I have this very simple function:

function calculateElectrostaticForce(distance) {            
        const charge1 = 1.602176634e-19; // Elementary charge in coulombs (C)
        const charge2 = 1.602176634e-19; // Elementary charge in coulombs (C)

        // Calculate the electrostatic force based on Coulomb's law
        const vacuumPermittivity = 8.854e-12;
        const electrostatic_force = (charge1 * charge2) / (4 * Math.PI * vacuumPermittivity * Math.pow(distance, 2));

        return electrostatic_force;

And lets say I use it to calculate force between proton and electron at 1.3 angstroms distance (1.3 / 10000000000 in meters). That would give me 1.3651267795407353e-8 Newtons. If divide that with, for example, the mass of Hydrogen atom to get acceleration, its 8110823001207866000 m/s^2 what is a very extreme number for atomic interactions. It rises the question, if Columb's law can be useful at such small/atomic scales?

  • 1
    $\begingroup$ These accelerations is what you indeed get for the keplerian orbits in the Bohr model and the old quantum theory. Quantum mechanics had to explain why these accelerations do not need to rapid loss of energy due to radiation by an accelerating charge. It was one of the crucial problems. The relativity corrections are not too important, hydrogen can be quite well modelled with the non/relativistic quantum mechanics. $\endgroup$ Commented Mar 17 at 9:12

2 Answers 2


The other answer correctly pointed out that using $F=ma$ and the mass of the hydrogen atom does not give the exact $a$, and even if one uses the correct relativistic formulas and the correct mass, one gets the expectation value of $a$, not $a$ itself. However, the OP's procedure should still lead to a correct order-of-magnitude estimate of the expectation value of $a$, and I believe it is a meaningful question to ask why the expectation value of $a$ seems so large.

Firstly, as for the qualitative correctness of OP's procedure: the average velocity of the electron in a ground-state hydrogenic atom with nuclear charge $Z$ is on the order of $Z\alpha c$, where $\alpha$ is the fine structure constant and $c$ is the speed of light. Plugging in $Z$ yields an order-of-magnitude estimate of $0.01c$. Therefore, the relativistic corrections to the ground state hydrogen atom wavefunction is on the order of $0.0001$. The relativistic corrections to the proton is much smaller due to the high mass of the proton, which gives the proton a much smaller velocity. Moreover, although using the mass of the hydrogen atom is not the correct procedure for calculating the acceleration of the hydrogen atom (which is 0 without external forces), using the mass of the proton gives the correct (expectation value of) acceleration of the proton, and the masses of the hydrogen atom and the proton are very similar. Therefore, one can make the OP's question rigorous by asking instead why the order of magnitude of the expectation value of the acceleration is so large. I believe this is what the OP really wants to ask.

Secondly, as for why the large acceleration should not be a surprise: atomic movements happen on extremely short timescales (a few tens of femtoseconds, i.e. $10^{-14}$ s), and under the SI units the timescales are even "smaller" in number than the length scales ($10^{-10}$ m). Since the acceleration has a unit of $m/s^2$, we should expect that the acceleration is on the order of $10^{18} m/s^2$, in qualitative agreement with the OP's value. The crucial thing to note is that atoms are much, much more resistant to huge accelerations that would tear us humans apart (humans find it difficult to withstand $20g \approx 200 m/s^2$ already). The extremely short timescales on which atomic movements occur, as well as the fact that the acceleration scales as the inverse square of time, both contribute to the surprisingly large acceleration felt by atomistic particles.


"It rises the question, if Columb's law can be useful at such small/atomic scales?"

Coulomb's law still plays a role in the Hamiltonian of an atom, but your discussion of "force" and "acceleration" along with your use of F = ma, is only valid for macroscopic objects that are moving much more slowly than the speed of light (i.e. classical physics). Correctly describing an electron interacting with a proton requires both quantum mechanics and special relativity (and more if you would like your numbers to be so accurate that they would also correctly account for the effects of gravity, but not even the most knowledgeable researchers in the world will be incorporating the effects of gravity for such quantum mechanical systems).

There is a lot of information about the hydrogen atom available here, including the "Hamiltonian" that I mentioned in the opening sentence of this answer (see the section about the Schroedinger equation and search for the word "Hamiltonian") and you will see:

$$\tag{1} -\frac{\hbar^2}{2\mu}\nabla^2 - \frac{e^2}{4\pi \epsilon_0 r}. $$

You can see that the formula that you used (except for the exponent of $r$, since the Hamiltonian uses the "potential" rather than the "force"), is used in this Hamiltonian. However, if we want to know the acceleration, we do not just divide this by a mass. In quantum mechanics, we cannot predict the "acceleration" at any given time, only the "expectation value" of the acceleration, $\langle \hat{a} \rangle$, which tells you what the acceleration would be on average if you were measure the acceleration multiple times (in the theory of quantum mechanics, it is predicted that the acceleration won't be the same every time that you measure it). Finally, in this case, the electron is moving at close enough to the speed of light, that I would not trust the expectation value of the acceleration, that is obtained from the above-mentioned Schroedinger equation, and this problem is improved by incorporating the effects of special relativity using the Dirac equation. If you want to know where the "Coulomb formula" appears in the Dirac equation, it might be better for you to ask that in a separate question.

Another point that I will make, is that dividing by the mass of the hydrogen atom, would probably not even be considered the correct way to get the acceleration in classical (non-quantum-mechanical, and non-relativistic) physics. The hydrogen atom is the entire system, whereas the force that you calculated is between the proton and electron inside that system. The hydrogen atom itself is stationary, but the electron and proton will move relative to each other, so if you want the acceleration of the electron you can divide by the mass of the electron, or if you want the acceleration of the proton you can divide by the mass of the proton, or if want the acceleration of the relative motion you can divide by the reduced mass of the system.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .