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We, as high school students have been taught that-because Bohr's model of an atom assigns specific orbits for electrons-that it is better than Rutherford's model. But what Rutherford failed to explain was why electrons don't emit EM waves & fall into the nucleus. I don't see how the introduction of 'atomic orbitals' overcame this defect. Can't it still radiate EM waves?

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Ron Maimon's answer is very good and I agree with him, but I would like to point out that Bohr's model is not completely consistent with classical electromagnetism. On the one hand, the electron does not radiate as dmckee and Ron have explained. But, on the other hand, in Bohr's model the electron is a pure particle in a closed orbit and it is therefore accelerating. So according to classical electromagnetism it has to radiate. Thus it seems to me that there is not goof agreement between Bohr's model and Maxwell electromagnetism.

In "modern" QM, the electron is however described by a probability wave function. And if one links the density current of probability to the density current of charge —what makes sense—, then classical electromagnetism tell us that the electron does not radiate because the density current is time independent in stationary states.

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Of course you are right, but when asking about history, I think it is good to be respectful of the historical authors. Bohr knew that the Maxwell theory had to be modified, and he had some ideas about how this works in QM with minimal damage to Maxwell. This was the failed 1924 Bohr Kramers Slater theory, which had classical EM waves causing distant transitions in atoms. It is essentially semiclassical QED without photons, one atom makes a field, another atom responds. Some of the ideas were unchanged in modern QM, but the lack of photons made it that energy couldn't be conserved. – Ron Maimon Aug 9 '12 at 8:44

Unfortunately, nobody reads Bohr nowadays so Bohr's arguments are not understood and transmitted. Modern quantum mechanics is more complete and superior as a physical theory to the old quantum mechanics, so the omission is perhaps understandable, but it is not forgivable.

Bohr's ideas explain the stability of H-atoms in a reasonable way, which is also correct in the modern theory. In modern quantum mechanics, it is obvious that there is a ground state for the electrons, because if you confine the electron to a box of radius a, it has a momentum of order ${1\over a}$, and therefore a kinetic energy of ${1\over a^2}$, while the negative potential energy is only ${1\over a}$, so there is a minimum distance, beyond which confining an electron costs energy. If you restore the factors of m and $\hbar$, this distance is a small multiple of the Bohr radius.

Bohr didn't have the uncertainty principle, but he made an argument which gives the same basic law. So Bohr explained the stability correctly in the context of semiclassical quantum theor, and his explanation is roughly isomorphic to the modern one.

Bohr's argument

The classical orbiting electron radiates EM waves. The frequency of the radiation emitted is equal to the classical orbital frequency (assuming small back-reaction, which is reasonable, since the electron is nonrelativistic), and the energy can be radiated in arbitrarily small increments.

In Bohr's model, the radiation emitted has to come in lumps, associated with discrete jumps between orbits of different frequency. The size of the energy lump is hf, where f is the frequency of the photon.

In classical mechanics (and for large orbits), there is only one frequency to emit with, and this is the orbital frequency (and integer multiples of the orbital frequency, corresponding to higher harmonics). But in Bohr's quantum mechanics there are two orbital frequencies associated with every jump, the initial frequency, and the final frequency, and the basic question is which frequency is the correct frequency of the emitted radiation?

The question has no real answer for small quantum numbers, since the semiclassical approximation breaks down. But Bohr assumed it is some average of the initial and final frequency. Let's say:

$$ f_0 + f_1 \over 2 $$

once you know the frequency of the outgoing radiation, you know the energy quantum, and therefore the energy spacing. Bohr used this rule to figure out the level spacings in the atoms, and he knew the rules were only correct for large quantum numbers.

The classical frequency of a classical orbit at radius R goes as Kepler's third law:

$$ f = {1\over T} \propto {1\over R^{1.5}}$$

The binding energy is of order the potential energy:

$$ E \propto {1\over R}$$

Note that E is the negative of the negative energy, so E is positive. Therefore the frequency f of the classical orbit obeys:

$$ f \propto |E|^{1.5}$$

Assuming we are at a high binding energy, and we make a transition from an initial frequency to the final frequency, the energy emitted is proportional to the average of the initial and final frequency, so the frequency of the emitted radiation is

$$ f= {1\over 2} (|E_i|^{1.5} + |E_f|^{1.5} ) $$

and this frequency average times h must be equal to the binding energy difference between the orbits:

$$ (E_f - E_i) $$

When $E_i $ is big enough, there can be no solution for $E_f$, because the frequency rises faster than the binding energy, and the two curves never meet. This means that there is a ground state, from which no radiation cannot be emitted.

