allows particles to only absorb quantized energy bits of $W=h f$ (for more details see Fermi's golden rule) while latter provides only quantized energy bits, which we call photons, with energy $E=hf$. For both cases the energy absorbed by one particle is only a function of $f$ and does not depend on E.

In reality we observe both quantization of matter and radiation.

$\endgroup$
-1
$\begingroup$

Classical electromagnetism has no way of linking energy of a wave to its frequency. In classical E-Mag energy density is usually given by $$ u = \frac{1}{2}( \epsilon E\cdot E + \frac {B\cdot B}{\mu} )$$ As you can see this has no frequency dependence in it at all. And in photoelectric effect experiment suggests that frequency affects the energy carried by the light.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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