3
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The hydrogen atom gets the spectrum it has because you analyze the Schrodinger equation in spherical symmetry with the potential given by $V=-\frac{1}{4\pi\epsilon_0}\frac{e^2}{r}$. Yet the same potential term can be given by 2 masses in a gravitational field, where you replace the charges with masses and the constant also changes.

Consider if we want to achieve an orbit radius equal to the Bohr radius. In that case, we need to make sure that the mass of the nucleus is corrected such that

$$GM_n*m_e=\frac{1}{4\pi\epsilon_0}e^2$$

Doing that calculation you get a mass of the nucleus of about 2.08e9kg. Using the formula for the Schwarzchild radius you get that this corresponds to a black hole with radius 3.01e-18m. So that is much smaller than the dimensions of a usual nucleus but that is not a major problem (or at least I think so)

Would such a gravitationally bound "hydrogen" atom be possible or are there other variables which play a role and make this configuration unstable?

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  • $\begingroup$ Define stable, because this "atom" will change as the black hole evaporates. $\endgroup$ – Triatticus Aug 22 at 21:14
  • $\begingroup$ From wikipedia the approximate time for black hole evaporation is given by $t=\frac{480c^2V_0}{\hbar G}$, and putting the values for the back hole in question you get about 6.94e11s. which is a fairly long amount of time to be taken as infinite. By stable I mean if there are any other mechanism outside of black hole evaporation which might make it collpase. $\endgroup$ – Francesco Aug 26 at 15:43
  • $\begingroup$ That's only around 21k years, I wouldn't call that stable...there are half-lives longer than that one and it will all evaporate rather than half at a time. But other than that I can't think of much else $\endgroup$ – Triatticus Aug 27 at 6:44

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