This argument is superficially different from the modern argument, using uncertainty, so it is important to check that it is essentially equivalent in order of magnitude to the modern argument.

The modern argument says that the size of the p for small x is roughly reciprocal, so that if you consider a classical orbit witht the same energy as the ground state, the momentum and the position in the orbit are such that the area enclosed by the orbit in phase space is order h.

The frequency argument says that the spacing between adjacent levels obeys the rule

$$ \Delta E = h {2\pi\over T} = h {\partial H \over \partial J} \approx H(J+h) - H(J) $$

where J is the classical orbit, so it says that J is quantized in integer steps of h. The statement that there is a lowest energy level is the statement that a orbit close enough to the nucleus encloses an arbitrarily small area in phase space. This is not completely obvious, because the momentum is getting large, but it is true, because the momentum is not getting large fast enough to compensate for x getting small, so that the area enclosed is convergent.

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You are right, and quantum mechanic does not explain it ether. You will have to study quantum electrodynamics to have the picture right. It is a common error found in introductory books to say that classical mechanics failed to explain the stability of the electrons while QM did. The true story is that in classical mechanics, if one introduces the radiation emitted by the electrons then the orbit is instable and in QM ones does not introduce the radiation emitted by the electrons. However, the results obtained by QM were so astounding that people knew they were on the good tracks.

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-1: This is ridiculous--- in QM there is no lower energy than the ground state. No matter what the coupling to photons (so long as it is small, as is clear from the nonrelativistic nature of the wavefunction), then the introduction of dynamical photons does not change this fact. – Ron Maimon Aug 9 '12 at 1:50
Keep it polite boy, read first my answer and try to understand it before voting me down. I have written that one can not invoque the instability of electrons orbit found in classical physics to justify QM, since QM does not take into account the coupling to EM field. Where do you read anything about a lower energy state ? If you do not fear to shame yourself, you can write to my professor Mr.Claude Cohen-Tannoudji , 1997 Nobel Laureate who wrote four books on the subject: Mécanique quantique. 2 vols;Introduction à l'électrodynamique quantique;Processus d'interaction photons-atomes. – Shaktyai Aug 9 '12 at 6:31
@Shaktai: I don't really care what Cohen-Tannoudji (or anybody else) thinks about this, because I understand it perfectly myself. I also wasn't rude, I just told you that you are wrong. The issue that ordinary QM does not couple to photons is absurd--- first it is not difficult to couple to photons, if you don't care about renormalization. But more importantly, if you couple to anything at all, photons, other atoms, you can't make transitions because you have a ground state. The existence of the ground state is the relevant thing: the question is why QM has a ground state when CM doesn't. – Ron Maimon Aug 9 '12 at 8:45
Do you happen to read the post (my) and the initial question. The point is that in introductory QM books, Qm is presented as solving the problem encountered in classical mechanics of electrons radiating energy. This is false because QM does not take the into account the coupling with EM field. Ground state is irrelevant if electrons in upper states radiate their energy until they reach ground state. – Shaktyai Aug 9 '12 at 12:32
This is confusing the issue--- if there is a ground state for 1 electron, there is a ground state for 2. Saying "QM" doesn't mean you don't include photon perturbations, it's completely obvious that if you do, there's no radiation from the ground state, because it's the lowest energy state, so there is no issue with introducing radiation coupling. You just replace p with p-eA and add the free photon Hamiltonian. – Ron Maimon Aug 9 '12 at 18:28

Classically emission is continuous and the electron would need to occupy a "in between" energy level for a while, and that is forbidden in Bohr's scheme, so the emission can't be allowed to happen.

This doesn't really explain why it can't happen, but that's phenomenology for you: you line keep lining up facts until your kludge (1) gets the right answer and (2) might be pointing at a better "real" theory.

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