# Tag Info

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I am answering with these quotes(and also my comment to probably_someone) Since opposite charges attract via a basic electromagnetic force, the negatively-charged electrons orbiting the nucleus and the positively-charged protons in the nucleus attract each other. Also, an electron positioned between two nuclei will be attracted to both of them. Thus, the ...

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You are asking why the formation of covalent bonds between two (or more) atoms causes attraction (bonding) between them. Now it is very important to understand that the more correct expression is the sharing of electron orbitals that causes this attraction (overlap is not even needed, when there is only one electron being shared). The sharing (between the ...

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When you put two atoms close enough together that both nuclei are important to the behavior of both sets of electrons, you have significantly changed the potential that an individual electron is moving within. This potential generates wavefunctions that are quite different from the wavefunctions in an isolated atom, with different energies and very different ...

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In this "overlap" explanation, the overlap is not an overlap between two electrons. You get a $\sigma$ bond in the H2+ molecule, where there is only one electron. There are two reasons why you get a lower energy in a covalent bond. One is that the kinetic energy is lower than in the unbound state, and the other is that the potential energy is lower. The ...

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This might be a bit helpful. The uncertainty in the momentum of an electron in an atom is defined as: $(\Delta P )^{2}= \langle P^{2}\rangle-\langle P\rangle^{2}$. An electron bounded by the nucleus, has an average momentum of zero, which means ($\Delta P)^2 = {\langle P^{2}\rangle}$, but again, Feynman's explain, as explained in the comment, was a really ...

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This has been done by many people. It is called the ac-Stark shift, and maybe you are able find programs in the internet, which are already for Lithium. Check out the phd theses in the ultra-cold atom community. I have done it for an atom which possesses only a $(LS)J$ coupling. In your case, you will have to adapt the methods, which will be painful -- I ...

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Systems in equilibrium have the energy of their atoms described by the Boltzmann distribution (https://en.m.wikipedia.org/wiki/Boltzmann_distribution) $p_i \propto e^{\frac{E_i}{k_BT}}$ $p_i$ is the probability of finding an atom in state $i$ $E_i$ is the energy of state $i$ $k_B$ is the Boltzmann constant $T$ is the temperature Like most other ...

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You can find this easily by counting atoms and bonds. In the diamond lattice each atom has bonds to 4 neighbor atoms, and each bond is between 2 atoms. Hence, in a diamond crystal with $N_{atom}$ atoms you have $N_{bond}=N_{atom}\cdot 4/2=N_{atom}\cdot 2$ bonds. Therefore the energy per bond is half the energy per atom: $7.37\text{ eV}/2 = 3.685\text{ eV}$

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Now suppose that we measure the the system and we find it in the ground state $\phi_1$. Is it then possible to have photon emission? No. The previous superposition becomes irrelevant once you perform a projective measurement that results in the system ending up in the ground state. The system is then in the ground state and has no energy to radiate away, ...

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The answer depends whether the hamiltonian of your system takes into account a photon field. From what I have seen, there aren't really great models (hamiltonians) of this in general. Quantum Field Theory (the formalism of the standard model) uses a hamiltonian with interactions between the atom and a field of light $A_\mu$. However it only describes the ...

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By a separate mark scheme, it appears that $b=\sqrt{2l}$ such that: $L_-|l\rangle=\hbar\sqrt{2l}|l-1\rangle$. Source: a similar question from a paper with newly found model answers: $L_-|3\rangle=\hbar\sqrt{6}|2\rangle$, $L_-|1\rangle=\hbar\sqrt{2}|0\rangle$. Note: There is not an explicitly stated $b$ value in the notes we have received for this ...

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If your Hamiltonian incorporates the photon field, then your eigenstate is a direct product of two Hilbert spaces. The atomic state ($\psi$) and the photonic state ($n$), $|\psi,n\rangle$. Here $n$ represents the number of photons present in the field. For simplicity we are considering monochromatic photons whose energy corresponds to the energy difference ...

