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The theory behind the trick is based on the Hellmann-Feynman (HF) theorem $$\frac{dE_{\lambda}}{d\lambda}~=~\langle \psi_{\lambda} | \frac{d\hat{H}_{\lambda}}{d\lambda}| \psi_{\lambda} \rangle,\tag{A}$$ which works with a single derivative, but not with a square of a derivative, cf. OP's failed calculation (5) for the expectation value $\langle\frac{1}{r^2}... 4 The short story is that the Hyperphysics article you link to is using classical and semiclassical heuristics to justify the numbers it present. As I'm sure you're aware, the hydrogen atom cannot be described in any rigorous detail using classical mechanics, and instead requires quantum mechanics for any appropriate treatment, particularly where the ... 3 In a comment elsewhere you write that you're interested in understanding how quantum-mechanical theory describes the radiation that a hydrogen atom does and does not emit. In your question you ask about another answer that suggests some significance to the electron having zero total momentum; I think that's a feature of the coordinate system choice rather ... 3 The time-independent Schrodinger equation$\hat{H} \psi = E \psi$only holds when the Hamiltonian does not depend explicitly on time. If you start with a time-independent Hamiltonian and make a time-dependent gauge transformation, then the new Hamiltonian will depend explicitly on time, and there is no reason to expect that the (time-dependent) eigenvalues ... 1 Electrons and nuclei both have spin. A spinning charged particle has a magnetic dipole moment. When a magnetic dipole is in a magnetic field, it experiences a force. This oversimplified description gives some brief intuition on the origin of the fine and hyperfine splittings. Fine structure is due to the interaction of the electron's spin with the ... 1 The most probable radius is found by maximising probability per unit radius, whereas$|\psi^2|$gives us probability per unit volume. To find the conversion factor from one to the other, we need to ask how much volume is there per unit radius near a radius$r$? The answer to this is$4 \pi r^2\$.