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It is from my homework question (my question is not about the answer of the question),

In the hydrogen atom $H$, an electron and a proton attract each other with the electric force and form a bounded system. The motion of the charges in the $H$-atom is quite similar to the satellite motion we have seen in the gravitational problem. Both forces obey the inverse-square law ($\text{Force}\; \alpha\; 1/r^2$). As a result, Kepler's laws also apply to the $H$-atom. Since proton is much heavier than the electron, it can be approximated to be at rest at its location and the electron is rotating around it. Consider a circular motion with radius r.

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Question: Derive the expression for the period $T$ of the motion as a function of $r$. Is the Kepler's Law of periods valid? Derive an expression for the orbital speed $v$ as a function of $r$. How does the speed scale?

My question:

Since there is only one electron in the orbit, how I can verify if Kepler's law is valid or not?

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Micheal Brain Hurts
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  • why downvote guys?? – Micheal Brain Hurts Feb 24 '18 at 18:59
  • Would Kepler's Laws of planetary motion apply if there was only one planet around the Sun? As a side comment, this is a stupid problem because the hydrogen atom is a quantum mechanical system and orbits are NOT defined. – Bill N Feb 24 '18 at 19:01
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    The question tells you exactly how to proceed: "Derive the expression for the period $T$ of the motion as a function of $r$." That is, figure out what the period would be for different radius circular orbits – dmckee --- ex-moderator kitten Feb 24 '18 at 19:18
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    @BillN while I don't fully disagree with your sentiment, note that highly-excited hydrogen in a non-stationary state can be very close to a classical system (i.e. very localized electron cloud orbiting a nucleus). This is already visible for states with $\langle n\rangle\sim 2000$. See e.g. this post of mine. In general, don't forget about the correspondence principle. – Ruslan Feb 24 '18 at 19:32
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    Wait, I don't get the hate on this question. Asking students to apply classical approaches to quantum problems is an excellent way to illustrate why we need quantum approaches, because classical methods fail. – Melvin Apr 25 '18 at 04:53
  • ... by comparing different hydrogen atoms. – Roland Puntaier Aug 24 '20 at 15:13

2 Answers2

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Let me kick off with a clarification: the problem you've been set has nothing to do with the hydrogen atom, which obeys quantum mechanics instead of classical mechanics and for which you cannot really ignore the radiation effects imposed by electrodynamics (as opposed to the electrostatics you're being asked to consider) on accelerating charges.

Nevertheless, the task you've been set is quite simple: consider a system of two particles that experience a mutual electrostatic attraction and which will therefore orbit each other in Keplerian orbits. Because the problem is identical to the Kepler problem, that means that the period of the orbits will depend on the semi-major axis of the ellipse: if you set them orbiting at some distance $a_1$ they will have some period $T_1$, and if you set them orbiting at some other distance $a_2$ then their orbital period will change to $T_2$, with a strict relationship on those four quantities imposed by the dynamics of the problem. You're being asked to describe that relationship, i.e. a two-body property that does not at all require multiple other electrons to show itself.

Emilio Pisanty
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    It's a bit unfair to say that this problem has nothing to do with the hydrogen atom. You can, for example, get a good estimate of the expectation value of the electron speed from classical arguments. – garyp Feb 25 '18 at 22:59
  • @garyp - I agree with your comment. You are probably thinking of the Ehrenfest theorem. – freecharly Feb 26 '18 at 02:27
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The question obviously wants a classical treatment of a negative charge orbiting an equal magnitude positive charge with assumed infinite mass in a circle. Classically, this single negative charge can orbit the positive in a circle for any given radius $r$ depending on the initial conditions. Thus you can apply Kepler's third law which relates the squares of the orbital periods $T_1$ and $T_2$ to third powers of the half-axes $a_1$ and $a_2$b of the assumed elliptical orbits: $$T_1^2:T_2^2=a_1^3:a_2^3$$ In the present case, your ellipses are circles and these half axes are just the radii $r_1$ and $r_2$ of two different circular paths. Thus Kepler's law relates the periods of the two orbits to the radii of these orbits: $$T_1^2:T_2^2=r_1^3:r_2^3$$ You can check this for your derived formula.

freecharly
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