worlov

Hello, http://www.spacedaily.com/reports/Worlds_most_sensitive_dark_matter_detector_completes_search_999.html I think the researchers are on the wrong way. The right way is https://sites.google.com/site/testsofphysicaltheories/English/dark-matter Rock and water are the best hiding place for dark matter. They have no internal energy source, so they emit no radiation. With earthly instruments they can not be detected. But they are much heavier than simple dust. Mathematical approximately in the solar neighborhood: The dark matter is to be concentrated in the spiral arms. We are in the almost empty room and therefore can not see dark matter. The right way of the research would have been close examination of the spiral arms of our galaxy, like here: http://earthsky.org/space/deepest-yet-look-into-orion-nebula "On July 12, 2016, the European Southern Observatory (ESO) released the infrared image above of this mighty cloud of gas and dust in space. They said it’s the deepest-ever look into the Orion Nebula, revealing some 10 times as many brown dwarfs and isolated planetary-mass objects as were previously known." Best regards Walter Orlov

I can read it anywhere. This is nothing new. But what I say is new. By intensity of light I mean generally accepted concept.

What has "very very very high fields" with simple linear photoelectric effect?

An honest man would now prefer to admit that Mr. Einstein was wrong

I would answer with "Yes " your last question. Because the scientists always avoid to mention the relationship I = const, v - variable and i = const. But it is no less important as the relationship between the frequency of light and kinetic energy of electrons. Here Einstein's photon hypothesis fails!

Another picture from http://www.cobalt.chem.ucalgary.ca/ziegler/educmat/chm386/rudiment/tourexp/photelec.htm "This graph shows the typical results of an experiment. For a given light frequency and a particular metal for the cathode, the photocurrent is constant until a large enough retarding potential prevents the ejected electrons from reaching the anode. This potential is called the stopping potential and just matches the kinetic energy of the ejected electrons. When a higher energy light source is used (greater frequency, shorter wavelength), the ejected electrons possess a greater kinetic energy and so the stopping potential is correspondingly larger." The text is not completely. It should be emphasized that the intensity of light is constant, otherwise, these curves have no sense. For modifying intensity is presented other graph: Here, however, stressed that the wavelength remains constant: "This graph shows how the photocurrent increases when the light intensity increases but the wavelength is held constant. The stopping potential is the same however, suggesting that the kinetic energy of the ejected electrons is the same and hence independent of light intensity. This is the opposite of what would be expected classically."

Only an excuse...

But 1) you haven't shown a graph where anyone has claimed otherwise... I presented three graphs, where I = const, v -> oo and i = const!

The intensity of light I = const. The number of photons per second n = I / hv. The number of electrons is equal to the number of photons, i.e. also is n = I / hv. Therefore, the photocurrent i = ne = eI / hv. The relationship: I = const, v -> oo => i -> 0! That's easy...

I always assumed that the intensity of light is constant in my view. You say it's not enough of the electrons. So I ask: where are the electrons when the intensity of light increased? In this way, I just wanted to show, there are always enough electrons. But this change of intensity has nothing to do with the original idea.

Per accelerating voltage all ejected electrons independently of their kinetic energy are driven to the anode and in this way the photocurrent reached the saturation

I see that you did not understand the photoelectric effect, the debate is pointless.

Perhaps the following computer simulation helps: http://phet.colorado.edu/en/simulation/photoelectric . But beware - it contains a serious error: if you in the saturation (U >> 0V) with the same intensity change the frequency (Wavelength) of light , the photocurrent also changes: But that should not happen!

"Figure 27.5 Photoelectric current versus applied potential difference for two light intensities. The current increases with intensity..." http://books.google.com/books?id=CX0u0mIOZ44C&lpg=PA872&ots=X3EtNX1HWT&dq=photoelectric%20effect%20Serway&pg=PA873#v=onepage&q&f=false

The picture I have from here http://www.uwsp.edu/...300/Lect26.html

Then, where do the electrons when the intensity of light increased?

It's about the metals. There are so-called conduction electrons, they are not bound to individual atoms. Because of the applied voltage - negative to cathode - there is always an excess of electrons.

Hi, I've discovered the following: If the metal surface is illuminated, some electrons can leave the metal. This phenomenon is called "Photoelectric effect". But the frequency of the radiation must be high enough. To explain this effect Einstein postulated in 1905 the existence of light quanta - photons. Each photon corresponds to the energy E = hv (h is Planck's constant and v - frequency of light). When a photon penetrates into the metal, it could all his energy transferred to an electron. Taking into account the work function W is then the kinetic energy of an ejected electron: Ekin = hv - W. And indeed, there is in this respect a good agreement with experiment. For the interpretation of the photoelectric effect Einstein received 1922 the Nobel Prize in Physics: "for his services to Theoretical Physics, and especially for his discovery of the law of the photoelectric effect". But there is an unpleasant shade: in the saturation the photocurrent depends on the intensity of the light, but not its frequency. This is in conflict with the light quantum hypothesis of Einstein. At constant light intensity I the photon flux density n must fall with frequency v: n = I / hv. Therefore, fewer electrons are knocked out (ideally, nelectron = nphoton). Higher kinetic energy of the electrons has nothing to do with it, because the photocurrent is determined as i = ne (e is the charge of an electron). The relationship is illustrated by the following animation Therefore, it is expected that the photocurrent curves would thus run: But this is not the case! - In reality, the photocurrent in the saturation for different frequencies is the same. This fact is usually not even mentioned. Therefore, I will present several sources to confirm this statement: Remarkable is the fact that Einstein for incorrect interpretation of the photoelectric effect was even awarded with the Nobel Prize. The division is still taught in first grade of school. Why members of Nobel Prize Committee could not solve simple arithmetic task - intensity divided by the energy of the photon? Anyway, it is clear that the photo effect still needs a proper interpretation.