Suppose we have a proton ($\require{mhchem} \ce{H^+}$). Suppose an electron comes to orbit the proton and starts to fall inwards. Suppose we know the exact position of the electron when it starts to orbit the proton. How long does it take until the electron is in a quantum state around the proton?

Does this change for larger atoms like fluorine and very large atoms such as plutonium?

  • 2
    $\begingroup$ Electrons are already in quantum states. High precision of position means high uncertainty of momentum. Effectively, by the very act of knowing its position you will fling your electron at unknown direction, and that with great force; most likely it will miss the proton and never settle around it. $\endgroup$
    – Ivan Neretin
    Commented May 9, 2016 at 21:34

2 Answers 2


Your question is null because electrons are always quantum objects and your question is based in classical mechanics.

The electron is a quantum object always. It does not behave like a particle when on the loose and then like a wave when captured by a nucleus. It has quantum weirdness all of the time. Thus, it needs to obey the uncertainty principle:

$$\sigma_p \sigma_x \ge \dfrac{\hbar}{2} $$

The product of the standard deviation (uncertainty) in momentum $\sigma_p$ and the standard deviation of the position $\sigma_x$ must be greater than or equal to the reduced planck constant $\hbar=1.054\times 10^{-34}\ \mathrm{J\cdot s}$ divided by 2. Since we know the position absolutely, $\sigma_x=0$. Let's determine the uncertainty in the momentum, but before we substitute in, let's do some symbolic maths:

$$\sigma_p \sigma_x \ge \dfrac{\hbar}{2} \\ \sigma_p=\dfrac{\hbar}{2\sigma_x}$$

Now we substitute in, and... $$\sigma_p=\dfrac{\hbar}{2\sigma_x}=\dfrac{\hbar}{0}=\mathrm{undefined}$$

However, we can use the limit approach and determine:

$$\sigma_p =\lim_{\sigma_x \to 0}\left( \dfrac{\hbar}{2\sigma_x}\right)=\infty$$

Thus, if we know the position of an electron super-specifically, we know nothing about the momentum. This is a problem because momentum is a product of mass and velocity. Since the mass of the electron is a constant, if we know nothing about the momentum, we know nothing about the velocity, which we need to determine a time lapse.


As the other answer has pointed out the electron is a quantum mechanical entity as well as the proton, obeying the Heisenberg uncertainty principle, HUP:


This does not mean that your question:

How long does it take for an electron to be described in a quantum state?

supposing it means the transition time from a free electron to a bound one, is meaningless. One has to use quantum mechanical equations whose solutions give the probability of the transition when an electron approaches a free proton, which depends on momentum, energy etc, as in classical physics. The difference is that quantum mechanics can only predict probability distributions, not tracks. So one has a probability for the electron to fall on a free energy level of the putative hydrogen atom.

In this link the Bohr model is used because particularly for the hydrogen atom its solutions in energy levels coincide with the solutions of the Schrodinger equation, but one has to keep clearly in mind that one has orbitals, not orbits, probability loci.


Cross-section of computed hydrogen atom orbital (ψ(r, θ, φ)2) for the 6s (n = 6, ℓ = 0, m = 0) orbital. Note that s orbitals, though spherically symmetrical, have radially placed wave-nodes for n > 1. However, only s orbitals invariably have a center anti-node; the other types never do.

This probability in energy width for the Bohr atom lines, gives a lifetime for the interaction electron - proton with the electron emitting a photon and settling at an energy level. Using the energy form of the HUP one can have a lifetime for the electron-proton binding.

Does this change for larger atoms like fluorine and very large atoms such as plutonium?

Each potential well of individual atoms has distinct energy levels for the electronic orbitals, and widths that will depend on the specific potential solutions. These will give different widths for the lines, but in general the order of magnitude is constrained by the electromagnetic constants which are common to the solutions of all atoms.


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