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When atoms and molecules interact with electromagnetic radiation, there could be some transitions between rotational, vibrational or electronic states. Whenever electronic transitions happens, the Coulomb interaction is changed due to the redistribution of the electrons. Hartree-Fock deals with the stationary states of many-electron systems, and do not ...

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It is not possible to reverse the structure of the atom. Electrons have a very low mass and (relatively) protons are quite massive. Because of the uncertainty relationship between position and momentum the more tightly confined a particle is the more certain its position and the more uncertain its momentum. As the uncertainty in momentum increases so does ...

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Compounds having even number of total electrons will be diamagnetic 10 and 16 are exceptions They Will be paramagnetic

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A magnetic moment $\vec \mu \propto g_l \vec L + g_s \vec S$ in a B field possesses the energy $E=-\vec \mu \cdot \vec B$. Thus, if the B field vanishes, $B=0T$, the energy levels of the states $|l, m_l; s, m_s\rangle$ are degenerated. Furthermore, since electrons with the same energy are indistinguishable, we are unable to tell which of the six $|l=1; s=1/... 14 The explanation of the force as due to oscillating dipoles is only an approximate description and should be regard only as a guide for students. The actual explanation is that the wavefunctions of the two atoms overlap and the system now needs to be described by a larger wavefunction that includes the electrons in both atoms. In principle this new ... 2 Nothing. The atomic cloud ( electron distribution) is probabilistic in nature. Untertainty in charge distribution causes polarization 0 I can answer this based on classical mechanics and thermodynamics. Someone else may have a quantum mechanics perspective. The baseball is a collection of atoms and molecules. The velocity of the baseball is the velocity of the center of mass of the collection with respect to an external (to the baseball) frame of reference (say, the ball field). As such, ... 5 Both$L$and$S$are quantum angular momenta. Angular momentum is different in quantum mechanics (QM). Almost everything is different in QM. In QM, angular momentum is complete if you give two numbers: the "angular momentum" and its "third component"$M$. So you give the pairs$|l,\ m_l\rangle$and$|s, \ m_s\rangle$So, any angular momentum is ... 0 Entropy can be understood as a ratio of number of microstates and macrostates, entropy increase in this picture looks like a shift toward macrostates that encode a wider set of microstates As the amount of total energy above base increases, loosely speaking there are more available microstates that can be described in the macro picture as being spread out ... 1 Yes, two photon photoelectric effect is possible. The first experimental observation of the two photon photoelectric effect was in 1964. They used Cs3Sb1 and laser irradiation, with 1.17 eV, which as you describe is far below the threshold (the work function of the electron), but the process is still satisfactory. You can see that this is a higher order ... 0 There is two-photon photoemission, for example on a silver surface where the electron is first excited to a long-lived image state on the surface. The second photon then excited this bound state to a state with sufficient energy to overcome the work function. Article from Franz Himpsel's group (1985) 1 The most important thing is to understand the difference between orbit and orbital. Originally the atom was imagined with classical orbits (like planets around stars), but eventually QM was developed and now we talk about electron orbitals. In atomic theory and quantum mechanics, an atomic orbital is a mathematical function that describes the wave-like ... 7 What we have is Quantum Mechanics supplemented by Quantum Electrodynamics. With the tools available you can calculate atomic properties to increasing accuracy. Neutral hydrogen can be treated by the Schrödinger and more accurately the Dirac equation. Then you can throw in perturbative QED radiative corrections and a finite size nucleus. This brings you to ... 4 It's impossible to have the final model of an atom. We'll have to do it with approximations. Ther are simply too many factors to take into account. Even a "simple" thing as the proton spin: See this PDF for example. And, depending on the kind of atom, there is at least 1 proton to find in every atom. And don't forget the neutrons (both nucleons). and the ... 