# Tag Info

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The probability should be very low that this would happen. For EC to happen, there needs to be a probability that the electron could be in the nucleus. An external electron is going to be interacting with the electron cloud of the atom. The experimental data of x-ray intensities from nuclei following decay by EC indicate a strong drop in capture of n=2 ...

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Yes. Since they are leptons, they both exhibit the weak nuclear force. However, the force is not to strong considering the protons and neutrons are made out of multiple leptons. They would probably be in a very short, near undetectably short binary orbit and than go back to their parent atoms, or back into space.

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Crystal field is not related to ionicity. It is any perturbation which has the symmetry lower than full rotational symmetry due to symmetry of crystal lattice. In diamond case, the symmetry is quite high $O_h$, but still lower than $SO(3)$. In the case of NV centers however, the story is even more tricky. It is not the point object: it is a pair of defects ...

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In the first equation you can replace the minus sign ($-$) by $i^2$, then factor that to be part of the $\alpha$.

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A non-quantam mechanic answer - would be electrostatic repulsions (b/w -vely charged electrons) would prevent it and give them definite paths which don't intersect. Even if they move with 0.3c they can't comes much nearer than 4 fm.

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The spin of the electron can change as long as angular momentum conservation is not violated. If a photon leaves carrying 1+ then an electron can flip spin. – anna v That is correct but oversimplified. Spin-flip transitions are dipole-forbidden, and they can only happen via magnetic-dipole interactions, which means that they are generally very weak and ...

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The answer would seem to be: How many atoms is needed to store the information of the whole Library of Congress? Only one, according to University of Michigan professor Philip Bucksbaum. Since electrons, like all elementary particles, are actually waves, Bucksbaum has found a way to phase-encode any number of ones and zeros along a single ...

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Just for a completeness from an experimentalist's pov: Is it true every sub-atomic particle can have a mathematical representation as a wave? Experiments have determined the fundamental particles that define the standard model of particle physics. These are mathematically represented as point particles, with the mass in the table and the other quantum ...

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There should be a negative sign in Equation 2. The exercise is a very simple one in substitution and does not require any sign cancelling.

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You seem to be asking two things, could a black hole's magnetic fields cool nearby matter, and could this cooling produce cold fusion. But maybe we should first ask whether "cold fusion" is a real thing. Nuclei contain protons and neutrons held together by pions. Fusion is when two nuclei become one. The barrier to this happening, is the positive electric ...

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You have identified the problem as the factor $2$ between your calculation and the book. You have read the problem correctly, so it seems the book is wrong.

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The most common type of imaging in these systems is absorption imaging- you just shine light on the atoms that matches a strong electronic transition, so the atoms scatter as much light as possible, and then get an image from seeing how much light is scattered. Essentially, you are looking at the shadow cast by the atoms. So this frequency is, naturally, ...

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One important thing to remember is that since the NV center is a defect, the way the electrons bond near it can be very different from the way they bond in the rest of the lattice. The negatively charged nitrogen-vacancy center defect in diamond's electronic states are understood to be composed of six electrons. Five of these are from unpaired electrons ...

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As a general rule adding thermal energy doesn't cause electronic transitions. That's because typical electronic transition energies are a few electron volts or around 100kT at room temperature. In a metal the electrons aren't in discrete energy levels but instead reside in a continuous band of energy levels called the conduction band. While thermal energy ...

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Visit http://www.science.uwaterloo.ca/~cchieh/cact/applychem/hydration.html You can see that the enthalpy of hydration is a two step proccess solvation and reverse crystallization, the ΔH(hydr) is actually positive so you have to give energy just to dissolve the NaCl in water, in order to seperate the water from the NaCl you need to account for the enthalpy ...

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An electron in a higher energy orbital does not necessarily decay directly into the ground state. If an electron in a hydrogen atom is excited into, say, the $n=4$ energy level, all of the following are possible decay paths: $4 \rightarrow 3 \rightarrow 2 \rightarrow 1$ $4 \rightarrow 3 \rightarrow 1$ $4 \rightarrow 2 \rightarrow 1$ $4 \rightarrow 1$ ...

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I don't know if I'm understanding your question right, but I think you are trying to pose a deeper question than it might seem at first sight... In ordinary quantum mechanics, when you study the hydrogen atom, you derive a set of solutions for the electron wavefunction using the Schrödinger equation (with different values of the energy). These are the ...

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The statement in the question is not fully correct. When an electron in a system is excited from a lower to a higher energy level the system becomes excited. That is to say, system reaches to a state which is not favourable. This is an unfavourable state for the system because the excited electron leaves a hole behind. There are several different mechanisms ...

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This is the hydrogen atom energy level solutions, as an easy example. The electron sits at the ground state, and can be kicked up to an excited state by the appropriate photon i.e. given that the photon has the quantized energy needed. For each energy level one can calculate using the solutions of the Schrodinger equation, the probability for the ...

