Can anyone explain the Grant Partition Function for electronic gas with Fermi particle?
489039
Grand stands for 'very open' or no constant total energy and no fixed number of electrons.\nThe partition function shows how the electrons have been divided over all the available energy levels.\nIt is a Fermi-Dirac distribution, because electrons have 1x the elementary spin.
653703
The only way to prove this formula is to conduct an experiment. For the theory, recall that current density j is the fundamental quantity in Ohm's law. Also you will require the Boltzmann energy distribution.
894053
A good starting point is wikipedia:\nhttp://en.wikipedia.org/wiki/Free-electron_theory_of_metals\n\nWhen you read this link, follow some of the referenced links for details on the specific applications you listed...\n\nHope this helps!
278842
Disregard butchtroll, he doesn't know anything and is obviously just guessing.\n\nA absolute zero, although practically impossible to reach, it is theoretically very easy to calculate. Just assume that all particles have a velocity of zero, and calculate the lowest potential structure according to the lennard-jones potential and possible electro-static potential. The result concludes a structure quite similar to crystals at living-room temperature.\n\nButch - practically as in cannot be obtained through anything but theoretical experiments. Although, temperatures can actually be achieved that is sufficiently low to allow, in some degree, to regard matter as having a temperature of absolute zero. Theoretically, if such matter should exist, it would affect the universe in any significant way. That the universe cease to exist if absolute zero was reached, would be a prerequisite, not a result. You can do that math yourself and see if it is possible - or borrow the particle similator program I've written.\n\nmore: I find it funny that you preach about humility, after calling me an idiot. Especially after mixing different types of energy together and calling the lot for temperature. Sure, if an atom contain absolutely no energy, it would collapse into a black hole. But temperature is NOT a measure of total energy.\n\nyet more: Still you call me an idiot, and still you attempt to call humiliary high ground. I'm not saying I know everything myself - but wheter temperature is all energy, or just momentum of "particles", is pretty far from'everything'. As a student in nanotechnology, I can honestly say that, you sir, are blinded by your own feeble knowledge of physics. Temperature of matter is usually described as the summation of vibrational, momentum, rotational energy and phonons. NOT electronic excitation or any kind of potential energy.\n\nYour question was inane, and I *did* answer it. Nothing would happen with the universe if a cloud of hydrogen would suddenly come into existence with a temperature of zero kelvin. If you don't understand what I'm writing, just say so, instead of stating that I didn't answer your question.\n\nWhat do you attempt to prove with your citation? You seem lost, sir?\n\nWhat do you need a value of? What the temperature of the solid hydrogen would be after 1 second? Sorry, but I don't care enough to calculate how much a certain amount of hydrogen crystal would increase in temperature from background radiation.
817099
Because the las energy lawyer is complete. The octet rule says that the last energy lawyer must be complete or have 8 electrons to be inreactive.
1171601
The conduction band is not really related to the energy levels of electrons in an atom. Instead, it is about how electrons behave in a crystal lattice.\n\nIf you think about it, when you try to get an electron to move through a solid, it is going to keep coming across nuclei that will stop it in its tracks or bounce it back. So how on earth does anything conduct?\n\nThe answer is that electrons behave like waves. So the crystal lattice behaves like a 3D diffraction grating. You will have seen 1D diffraction gratings  splitting light into a spectrum. What this tells you is that photon of a given energy can go easily in one direction but not at all in another (the dark bands). The same applies - but more complicated - to the electrons in a crystal (metals are crystalline in structure).\n\nYou can think of it this way - in some directions the electron can wiggle around all the nuclei because the spacing is a multiple of its wavelength, in others it cannot.\n\nBecause electrons are no free to move as they please in a metal then the usual rule that their energy will vary with the square of their momentum does not apply. The band structure of a material shows how energy and momentum are related. It varies in different directions in the material.\n\nImportantly, the band structure has a gap in it. This gap is the area where there is no momentum an electron can have to have a given energy (you can see why this exists above). The band below this gap is called the valence band. The band above is called the conduction band.\n\nBecause of the Pauli exclusion principle. no two electrons can have the same state (i.e. the same momentum/energy/spin relation). So the bands fill up from the bottom.\n\nTo conduct, it must be possible to add some momentum to an electron to make it move. You can see that if the valence band is full, then you have to give a lot of energy before you can change the momentum because there is a gap before you find a vacant state. \n\nMaterials with a full valence band and empty conduction band are insulators.\n\nMaterials with some electrons in their conduction band are conductors.
