In the Bohr's Atomic model, we have assumed the centripetal force to be provided by the electrostatic force between the proton and electron and derived the radius, energy of orbit and the velocity of the electron.

However if the electron is not stationary, how have we used Coulomb's Law for finding the electro 'static' force?

  • $\begingroup$ This is an obsolete, useless and semiclassical model. Who cares about the consistency of its assumptions? They were no longer relevant, once Pauli's article on the spectrum of the Hydrogen atom came out. $\endgroup$
    – DanielC
    Commented Sep 6, 2017 at 12:52
  • 2
    $\begingroup$ @DanielC Obsolete and semiclassical, yes, but certainly not useless. The fact that it has been superseded by better theories does not change the fact that it is a useful pedagogical step en route to quantum mechanics, it's a nice mental picture when considering phenomenon of diamagnetism, and it's my go-to when I forget the formula for the Bohr radius. $\endgroup$
    – J. Murray
    Commented Sep 6, 2017 at 14:36

1 Answer 1


For the case of the Bohr model of the hydrogen atom, the use of electrostatic forces is well justified. Electrostatics does not demand that all charges remain completely stationary, or it would be entirely useless as a theory (given how it describes forces, which are only meaningful when they impart accelerations). Instead, it only requires that any charges move slowly enough, which for the case of electrostatics essentially means that

  • all velocities need to be a good deal smaller than the speed of light, and
  • any movement needs to be slow enough that retarded fields can be replaced with their instantaneous versions; i.e. one must be able to assume that the propagation of any changes in the fields, at the speed of light, is instantaneous.

The two requirements are slightly different, and therefore independent: as an example of a situation that breaks the latter condition, consider two charges oscillating on Hookian springs, with period $T$, held a distance $L$ apart. If $L$ is comparable to, or bigger than, $c\,T$, then the electric forces from either charge will be appreciably retarded when they get to the other one.

In the Bohr model, both requirements are fulfilled. The electron's velocity at the fastest orbit is $v=\alpha c$, where $\alpha\approx 1/137$ is the fine-structure constant, so $v$ is a good deal slower than $c$, and retardation effects are negligible.

That said, it's important to remark that the considerations above are not exclusive to the Bohr model, and they are also relevant in quantum mechanics proper. The Bohr model is normally held to be obsolete, since the eigenvalues of the Schrödinger equation $$ \left[ -\frac{\hbar^2}{2m_e} \nabla^2 - \frac{e}{r} \right] \Psi(\mathbf r) = E \Psi(\mathbf r) \tag1 $$ fully describe the observed spectrum, within a coherent and expandable formalism. However, the critique over the use of electrostatics is equally valid of the Schrödinger equation as phrased in $(1)$ above: why can we take that far-off-relativistic stance for mechanics where the electron moves? The answer is the same as for the quasi-classical Bohr model: because it doesn't move too much. There are relevant relativistic corrections, known as the fine structure of atomic spectra, that account for the fact that electrostatics isn't a perfect description of the physics, but they remain corrections on top of a largely correct theory.


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