# Tag Info

## Hot answers tagged atomic-physics

5

These states represent intermediate coupling schemes that are halfway between the usual $LS$ coupling and the more extreme $jj$ coupling that happens in heavier atoms where relativistic effects mean that the spin-orbit coupling for each individual electron can match or exceed the orbit-orbit coupling between different electrons. The intermediate coupling ...

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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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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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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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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.

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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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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 ...

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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, ...

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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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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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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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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 ...

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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 $KE = \frac{1}{2}m v^2$: ...

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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 ...

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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 ...

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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 ...

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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.

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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 ...

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