swansont Posted December 13, 2011 Share Posted December 13, 2011 Are these forbidden energy values for a given bound system absolute, or are they just the most probable energy values? In other words, can an electron have any energy value in a bound system, but the probability is the highest at the so-called permitted energy values? Both, kind of. The permitted energies are discrete but the transitions are not infinitely narrow — any excited state has an energy width dictated by the Heisenberg Uncertainty Principle where the uncertainty is the lifetime of the state (e.g. in alkali atoms the first excited states have lifetimes of around 20-30 ns, and the frequency widths are around 5 MHz). So one could say that there is some uncertainty in the energy of that state, but the energy of that state is orders of magnitude larger than the uncertainty. 1 Link to comment Share on other sites More sharing options...
questionposter Posted December 13, 2011 Author Share Posted December 13, 2011 In principle, yes. That's not something that is restricted — you have quantized energy levels in a bound system, so there are forbidden energy values in each, but there is nothing that keeps that energy from being allowed in another system. That's why you have various materials that are useful in different situations, depending on the application. e.g. semiconductor lasers that emit red light are a different material than those that emit blue light. Ok, I think I'm getting a picture that the energies are different, but because of the orbitals and the different numbers of electrons, some atoms are bigger than others. However, I'm still not completely sure whether or not any energy actually IS possible, just not within one system. Link to comment Share on other sites More sharing options...
swansont Posted December 13, 2011 Share Posted December 13, 2011 Ok, I think I'm getting a picture that the energies are different, but because of the orbitals and the different numbers of electrons, some atoms are bigger than others. However, I'm still not completely sure whether or not any energy actually IS possible, just not within one system. If you looked at all the types of atoms there would still be some energies that are not allowed. Energies below the tightest bound electron would not be possible, since there is no state below the ground state, for example. Link to comment Share on other sites More sharing options...
questionposter Posted December 14, 2011 Author Share Posted December 14, 2011 If you looked at all the types of atoms there would still be some energies that are not allowed. Energies below the tightest bound electron would not be possible, since there is no state below the ground state, for example. I thought electrons could appear inside hte nucleus without interacting with it? Shouldn't that technically be energy 0? Link to comment Share on other sites More sharing options...
swansont Posted December 14, 2011 Share Posted December 14, 2011 I thought electrons could appear inside hte nucleus without interacting with it? Shouldn't that technically be energy 0? No. An electron close to a bunch of protons is going to have a large negative potential energy, and that would boost its kinetic energy, but as with any bound system the total energy (KE + PE) is negative. There's no reason to expect the energy to be zero. Link to comment Share on other sites More sharing options...
IM Egdall Posted December 17, 2011 Share Posted December 17, 2011 Both, kind of. The permitted energies are discrete but the transitions are not infinitely narrow — any excited state has an energy width dictated by the Heisenberg Uncertainty Principle where the uncertainty is the lifetime of the state (e.g. in alkali atoms the first excited states have lifetimes of around 20-30 ns, and the frequency widths are around 5 MHz). So one could say that there is some uncertainty in the energy of that state, but the energy of that state is orders of magnitude larger than the uncertainty. Thanks for the clear and excellent explanation. Link to comment Share on other sites More sharing options...
questionposter Posted December 19, 2011 Author Share Posted December 19, 2011 Wait, in helium the electrons are closer, but how much are they closer if they aren't half as much as hydrogen and why? Shouldn't the distance of the ground state electron(s) be somewhat exactly proportional to the amount of protons in the nucleus? Link to comment Share on other sites More sharing options...
swansont Posted December 19, 2011 Share Posted December 19, 2011 Wait, in helium the electrons are closer, but how much are they closer if they aren't half as much as hydrogen and why? Shouldn't the distance of the ground state electron(s) be somewhat exactly proportional to the amount of protons in the nucleus? Electrons interact. We have already seen that the ionization energy of He in less than twice the ionization energy of H (24.6 eV vs 13.6 eV). The second ionization of He is 54.4 eV, which is exactly what you'd expect of a single-electron system, and in that system you'd expect the radius to have been halved (following the Bohr model). That shows there is significant screening going on by having another electron around. http://en.wikipedia.org/wiki/Bohr_model http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/helium.html Link to comment Share on other sites More sharing options...
immijimmi Posted January 5, 2012 Share Posted January 5, 2012 If all atoms have the same energy levels, as in the electrons in the ground state electron in lead wouldn't be closer than in hydrogen, and those energy levels can absorb only specific energies of light, why are there many different shaped orbitals? Firstly the different atoms dont necessarily have the same energy levels as each other. Proton count in lead is higher so the inner electrons would be closer. But that's irrelevant to the question. There are different shapes of orbitals because of the different relative energy levels of the electrons in that atom. For example carbon. It has 6 electrons. The first two are the closest so they have the lowest energy level. These inhabit a sphere shaped orbital. The next two electrons do the same, but farther from the nucleus because the inner electrons repel them. The last two inhabit a different shaped orbital: this one has a dumb-bell shape. The different shape is due to the fact that these two have more energy and so are confined to the atom differently. I don't know enough about it to go much further than that, but I hope I answered a few of your questions. Link to comment Share on other sites More sharing options...
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