Why the cycles completed by standing wave completed by electron in a certain orbit is same as principle quantum number?

Question

According to the bohr model and de-broglie hypothesis why the cycles completed by standing wave completed by electron in a certain orbit is same as principle quantum number? When we derive the angular momentum we use $2\pi r =n\lambda$ where n is obviously the cycles completed by electron's standard wave . Whereas we do $\lambda = h/mv$ then we conclude both $\lambda $ should be same to obtain $L=\frac{nh}{2\pi}$ now we call n as the principle quantum number. Why the both electron wave cycles be identical with principle quantum number?

Answer 1

The reason is that the Bohr model was built that way. The quantization of the energy in an atom, which is what gives the notion of a "principal quantum number", was an empirical observation - after all, the quantum theory had to be developed to explain something, as any scientific theory is. And the hypothesis was that a standing wave pattern had to be formed around a circular orbit, like a plucked guitar string, because primarily of two things:

1. the phenomenon of the standing wave was a long-known one (ever since Melde in 1860 with the famous "cord of Melde" experiment that is now often used in intro phys demonstration labs though perhaps its discovered not named) in which a similar phenomenon of "quantization" - i.e. the restriction of a physical quantity to taking on only values from a discrete set as opposed to a continuous range - was known,

2. other experiments with electrons revealed that their propagation in some ways resembles that of a wave, e.g. that they can undergo diffraction.

Naturally, given these two things, it is very easy to hypothesize that, then, if the electron in the atom somehow is, or is such that its behavior can be described with, waves, the observed quantization of energies may be due to a similar standing-wave effect as in the cord experiment. And thus, by design of the model, one would naturally hypothesize that the lowest-energy state corresponds to the lowest standing wave pattern, i.e. the fundamental mode, and then the next higher to the first overtone, and so forth.

And in fact, the modern quantum theory of the atom does effectively that, however the relevant waves are not physical waves in fluid, but (though the interpretation is still debated) rather a more subtle kind of wave that appears within a probability distribution reflecting the fact that the electron's position is not determined with infinite information, and moreover their standing-wave patterns set themselves up in full three-dimensional space, not the one-dimensional "wrapped string" oscillation envisioned in the old model. And due to these points, this model can account for some things the Bohr model could not: for example, due to the second point, the Bohr model is unable to account correctly for angular momentum, while the modern model allows for a range of possible angular momenta "suggestions" at each energy step, to the extent of the informational limits imposed by the quantum laws on them just as much as on position.

Answer 2

The full solution to Schrodinger's equation for the hydrogen atom potential involves three quantum numbers usually labeled $n, \ell,$ and $m$. $n$, as in your notation, is the "principle quantum number" here. The full solution also tells you that the exact angular momentum is $L=\sqrt{\ell(\ell+1)}h/2\pi$. The solution only tells you that $\ell<n$, so your equation, the result of the bohr model, isn't always true. However, for $\ell=n-1\gg0$, we have $\ell\approx n$ and $\sqrt{\ell(\ell+1)}\approx\ell$. So you see, this result from the bohr model is good for large and maximal $\ell$. In fact, these atoms in these states have a special name, they're called "circular state Rydberg atoms", since if you look at the wavefunction the wavefunction looks like a big donut.

This is all well and good, but maybe you're looking for something more intuitive? i.e. Why are $\ell$ and $n$ related in this way? Well, $n$ is the primary contribution contribution to the energy of the atom, and accordingly higher $n$ wavefunctions have the electron farther away from the nucleus (the minimum of the potential). If you want to increase $\ell$ and the angular momentum, beyond a certain point you can't do so without causing the centrifugal force to push the atom farther from the nucleus, and hence you can't increase $\ell$ beyond a certain point without increasing $n$. Similarly, in the bohr model, in order to get an extra cycle of the wavefunction in the circle, you have to increase the diameter of the circle, which is directly related to the principle quantum number.

Hopefully that gives you some more intuition for why the principle quantum number shows up this way. IMHO the Bohr model is over-emphasized in undergraduate physics. I've never used it except as a heuristic to remember some scaling relations, and it only really starts making sense when you reach real quantum mechanics anyways.