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I'm basically at a loss how to simplify the integral. I am not sure this needs to be done. In spherically symmetric cases, the volume element, $dV$ is quite simple to deal with. In your case, you have: \begin{align} W & = \frac{1}{2} \int dV \ \rho\left(r\right) \ \Phi\left(r\right) \\ & = \frac{4 \pi \ C}{2} \int dr \ r^{2} \left( r^{-2} ... 1 You are confusing the energy of a single photon with the total energy (intensity) in a differential wavelength band  \lambda  to  \lambda + \epsilon . Yes, the shortest wavelength corresponds to the photon with greatest energy, but the X-ray machine generates far more photons at longer wavelengths. Thus the peak intensity occurs where the product of ... 1 The atom absorbs the photon that kicks up an electron to an excited state, and it is the atom that will emit a photon when it de-excites. Not the electron. Is the invariant mass of an atom higher when the electron is in an excited state? Take the hydrogen atom. The ground state energy is at -13.6eV. This means that the mass of hydrogen is less than the ... 1 Your graph is a standard one to show the spectrum of wavelengths emitted from an X-ray tube. The X-rays are produced by getting energetic electrons hit a metal target. The electrons are first accelerated by being attracted to a positive anode which is at a high potential V relative to the negative cathode from which they are emitted. The kinetic energy of ... 0 My short answer comes from reading this page on the web: https://www.4physics.com/phy_demo/at_clock/at_clock.htm For more details, please go to that page. It is very clear. The Cesium atom itself is not oscillating. The oscillation being measured is the frequency of the photon that is emitted by the Cesium valence electron when it jumps from an excited ... 1 In the Wikipedia article about rutherford scattering the derivation of the scattering cross section \frac{d\sigma}{d\Omega} =\left(\frac{ Z_1 Z_2 e^2}{8\pi\epsilon_0 m v_0^2}\right)^2 \csc^4{\left(\frac{\Theta}{2}\right)}$$is given. Let's rewrite that in your notation: Z_1 = Z, Z_2 = 4, k = \frac{1}{4\pi\varepsilon_0} and KE = \frac{1}{2}m v^2: ... 0 Your mistake is that the other particle has energy too. In the hydrogen atom, the nucleus is just sitting there, but in positronium, the positron has an equal amount of kinetic energy. (You are also cutting the potential energy in half.) Doubling all energies will give the right result. As other people have said, it's much much easier to do this with ... 1 Your proposal will work as long as mass and charge enter into the expression in a ratio 1:-2. Find an expression with a different ratio (like electric potential, or radiated power ...) and you're sunk. The fine structure constant$$ \alpha = \frac{e^2}{\hbar c} \tag{Gaussian units}$$is an example of such an expression and controls the fine structure of ... 0 I do not think many people attempt to find the absolute total internal energy of a system. Rather they will pick and choose what needs to be counted to suit their purposes. If interested in statistical mechanics then treated atoms as billiard balls which can rotate and vibrate, have translational kinetic energy and potential energy due to the inter atomic ... 0 It’s all about the electromagnetic force. I’ll give the answer in two ways- 1- Imagine a satellite. It also revolves around the earth and is also attracted by the earth but it does not fall towards it. WHY? Because the earth does ZERO work on it. 2- The uncertainty principle. IF the electron comes closer to the nucleus ( i mean really close ) then it’s ... 0 A better way to analyze this is to realize that in the photoelectric effect, the electromagnetic wave couples two electron states (bound and excited) via the frequency difference which those states share with the electromagnetic wave. In the Compton effect, there is also an electromagnetic wave and two electron states (in a center-of-mass system we can ... 3 Photo electric effect occur in bound electron, while compton effect occur in free electron. In photo electron effect, the photo and hence energy of the photon is absorbed by the electron. While in compton effect photon is scattered. 1 Consider the following model of an atom: Keep in mind that it is only a model and while it is a good model that elevates our understanding of the subatomic world, it is still just a model and reality will look different. How exactly? We don't know. The model is good enough, though, to understand what an excited atom is. With this caveat out of the way, ... 2 Introduction In general, a Physical System State is described by a set of variables Let’s consider the “System Internal Energy” variable System States A System is said to be in its “Ground State” when it is at the lowest possible energy level Any other State is then an “Excited State” and they would correspond to energy level greater than the ground ... 0 Basically electrons prefer to stay in least energy level in an atom. If certain amout of energy is given to it then it jumps to a higher energy level. There are discrete enegry levels so e- would accept only some particular energy to get exicted to higher energy level. When it returns to a lower state it gives out the energy in form of photons. Search ... 1 Excitation is an elevation in energy level above an arbitrary baseline energy state. "In English, please!" So what this is effectively saying is that an atom is considered "excited" when its energy level is higher than the rest. This can be manifested as heat, light, etc. For example, the Aurora Borealis. The Aurora is when radiation from the sun ... 1 Your equation k*exp(-r/a) is the wavefunction(n=1,I=0,m=0), so n=1 = ground state. So while n does not appear explicitly in the equation, it’s really there and it’s equal to 1 in this case. The equation should really be written H x wavefunction(n) = En x wavefunction(n), En = E(0)/n^2. 0 \langle z\rangle=\int_0^\infty r^3dr[\int_0^{2\pi}d\phi\int_0^\pi \sin \theta \cos \theta d\theta]=0 As the \theta  integration is zero. But the symmetry argument is clear if the integration is written is Cartesian coordinates.In that case$$\langle z\rangle=\int_{-\infty}^\infty \int_{-\infty}^\infty \int_{-\infty}^\infty z dz |\psi|^2 dx dy As you ...

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Perhaps you could fill us in on some more details about what your application is. There are a whole bunch of ways to produce plasmas and the techniques vary largely by the application. For example DC or RF fields applied across a very small gap are used for plasmas that need to be produced at atmospheric pressures (DBDs). At the other end of a scale, a ...

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These atoms would not have a negative charge, they would have zero charge. They would have no electrons at all. You could imagine (if they were far from anything else) an electron orbiting a neutron under gravity. The attraction is so weak that any other matter anywhere close would disturb it. The real question is whether a clump of neutrons could be ...

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