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Following Garyp's comment; "r , being the radial coordinate, has no knowledge of direction. Another way to see this is to note that the dipole matrix element has to be a vector. Calculating the rr matrix element gives you just one number. Perhaps more to the point: you want to find the matrix element of r⃗ r→, not rr. – garyp" The dipole element should be; ...

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There exists something called Kramers's recursion rule and I think it is what are you looking for. $\frac{k+1}{n^2} \left\langle r^k \right\rangle - \frac{a_0}{Z} \left(2k+1\right)\left\langle r^{k-1} \right\rangle + \frac{k a_0^2}{4Z^2} \left( \left(2l+1\right)^2 - k^2 \right) \left\langle r^{k-2} \right\rangle$ , where $k$ is integer and $a_0$ Bohr ...

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It's possible to imagine living in a different universe where most nuclei of the element with charge 1 were deuterium, and the lighter protium was the rare outlier. However, we don't live in that universe. Most of the ordinary matter in the universe is hydrogen (75% by mass) and helium (25% by mass) which has been unprocessed since the Big Bang. Deuterium ...

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The interesting thing is that the electrolysis of water to produce hydrogen is a possible way of storing energy when the electricity is produced by renewable energy sources which have a variable output eg wind turbines and solar cells. There is currently a lot of research being done at the National Renewable Energy Laboratory (NREL) and many other sites ...

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First, the idealized clean non-relativistic problem with the $1/r$ Coulomb potential is integrable: the exact wave functions and energy eigenvalues may be precisely written down using elementary functions. It is exactly true that the eigenvalues of the bound states are $-13.6\,{\rm eV}/n^2$ etc. Now, the first simple correction is the proton motion. This is ...

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It is perfectly possible to produce hydrogen from electrolysis. It's theoretically simple and easy to understand and remember. The only reason why it's not used most of the time is that even more efficient methods exist. Electrolysis has a very low efficiency, most of the energy ends up being converted to heat. These days, 95% of the hydrogen is produced by ...

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The lamp does indeed contain molecular hydrogen but: the discharge dissociates (a small proportion of) the $H_2$ molecules to produce excited hydrogen atoms the emission from molecular $H_2$ is quite weak under these conditions, though it can can sometimes be observed as braod bands between the blue, green, and red lines hydrogen atom lines. So the ...

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