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If we solve the time-independent Schrödinger equation for any atom by considering only the electrostatic potential, an electron has the same probability of occupying the $p_{x}$ orbital as it does the $p_{y}$ and $p_{z}$ orbitals, because the electrostatic force is a central / spherically symmetric force which depends only on $r$, and the wavefunctions of the $p_{x}$ , $p_{y}$ , $p_{z}$ orbitals only differ in terms of the angular coordinates $(\theta,\phi)$.

However, a nucleus has a quantum spin which introduces a magnetic field, and the symmetry is broken. If we take into account the magnetic field of the nucleus, then the electromagnetic force which the nucleus exerts on electrons occupying the $p_{x}$, $p_{y}$, and $p_{z}$ orbitals is no longer central. So there should be a difference between the energies of the wavefunctions of the $p_{x}$, $p_{y}$, and $p_{z}$ orbitals.

How significant is this effect? In atoms with a small atomic number or with a small nucleus spin the effect is negligible and can be only observed near absolute zero. However shouldn't it become significant for atoms with bigger atomic number or with a large nuclear spin?

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  • $\begingroup$ The degeneracy is only valid for one-electron atoms/ions. $\endgroup$
    – ananta
    May 16, 2023 at 21:00
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    $\begingroup$ As I understand it, when you start taking interactions like these into account (what you're describing here is hyperfine splitting), it's no longer valid to refer to $p_x$, $p_y$, $p_z$ orbitals anymore (which, as you note, are solutions to the Schrodinger equation when you neglect such factors). But yes, my understanding is that there is a lifting of degeneracy. (See the linked article.) || @ananta That's not true, what you're describing sounds more like degeneracy within a given $n$, and OP is asking about degeneracy within a given $\ell$. $\endgroup$ May 16, 2023 at 22:09
  • $\begingroup$ Should we be asking questions about $p$ orbitals when all we have in one-electron system is an $s$ orbital? $\endgroup$ May 17, 2023 at 9:33
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    $\begingroup$ You can excite the electron of a one-electron atoms into a $p$ orbital or any orbital at all. We can distinguish all the subshell states in the hydrogen spectrum when fine splitting (relativistic, nuclear spin interactions, etc) is taken into account. $\endgroup$ May 17, 2023 at 17:32

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