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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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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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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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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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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 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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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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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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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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