If electrons make no transition to any of the other orbital or a shape, then what is the shape pf the path it makes while revolving around the nucleus?

Does it have to do anything with the equipotential surfaces.

My approach to the question is in this way: If electron does not change chape or orbital it means that it doesn't release energy or the nucleus does no work on the electron. So it means electron is revolving on an equipotential surface of the shape of a sphere. Am I correct?

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  • $\begingroup$ "If electron does not change shape or orbital it means that it doesn't release energy" This is correct. However an electron is also a charged particle, so if it accelerates (as it must in order to follow a curved trajectory) then it does release energy in the form of EM waves. This paradox was one of the major clues in developing QM. $\endgroup$
    – The Photon
    Jun 22 at 15:14

1 Answer 1


While it is a useful concept for to use in the introduction to quantum mechanics, the planetary model of the atom is quite false.

The electron is not a point like particle literally orbiting the nucleus in an atom, but rather should be thought of as a wave $\psi(\boldsymbol r)$ spread out over a certain region around the nucleus. The probability of finding an electron in a volume $dV$ is given as $|\psi(\boldsymbol r)|^{2} dV$.

The actual form of the wavefunction $\psi(\boldsymbol r)$ is found by solving the Schrodinger equation for that particular atom. Now, if the planetary model of the atom was true and we could think of the electron existing in a particular shell at some constant radius from the atom then sure, we could say that the orbital was across an equipotential region. However, the wavefunction $\psi(\boldsymbol r)$ does not strictly occupy one region and is spread out across more of space. Therefore, orbitals are not equipotential surfaces.


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