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If you treat the 1s ground state's probability distribution as a classical charge density distribution (not really accurate, but I think the simplest way to interpret the problem), then there isn't one. This state is spherically symmetric, so the electric field is always radial and depends only on the radial coordinate and not on the angular coordinates. So ...

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One has to distinguish between, on one hand, an orbit and an orbital motion, which are classical notions; and on the other hand, an orbital, which is a quantum mechanical notion, cf. above comment by dmckee. If the question is really Why quantum mechanics?, then have a look at e.g. this Phys.SE post and links therein. Here we will assume that OP accepts ...

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Electron in a ground state hydrogen atom has zero angular momentum $L^2$, l=0. Moon has a huge angular momentum. Therefore it is a poor comparison. If moon would have zero angular momentum, in classical physics, it would fall down and hit earth. Electron in an hydrogen atom, in l=0 state gets constantly pulled to the center, but this is countered by the ...

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I was looking through some NIST atomic data for the Balmer Series from here: http://www.nist.gov/srd/upload/jpcrd382009565p.pdf It lists the spontaneous emission rates for the Balmer series as follows: $\lambda = 656 \text{ nm}$, $A_{32} = 4.41\text{e+}7\text{ s}^{-1}$ $\lambda = 486 \text{ nm}$, $A_{42} = 8.42\text{e+}6\text{ s}^{-1}$ $\lambda = 434 ... 0 Note that$\Delta p_x \Delta r$does not satisfy the uncertainty principle in the strict sense since$r$is not conjugate to$p_x$(or$p_y$and$p_z$). Instead you can consider$\Delta p_x \Delta x$. The ground state of the hydrogen atom is $$\psi_0(r) = \frac{1}{\sqrt{\pi a^3}} e^{-r/a},$$ where$a$is the Bohr radius. First of ... 1 One has to keep in mind 1) that it is the complex conjugate square of the eigenfunction that gives the probability of finding the electron with energy E at a specific radius. 2) There are no fixed orbits in the quantum mechanical solution, only a locus of probability called orbital 3)orbitals overlap in space, it is the energy that is keeping the electron ... 0 So the matrix element I tried to calculate is indeed zero for the dipole moment. In order to find the hyperfine splitting, one must calculate$|\langle a_f|\mu|a_i \rangle|^2$. 1 This answer gives an analytical approach for diagonal matrix elements. First of all, since$r^k$is spherically symmetric you can immediately integrate the angular parts: $$\left< n' l' m' | r^k | n l m \right> = \left< n'l \right\| r^k \left\| nl \right> \delta_{l',l} \delta_{m',m},$$ where$\left< n'l \right\| ...

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This looks to be a messy calculation for the most general case, and a direct attempt based on special-functions properties of the Laguerre wavefunctions is likely to simply falter and die in the not-quite-right forms of the integrals. These matrix elements are calculated in terms of recursion relations in Matrix-element calculations for hydrogenlike ...

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Given that the eigenstates of the hydrogen atom can be separated into a tensor product of angular momentum eigenstates and a radial solution, the Wigner-Eckart theorem helps to calculate the matrix element of any tensor-like operator (a scalar as $r^k$ is a trivial case) between states of the form $|lm\rangle$. The rest is an integral over the radial ...

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The presence of gamma rays and positrons associated with lightning strikes suggests that the strikes are capable of accelerating particles with enough energy for D-H or D-D fusion. However I'd expect that even in a pure deuterium atmosphere the energy released by D-D fusion would be negligible compared to the energy released in the lightning strike itself.

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