Timeline for What are thermal energy distributions?
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Jan 8, 2013 at 4:31 | vote | accept | hhh | ||
| Jan 8, 2013 at 12:23 | |||||
| Jan 8, 2013 at 4:31 | comment | added | hhh | ...have to take pen-and-paper to really dig into this, takes some time to process this material. Thank you for helping. | |
| Jan 8, 2013 at 4:30 | vote | accept | hhh | ||
| Jan 8, 2013 at 4:31 | |||||
| Jan 8, 2013 at 4:17 | comment | added | hhh | Can I use somehow the propability function with partition function $P_s=\frac{1}{Z} e^{-\beta E_i}$ to make the deduction with less steps? en.wikipedia.org/wiki/… | |
| Jan 8, 2013 at 2:59 | comment | added | Michael | "can you see it somehow form the distribution" - Yes. en.wikipedia.org/wiki/… en.wikipedia.org/wiki/… | |
| Jan 8, 2013 at 2:57 | comment | added | hhh | My lecture slides deduce $P=\frac{g_1^{n_1}g_2^{n_2}g_3^{n_3}...}{n_1! n_2! n_3! ...}=\Pi_i \frac{g_i^{n_i}}{n_i!}$ for Maxwell-Boltzmann statistics. | |
| Jan 8, 2013 at 2:50 | comment | added | hhh |
"Many Bose particles can be in the same state, whereas only one Fermi particle can be in a given state." -- can you see it somehow form the distribution? (sorry I used the words distribution and statistics interchangeable)
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| Jan 8, 2013 at 1:38 | comment | added | Michael | Distribution is used in the sense of probability distributions en.wikipedia.org/wiki/Probability_distribution. An energy distribution is a function which tells you the probability that a particle has an energy in a given range. Statistics in this context refers to Bose-Einstein or Fermi-Dirac statistics, which just tell you that the particles are indistinguishable, and how many particles can be in the same state. Many Bose particles can be in the same state, whereas only one Fermi particle can be in a given state. | |
| Jan 8, 2013 at 1:33 | comment | added | hhh | If u need to deduce the statistics, how would you proceed? | |
| Jan 5, 2013 at 1:47 | comment | added | Michael | Classical here refers to the classical theory of electromagnetism, the complete description of which is found in Maxwell's equations: en.wikipedia.org/wiki/Maxwell%27s_equations This theory describes electromagnetic waves which can carry an arbitrarily small amount of energy in proportion to their intensity. There is no classical theory of light which involves photons. Photons are intrinsically quantum. | |
| Jan 4, 2013 at 6:34 | comment | added | Michael | If you have a more intense light, that means more photons -> more electrons. But the electrons don't have any more energy. More electrons yields a greater current, but the voltage required to stop the current doesn't change. But the experiment also works with an extremely dim light, so low that you only get a single photon at a time. Then the current is so low that you can detect individual electrons getting kicked out, one at a time. But their individual energy is still the same. In this extreme situation it is very obvious that the energy is being transferred in discrete lumps - quanta. | |
| Jan 4, 2013 at 6:32 | comment | added | Michael | Light comes in discrete units - photons. Photons of a given color have a definite energy given by $E=h\nu$. If a photon strikes an electron, transferring its energy to the electron and ejecting it from the metal. The electron then has an energy $E=h\nu-W$, where $W$ is a characteristic of the metal that measures how much energy it takes to just barely remove an electron. The point is this energy doesn't depend on the intensity of the light, only the color. One photon -> one collision -> one electron ejected -> same energy every time. | |
| Jan 4, 2013 at 6:26 | comment | added | hhh | When a photon hits a surface, the surface may emit an electron or a phonon i.e. heat. I have heard this statement many times: higher intensity should classically result in larger energy per electron. But in reality, it does not and the energy of electron depends on the stopping voltage. I cannot understand how this is a proof of QM phenomenon -- which mathematical formulae contradicting...? | |
| Jan 4, 2013 at 6:07 | history | edited | Michael | CC BY-SA 3.0 |
spelling of Planck - derp
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| Jan 4, 2013 at 6:07 | comment | added | Michael | @hhh Ahh, I see where you're getting confused. Planck's law - meaning the blackbody distribution function - is not directly connected to the photoelectric effect. Two different things. You're after "Planck's relation" en.wikipedia.org/wiki/Planck%27s_relation, the energy of photons is related to their frequency by $E = h \nu$. Planck proposed and used it to derive the blackbody distribution. Then Einstein used it to explain the photoelectric effect. | |
| Jan 4, 2013 at 3:11 | history | answered | Michael | CC BY-SA 3.0 |