Bohr's model of an atom doesn't seem to have overcome the drawback of Rutherford's model

Question

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?

Answer 1

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 keep lining up facts until your kludge (1) gets the right answer and (2) might be pointing at a better "real" theory.

Answer 2

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.

Answer 3

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.

Answer 4

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.