The Bohr's postulate of quantisation of angular momentum can be written in a formula as $mvr = nh/(2π)$

where $m$ is mass of electron

$v$ is its velocity

$r$ is the radius of that shell

$h$ is Plank's constant

So, I was wondering that if, I want to find where approximately the thing with $n=2$ exists, or in other words, what is the velocity of electron and radius of its circular path in the second shell, I could rewrite it as

$vr = nh/(2mπ)$


$vr = constant$

So if I was to just increase the radius by a factor of say $z$ and velocity by a factor of $1/z$ it would give me the exact same value for $vr$ every time. Doesn't that mean I could indefinitely increase the radius and keep reducing the velocity and still stay in second shell forever?

  • 2
    $\begingroup$ There are (were) additional constraints on the velocity because of classical mechanics: en.wikipedia.org/wiki/Bohr_model#Electron_energy_levels Anyway, I'd really discourage reading too much into the Bohr model. It is severely outdated; perhaps useful to know because it provides some historical perspective for the development of quantum mechanics, but not useful to analyse in a serious fashion beyond the fact that it didn't work. $\endgroup$ Sep 20, 2019 at 8:38
  • $\begingroup$ If the parameter $n$ is chosen, the velocity $v$ and the radius $r$ are defined and cannot be changed $\endgroup$
    – Maurice
    Sep 25, 2021 at 6:41

1 Answer 1


What you're missing is that the electron (or rather, Bohr's version thereof) moves in a curved path because of the centripetal force provided by electrostatic attraction to the proton. You therefore have

$mvr=nh/2\pi$, quantized angular momentum


$mv^2/r=\epsilon_0e^2/r$ centripetal force from electrostatic attraction

For each quantum number $n$ we have just one shell radius $r$ and one speed $v$ meeting both requirements.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.