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This is from The Nuclear Weapons Archive:2.1.4.1.2 Gun AssemblyAssembling a critical mass by firing one piece of fissionable material at another is an obvious idea and was the first approach developed for designing atomic bombs. But it is probably not obvious how you take two subcritical masses and obtain the equivalent of three critical masses by bringing ...

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The atom has some charge distribution $\rho(r)$. We don't don't know what form the function $\rho(r)$ has, but we do know it depends only on $r$ because an atom is spherically symmetric. When you have a spherical charge distribution the potential at a distance $r$ is simply due to the total charge inside the distance $r$: $$V(r) = ... 4 For a 6 year old, you might want to focus on thickness instead of length, as the numbers get too big with length. A ream of paper (500 sheets) is a bit over an inch thick, say 3.5 \, \text{cm}, so one sheet is 3.5/50 \, \text{mm}, or .07 \, \text{mm}, which is 7 \times 10^{-5} \text{m}. An atom has diameter 0.1 \, \text{nm} to 0.5 \, \text{nm} ... 5 A typical atom is roughly a few times 10^{-10} \text{m} wide. A piece of paper is say (1/4) \text{m} wide. Therefore the ratio of the width of an atom to the width of a piece of paper is around 10^9. A piece of paper is roughly the same width as a human, so 10^9 is also a rough guess for the ratio of the width of a human to the width of an atom. The ... 6 The total angular momentum of a closed shell is zero because for fixed l, we have the possible states labeled by eigenvalues of L_i as m_{l,i} = l,\dots,0,\dots,-l in integer steps. The sum over all m_l inside a shell is always zero, so total angular momentum of a shell is zero. This is just the generalization of the argument with "up/down" for ... 0 Balmer Series requires electrons to jump from n=2 to n=3,4,5.... and again back to n=2. Not quite. The Balmer series [emission] does not require any particular path for the jump up. The atoms do have to be excited to n=3,4,5, or higher but it doesn't matter how. In a laboratory, the Balmer series can easily be created in a plasma which has ... 1 If you were considering a theoretical system, at T=0K, you would be correct in assuming a Hydrogen atom has no n=2 electrons. However, at high temperatures (Or any finite temperature actually), there is a chance that the atom would be excited. (Meaning there are n=2, and there could be n=3, and so forth also) In fact, the Balmer series is very ... 2 Electron capture: p+e^-\rightarrow n+\nu_e Beta plus decay: p\rightarrow n+e^++\bar{\nu_e} Let's check the masses of both sides of the processes: Electron capture, initial state: m_p+m_e=938.78 \frac{MeV}{c^2} Final state: m_n=939.56 \frac{MeV}{c^2} The difference: (Final minus initial) 0.78 \frac{MeV}{c^2} Beta plus decay: (Positron emission) ... -1 The stronger positive nucleus of one atom firstly pulls the electron(s) of the other atom. This makes one atom positively charged and one negatively charged causing an ionic bond to be formed 0 So, the answer is basically, you need a rather complicated model before you can really do this. We can start by describing the Hydrogen atom with number operators for the different states; it takes the form:$$\hat H_0 = -\frac{m_e c^2 Z^2 \alpha^2}{2} \sum_{\sigma\in\{\uparrow,\downarrow\}} \left[\hat s_{1\sigma} + \frac 14\left(\hat s_{2\sigma} \hat ...

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Let's take these separately Are all electrons always in pairs except the final single one if odd number electrons are considered? For a ground state atom, then this depends on the sort of shell you're looking at. If you have an odd number of electrons, the simplest sort of ground state will look like this. In this case, if you have unpaired ...

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Yes, there could be. Yes, in general unstable electron configuration it could be. No, we can hardly say anything about "force" between two electrons. Two electrons are undistinctable - one cannot number them up, no paint exist to do so. We distinct these electrons by quantum numbers they hold to. So we cannot say which one electron exactly was moved to ...

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All "rules of filling" are empirical and there are many violations. What is not violated is Pauli principle - you will not find two electrons with same wave functions (i.e. same quantum numbers). To find out what is happening there you need to build numerical solution for that "spherical wave function problem" of atom. But as far as I can say, noble gases ...

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Indeed, this is a very good question and far from obvious. From a classical point of view one can only argue using charge distribution and atom radius/ion radius. The more spherical the charge distribution in the atom, the lower its energy in an electric potential, because the more spherically it is, the more neutral the atom in respect to the potential. ...

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The typical way to handle such things is the "sudden approximation". The time scale of the decay/capture process is assumed to be much smaller than the time scale of the evolution of the electron shell. The probabilities of the new states will then just be the projections of the old state to the new stationary states. (The typical analytically solvable ...

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Experiments with trapped ions generally use fluorescence for detecting the ions. This means that they use a strongish pump to take the ion from its ground state to a dipole-allowed excited state and wait for the ion to decay by emitting a photon in a random direction, and then re-run the cycle over and over. This means that each ion essentially emits one ...