50 Electrons do not move inside atoms. If an electron is in a given energy level$E$, the wavefunction is given by$\psi(x,y) = \phi(x)_{n\ell m} \,\mathrm{e}^{-\mathrm{i}E t/\hbar}$. The time dependence is a pure phase factor, hence the real-space probability density of the electron is$|\psi(x)|^2 = |\phi(x)|^2 \neq f(t)$, not a function of time. These are ... 24 There are no final models in science, there's always room for improvement. And major paradigm shifts cannot be totally ruled out. However, we can be quite confident in our current model of the electronic structure of the atom, which is based on quantum electrodynamics (QED), which has been validated to very high precision. Wikipedia has numerous orbital ... 4 The linked video shows a TEM image. In transmission electron microscopy, you have a parallel beam of electrons incident on the material, which gets scattered by the atoms in the material. Note that, it is not just the electron cloud, but the potential formed by the atomic core (nucleus+ core electrons) and the valence electrons together, that scatter the ... 3 Yes and no. It is indeed true that for every emission line, there is a corresponding absorption transition that, under suitable conditions, can be observed in an absorption spectrum. However, it is not true that if you take an emission spectrum and an absorption spectrum from the same atom, the lines in the two must always match. The reason is simply that ... 0 Where did you get this idea? I watched the video and nothing like that was even mentioned. All platinum atoms are about the same size and shape unless an atom is excited. So there is no answer to your question since the atoms do not all have different shapes and sizes. 1 Not in general (though it is very common). Under some conditions you can have states that are reached by a single absorption event (creating a high energy absorption line) that decay (with a non-trivial branching ratio) through several steps (creating several lower energy emission lines). Indeed this kind of process is the usual scheme for setting a ... 1 The Hydrogen atom model is a very basic one. It consists of a an electron in a central potential (which represents the proton), and assumes nothing else exists in the universe. This is of course a simplification, but a necessary one. In this model, the electron has bound states, which are states characterized by energy less than zero, and more importantly, ... 2 So roughly speaking this is the contents of Newton's first law. Newton was living in an era when people wanted to think that the natural state was things standing still, and that all motion required a force to explain it. But but during Newton's life, Galileo had observed something different: when he rolled a smooth metal ball over a large flat floor, it ... 1 "Forces" don't keep things in motion - that was the old idea of Aristotelian mechanics, and it is not a good principle for understanding the rest of the Universe, already dated by the time of Sir Isaac Newton. "Forces" influence motion, but "motion" exists of its own accord. So what I suppose you are really asking is "what force confines the electrons to ... 0 Classical Picture There is no need for a force to cause motion, it's the presence of them that disturbs motion! (Cause: Inertia) But if you are asking for considering the circular motion of electron (note that this model was used by Bohr but isn't currently in use) then yes the Electromagnetic Force is the cause of it. 0 Strictly speaking, no. The hydrogen nucleus contains only one proton, there is nothing there to decay, except for the proton itself, and we do not know whether protons decay at all. Even if we knew that any other isotope of any elements could decay, it would not change to know whether all isotopes can decay. 1 Yes, mostly. The Coulomb force is described by but one of the four Maxwell Equations, acting between the (mostly) stationary protons and neutrons in the nucleus and the fast-moving electrons that can be found, each in its orbital, though protons, neutrons, and electrons all also have magnetic moments. Coulomb's Law is the only one of the equations that is ... 1 The potential term in the Hamiltonian is based only on the Coulomb interaction. They are the causes of chemical bonds. 0 To me the key takeaway from https://en.wikipedia.org/wiki/Spontaneous_emission is that: Spontaneous transitions were not explainable within the framework of the Schrödinger equation, and: The first person to derive the rate of spontaneous emission accurately from first principles was Dirac in his quantum theory of radiation I haven't learnt Dirac or ... 0 First of all, you should not quote the$\ell$selection rules but the$F$and$m_F$ones, see below. This is an adaptation from the wikipedia table: So the$\Delta m_F = \pm 1\$ transitions are allowed. What you should really be asking is how are these transitions allowed within the same principal quantum number n, so that parity is unchanged? Electric ...

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