960070
Two dimensions should be explanation enough. Assume the de Broglie wavelength of an electron is some distance in nanometers. If we draw the nucleus as a dot and draw the electron as a continuous wave around the nucleus, the wave has to be the length of some multiple of its wavelength. If the wave were not the length of a discrete multiple of its wavelength it would not be continuous because a full circle's rotation would cause the wave to jump. If we constrain the atomic model to behave like this, then the answer is simple. \n\nConsider the electron wave that is four wavelengths long. This wave is a function which describes electron density. The next energy level has to correspond to a wave that is five wavelengths long. In this way, the wave cannot increase gradually, but rather the levels are split into discrete shells, one for each number of wavelengths in the wave.
272822
One cool way that I have seen it derived is based on Jayne's Formalism.\n\nAssume we sample process, and measure the mean and standard devation of the samples. There are many possible distributions that will produce those exact mean and standard deviation. \n\nJayne's Formalism chooses the distribution which maximizes the entropy of the distribuiton. This is viewed mathematically as the distribution that makes the minimum number of assumptions regarding the distribution.\n\nI always thought that was kinda cool.
501994
http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/posfe.html#c3\n\nThe link above may be of some help. Input the values you want or need.
797567
Hi. From the web : "The quantum nuclear-spin relaxation rate (T-11) is computed as functions of magnetic field and temperature. A possibility that T-11 can be used as a magnetic probe in studying the electron-electron interaction in high-mobility 2DEG systems is discussed." I would say there was an effect.
569322
Use the ideal gas law \n\nPV/T = constant for the final and initial states.
144611
You might find this paper useful\nhttp://arxiv.org/abs/math-ph/0106027
887107
B.the electron subshell
1151044
http://www.allaboutcircuits.com/vol_6/chpt_7/5.html\nhttp://hyperphysics.phy-astr.gsu.edu/Hbase/electronic/nandlatch.html
556249
In Be and Mg only the s subshell is full while the p is completely vacant. Moreover for every element the most stable state is tht of the nearest noble gas configuration. So for Be and Mg it will be quite easy to give to 2 electrons to get the nearest noble gas configuration to make stable cations. Hence they are not inactive/inert. Also they can gain electron pairs in the p-subshell (although it is energetically not favourable) so they can not be inert.
901447
Note that we're not taking energy differences into account.  In cases where the difference in orbital energy is greater than 15eV you get negligible overlap.  Now onto the symmetry-allowed overlaps.\n\n1s and 2s are both spherical and overlap to form a sigma bond.\n\n1s and 2p(x) form a sigma bond.  You can tell because if you draw the 2p(x) lobe you'll see it overlap with the corresponding 1s.\n\nThe 2p(x) and the 2p(y) do not overlap.  There is no possible bond.  Draw the 2p(x) extending toward the other atom as positive or shaded (remember: positive does not denote charge).  The other atom has its 2p(y) pointing up and down (perpendicular) and there is equal positive and negative overlap, thus canceling both of them out.  No net overlap.\n\n3p(y)-3p(y) form a pi bond.  Once again, draw these out and see it for yourself.  Draw the 3p(y) orbitals as rather fat so you can see how they overlap.  Since they are not cylindrically symmetric they are not a sigma bond.  Pretend you were to rotate one of the atoms about the x-axis... you would change orientation and there'd be less overlap.  For the sigma bonds above, rotation does not affect the overlap.\n\n2p(x)-2p(x) have positive overlap.  Draw them and see why.  They point toward one another.  Moreover, if you rotate one of the 2p(x) orbitals about the x-axis you keep the same structure/orientation since the 2p(x) is along the x-axis.  Thus, you can tell that this gives a sigma bond.
918910
See if this helps:\nPV = nRT = NkT, where P is pressure, V is volume, n is the number of moles, N is the total number of molecules, T is temperature, k = 1:38 £ 10¡23 J/K is the Boltzmann constant and R is a universal constant, R = 8:314 J/mol.K.