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Let $E = KE +U$ be the total energy . We know the momentum operator & total-energy operator are $$\hat{p} = \frac{\hbar}{i} \frac{\partial}{\partial x} \\\\\\\\\\\\ \hat{E} = i\hbar\dfrac{\partial}{\partial t}$$ . This prompts us to write $$\hat{E} = \hat{KE} + \hat{U} \implies \hat{E} = \dfrac{\hat{p}^2}{2m} + U \implies \hat{E} = ... 0 How to explain what an electron is to someone new to physics? I think you can come up with an easy-reading explanation that's physically correct. There's plenty of clues in the literature if you're willing to play detective. For example see pair production. We quite literally make an electron (and a positron) out of light. And then when we annihilate ... -1 Hey I'd it hard to grasp a person that what an electron is just relate it to real world.like this Protons are positive charged. Let's say they are beautiful girls now they attract the negative charged electron. Let's say they are the boys wandering around. They both attract each other. But father the neutrons don't allow them to meet and bind protons in the ... 0 Here is a more complicated answer: I am going to try my best, ok? An electron is a negative elementary charge subatomic particle of an atom. It was once known as a beta particle, but it is now an electron. It takes electromagnetic, gravity, and weak interactions into play. In Coulombs, it's charge is approximately -1.6 * 10^{-19} C and it's mass in ... 2 Electron is a particle with mass and a certain probability of being found at a given distance around the nucleus of an atom at a certain time. It caries a negative charge which makes chemical reactions possible since chemical reactions are driven by the electrostatic forces between electrons and positively charged protons which reside in the nucleus of the ... 1 Hmmm... I'll take a crack at it. An electron is a negatively charged subatomic particle that orbits the nucleus of an atom, which contains a positively charged subatomic particle called a proton and a neutral subatomic particle called a neutron. 2 The quantization of energy levels appears both in quantum and classical mechanics, and it is not a consequence of the Schrödinger equation. It is a consequence of confinement. In fact, anytime that a wave equation (any quantum equation for the wavefunction, or a classical equation for a classical field, e.g., EM field) has periodic boundary conditions in ... 2 Quantization is an experimental fact that forced physicists to consider theories that could explain the data. This happened in the beginning of the twentieth century. 1) black body radiation could only be explained by assuming that the radiation came in quanta, i.e. not in a continuous spectrum.) 2) The photoelectric effect showed that light behaved as a ... 0 Consider a Hamiltonian which is translationally invariant. For example, H = \frac{ \hat{p}^2}{2m} = - \frac{1}{2m} \frac{ \partial^2}{\partial x^2}. There are other options (any Hamiltonian which does not contain the operator \hat{x} for example). Such a Hamiltonian forms quantum states which have definite momentum, i.e. they are eigenstates of the ... 0 Bound states have quantized energies, while unbound states have continuous energies. This can be understood by thinking of, for example, the 1D infinite square well. You can think semiclassically of the particle "bouncing back and forth" between the walls of the square well potential. At most wavelengths, the reflected particle will interfere with itself and ... 4 Technically, both solids and gasses have quantized energy levels. The difference is that molecules of a gas interact with other molecules very weakly, so the energy levels observed in emission or absorption of a collection of gas molecules are almost exactly the same as the energy levels that would be observed if you had a single gas molecule in isolation. ... 2 In quantum mechanics the equation of motion is the Schrödinger equation$$ i\hbar\,\frac{\partial}{\partial t}\,|\psi\rangle = H|\psi\rangle  where the (self-adjoint) operator $H$, the Hamiltonian, determines its evolution. The energy levels are, by definition, the eigenvalues of such operator in its domain of definition $\mathcal{D}_H$. Spectral theory ...

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If the energy levels are continuous (within a given interval of energies), then a particle (or system) can in principle have any energy in that interval. If they are quantized into say $E_1,E_2...$, then a particle (or system) can have only one of those energies, and not anything in between them.

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Would a photon be released due to changes in the electron orbital energy in this case or would it be transparent or absorbed elsewhere in other state reconfiguration? If something causes a vacancy in an electron inner shell, there can be a photon emitted. Two processes related to nuclear decay in which this often happens are electron capture and ...

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This would certainly result in a change in energy levels if the nucleus changes charge; in addition, you would expect a electron ejected as well to remain overall neutral (though there are some stable ions). The simplest quantum model that would give you an idea of the energy changes for an atom are described by so-called "Hydrogen-Like" atoms. The ...

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The obvious Google search finds various articles on the subject, including this one that has a graph of excitation lifetime against temperature: The lifetimes vary from about 600$\mu$s to about 3ms, so a 5 kHz signal (200$\mu$s) would indeed appear steady.

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The accepted airglow answer might be technically true, but it does not answer the question! The existence of an additional and very faint source of green light in the atmosphere does not explain the absence of the green light in the sunset sky gradient. I wasn't satisfied with other answers either. The only satisfactory answer I could find is this one. ...

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Soft x-ray optics typically uses grazing reflections on suitable mirror materials and gratings (e.g. Pt coated optics) or crystals and the detection can be done with scintillators using CCDs or PMTs. Commercially available instruments have approx. 0.2eV resolution (2000 lines) while research grade instruments can achieve much higher resolutions e.g. 30000 ...

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It is possible to measure wavelengths of light to many decimal places. When you see accurate determinations of atomic energy levels, they were done spectroscopically, looking at absorption or, more commonly, atomic fluorescence. Since $E=hc/\lambda$, one can accurately convert between wavelength and energy. When excited, atoms emit many wavelengths of ...

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