0
$\begingroup$

Why don't the electrons revolving around the nucleus in the Bohr's atomic model lose energy? I mean don't they accelerate because of which they should emit radiations. I don't get this concept. can someone help me out with this?

$\endgroup$

marked as duplicate by John Rennie, ACuriousMind, Qmechanic Sep 6 '15 at 11:56

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 3
    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/20003/2451 , physics.stackexchange.com/q/9415/2451 and links therein. $\endgroup$ – Qmechanic Sep 6 '15 at 8:16
  • $\begingroup$ The difficulty with the proposed duplicates is that most of the answer don't really address the misconception here. That is, they don't talk about the importance of the vanishing second derivative of $\langle X \rangle$ with respect to time (mine is as guilty as any other in that regard). $\endgroup$ – dmckee Sep 6 '15 at 20:51
0
$\begingroup$

The electron according to the bohr model loses energy but its not continuous. The energy is lost in discrete quantities also known as a quanta.

$\endgroup$
  • 2
    $\begingroup$ This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post - you can always comment on your own posts, and once you have sufficient reputation you will be able to comment on any post. $\endgroup$ – Hritik Narayan Sep 6 '15 at 9:56
  • $\begingroup$ I guess i have given a pretty precise answer and if the op needs further clarification i am in for that too. You wont be the one deciding whether i am right or wrong and secondly i will be reading the guidelines of the forum and will follow them. $\endgroup$ – Tajammul Saleem Sep 6 '15 at 10:08
  • $\begingroup$ Tahammul, no you haven't said why. Neither did Bohr. He simple said if we assume they have to obey this discrete energy loss relationship (for some unexplained reason) then we get a model with some features that agree with certain other observations. The "Why?" question can be answered with reference to a more complete quantum mechanical theory, but you have to abandon much of the thinking that goes into the Bohr model in the first place (no great loss there: the Bohr model gets most things wrong anyway). $\endgroup$ – dmckee Sep 6 '15 at 20:46

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