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The simplest reaction deuterium and tritium. Tritium is common in big labs (like NIF, JET, Omega) [1]. Tritium sucks - practically speaking. It is expensive, radioactive, and hard to stockpile. Omega spent millions and years on a tritium facility. It may even never be used in fusion power [2]. The next easiest reaction is deuterium with itself. This ...


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The expression you have there looks like that of the electron relative to the proton.


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The problem here is that you're looking at the magnetic field $\textit{at the proton}$. Using this approach you can't derive the spin-orbit coupling that you want. Because of the symmetry of the system you would expect the magnetic fields at each particle to have the same magnitude, but the energy of spin-orbit coupling comes from the $\textit{electron's}$ ...


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Note: ChocoPouce's answer is the same as mine but is more mathematical. You have a (spherically symmetric) probability density distribution $\rho$ in space (which we get from the square of the amplitude). The "radial probability density" is roughly the chance that the electron is at a given radius, say $r = 0.1\mathrm{nm}$? In other words, how much of this ...


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The link uses a spherical shell element which is $4\pi r^2 \mathrm{d}r$ and has the dimension of a volume ($r^2$ is a surface and $\mathrm{d}r$ is a length. The wavefunction of the ground state is spherical, if it weren't the calculation should have been made using a spherical volume element such as $r^2\mathrm{d}r\mathrm{d}\theta\mathrm{d}\varphi$. There ...


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This calculation agrees with experimentally measured spectral lines, but why would we expect it to be true, even if we accept that the electron moves according to the Schrodinger equation? Your puzzlement arises because you are putting the cart in-front of the horse. The cart is the theoretical model of quantum mechanics and the horse is the data. As ...


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This calculation agrees with experimentally measured spectral lines, but why would we expect it to be true, even if we accept that the electron moves according to the Schrodinger equation? After all, there's no particular reason for an electron to be in an eigenstate. Good question! The function $\psi$ does not need to be Hamiltonian eigenfunction. ...


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To have emission (or absorption) of photons you must have a Hamiltonian that includes those degrees of freedom also. If your system consists of (a) the electromagnetic field and (b) a hydrogen atom, you can specify the state with (a) for each frequency, the number of photons with that frequency and (b) the state of the hydrogen atom, in your favorite way, ...


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Conservation of energy. If we measure the energy of an atom, we will always report an eigenvalue, because we are forcing it into an eigenstate (this is something like the quantum mechanical definition of measurement). Now suppose that we measure the energy of an atom twice, before and after it emits a photon. For conservation of energy to hold, the energy ...


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The idea here is increasingly complex depending on how deep into modern physics you want to delve, but also key to understanding quantum mechanics. So, I'll give a bit deeper explanation than it seems you've seen, but there's plenty more. It's understood that a photon acts both as a particle and a wave. As a particle it has an amount of energy associated ...


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Firstly, fusion doesn't happen in the way depicted in the question. Four protons don't participate in a 4-body reaction. Instead there are many intermediate steps: Each step has its own reaction rate. The overall reaction rate is determined by the rate limiting step. The proton-proton reaction is the rate limiting step in this case. It is important ...


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Three different ways (at least). 1) Use a detector, like a photodiode or photomultiplier, which responds to UV by producing electrons when hit by photons even if those photons are not visible, 2) Use a detector, like a bolometer, which measures the temperature rise produced when UV energy is absorbed, or 3) Coat a visible-only detector with a phosphor ...


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It's common to measure spectra in the non-visible region and UV or infra-red spectrometers are available off the shelf. Google for many, many manufacturers of the kit. Also see the Wikipedia articles on UV spectroscopy and infrared spectroscopy. Typically UV and IR spectrometers will use a diffraction grating to disperse the light and a photomultiplier tube ...


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Muon mean lifetime is 2.2 ┬Ás. There's your problem. Muons mass 105.7 MeV/c2, about 200 times that of the electron. If you wanted to ionize a hydrogen atom, you would need 13.6 eV. If you wanted to ionize a muonic hydrogen atom, you would need about 2813 eV or about a 0.441 nm photon. Start building your laser.


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The numbers are not measured, they are part of a model that explains why the hydrogen atom emits/absorbs only at those wavelengths. The first series of Hydrogen lines to be discovered was the Balmer Series, and nobody knew why they were discrete lines instead of a continuous spectrum. Johann Balmer discovered that the lines all had wavelengths equal to, ...


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How did Lyman discover his series in hydrogen atom? He was directed into spectroscopy by his advisor. At the time the equipment was pretty poor for spectral measurements and much of his time he spent trying to get good spectral wavelength measurements. Part of the measurement error ended up coining the term "Lyman ghosts" in the spectral lines due to ...


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The Wikipedia article answers most of your questions. What are the requirements for hydrogen atoms to go through fusion? Two atoms must overcome the coulomb barrier, which can be done by forcing two atoms very close together, or by leaving them moderately close for long periods of time, which allows them to tunnel through the barrier. Is it a ...


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A solid material is a lot more complicated than the hydrogen atom, but you can imagine a larger atom as being similar to a hydrogen atom with many electrons occupying the energy levels. The hydrogen energy levels are $E_n=-{13.6\,\text{eV} \over n^2}$, so you can see that as you go to higher energies, the difference between the levels gets smaller. For ...


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The equilibrium position of the molecule from the surface will be due to the tradeoff between a (relatively) long range Van der Waals force (or electrostatic force if the molecule is charged or dipolar) and the short range exchange repulsion. You would expect the minimum to be when the molecule touches the surface, so it would around the Van der Waals ...


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Solution of an initial value problem can be written as integral of the initial function $\psi_0$ multiplied by the propagator of the Schr. equation. Depending on the function $\psi_0$, the integral may or may not be calculable in terms of simple functions. I do not know of any initial function $\psi_0$ and potential $A(t)$ that would admit simple exact ...


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Express V and E as explict functions of r. $$V = \frac{-q^2}{(4\pi\epsilon_0)r}$$ $$E = \frac{-q^2}{2(4\pi\epsilon_0)r}$$ Also, the previous page of Sah emphasizes that 2E = V.



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