@Remi

Thanks, for the answer

But I have since worked it out. It is the frequency and in turn the wavelelght that is determined by the numerical powers of c. Based on the constancy of the speed of light.

To give an example:the Bohr atom for instance has a radius (the Bohr radius) which is determined by the completion of exactly 1 wavelength, at a given velocity, which happens to be c/alpha.

In none. The speed of light has dimensions, length/time. It's possible to choose *units*, such that it takes the value 1, or any, finite, non-zero, value in those units. But it's not possible to alter the fact that it's a velocity. That's the difference between units, that are conventional and dimensions, that aren't.

If alpha is the fine structure constant, the statement is wrong, because the fine structure constant is dimensionless, and a length can't be set equal to a velocity. There is a numerical coefficient, that has dimensions of time, that must be specified.

@ Stam

Read the answer more carefully then you might understand the basis of quantum physics.

Alpha is dimensionless  of course but when multiplied by c the whole term clearly has dimensions of length /time.

Let see if you can derive the Bohr radius from first principles, do that and I will be impressed.

The Bohr radius is derived from first principles in any quantum mechanics textbook: it's the average value of the position, computed with the probability density, equal to the square of the wavefunction, solution of the Schrödinger equation in the Coulomb potential, that corresponds to the ground state, that quantum effects create. 

The correct treatment of this problem is known at least since Pauli and Fock in 1925-1926. 

@ stam

I am well aware of the standard derivation 

But can you show how its is worked it out from first principles?

Give an exact derivation please.

That is the challenge, if so I would be very impressed.

You can use the standard derivation if you wish which is very lengthy, or is there a shorter one?

This isn't any challenge, there's nothing original in it. One solves the Schrödinger equation, finds the groundstate wavefunction, which is real, takes its square, computes the integral-that's the denominator-multiplies the square of the groundstate wavefunction by the radial coordinate and integrates that-that's the numerator. Divides the two. That's the Bohr radius. The reason it's the Bohr radius is that it is the average radial distance of the electron in the Coulomb potential, by construction. 

One way is mentioned here: http://www.phys.spbu.ru/content/File/Library/studentlectures/schlippe/qm07-05.pdf

There are shortcuts to finding  the wavefunction, depending on the fact that the Coulomb problem is integrable, that's what Pauli and Fock noticed. And, of course, it's good to use dimensional analysis to extract the combination of constants,that has the dimensions of length-so the non-trivial information is in the numerical coefficient. (However the derivation, often proposed, that equates the centripetal force mv^2/r to the Coulomb force e^2/(4πε_0 r^2) and uses the relation mvr = h=Planck's constant, to deduce an expression for r, while providing the ``right answer'' is, in fact, wrong-it's just a coincidence, that the numerical factor is correct-only  the dimensional analysis is, of course, correct, since all dimensionful constants are included-just why the result can be identified with the Bohr radius remains open and the answer to that is provided by solving the Schrödinger equation, expressed in dimensionless quantities.   The reason this derivation is wrong is that the total energy is the eigenvalue of the Hamiltonian, that's a sum of two, non-commuting, operators. Since they don't commute, this means that the electron in a state of given energy, doesn't have either a well-defined position, or a well-defined velocity.)  The Bohr radius is an average value and has a non-trivial distribution, that of course, is known. 

@stam

Thank you for your detailed answer.

Apologies it must be my mistake, but I cannot find the numerical answer for the Bohr radius of the electron in your answer or your link only that of protonium in the latter.

Please show me the actual eqautions for the  derivation for your first paragraph, including of course the numerical answer

This isn't a forum for ghostwriting homework. The link provides the calculation, that is to be read by humans, not parsed by a computer program (or a student looking for canned answers).  From the equations provided it's possible to obtain the expression for the Bohr radius, up to a numerical coefficient, that's given by the calculation of the Laguerre polynomial. Protonium isn't a term used to describe hydrogen, but, apparently, antihydrogen-though it's certainly not standard. However it's trivial to show how the Bohr radius of hydrogen is related to that of anti hydrogen. However the link doesn't mention antihydrogen. 

@Stam.

As I thought you are unable to derive the Bohr radius from first principles.

If you have started with Schrodinger, which cannot itself be derived from first principles - then it would be impossible to do so. Plus you have all those extra tortuous mathematical derivations

The real answer is as follows: to complete a single wavelength lambda, find the radius r

r = lambda/2pi

lambda = vw/f

where vw is the phase wave velocity, and f the frequency

vw = c/alpha

E=hf =mc^2, so f = mc^2/h

Putting the two together

r = h.c/m.c^2. alpha.2pi =  h/m.c.alpha.2pi = 5.292 x 10^-11                                                                                                                                                                                                

You can of course substitute alpha =e^2/2.e0.h.c

and you get the standard equation (which hides the proper derivation).

r= h^2. e0/pi.m.e^2 = 5.292 x 10^-11.

The statement about a single wavelength is wrong, because a particle in a bound state doesn"t have a well-defined position. The identification hf = mc^2 is meaningless, beyond dimensional analysis, since the mass isn't the mass of the particle in the bound state, since it isn't at rest, which is what the equation E=mc^2 means. And, while the mass is a Lorentz invariant, the frequency isn't-it's the time-like component of the 4-wavevector. So, once more, these expressions are empty of physical content, they are just dimensional analysis. The derivation from first principles is from quantum mechanics, that introduces a new constant, Planck's constant and shows how the Coulomb potential, that classically, doesn't have a ground state, does acquire one, when quantum effects are taken into account. Quantum mechanics *provides* the ``first principles'', that's the point.  It's not possible to deduce quantum mechanics from something else, unless it's quantum field theory, when the fluctuations of the electromagnetic field are neglected. But Planck's constant is the input, not the output. However quantum mechanics provides the framework for calculating the numerical coefficient to the expression, that has dimensions of length and contains Planck's constant, as well as its fluctuations-these are the *predictions*  of quantum mechanics. And it can be checked that the Bohr radius does vanish if Planck's constant were zero and quantum effects could be neglected.

@Stam

Your objections are very readily addressed but I really don't have the time to explain.

I note that you are completely stuck in your attitudes, which is why there is no point.

Sadly that means you will not be able  understand the rest of quantum electrodynamics and you will be stuck in this current intellectual quagmire, where there are so many paradoxes and intrepretations of quantum physics

Quantum electrodynamics is, for decades, part of the Standard Model, which, in turn, is a special case of a quantum field theory. It's interesting how many people today seem to be living in the 1930s in this regard, when quantum electrodynamics wasn't understood  (for quantum electrodynamics was completely understood by the mid-1950s, even though precision measurements continue, this time to discover effects beyond the Standard Model).

There aren't any paradoxes in quantum physics and, while the term ``interpretation'' is, often, used, it's wrong-there isn't any ``interpretation''. Quantum physics, like all of physics, involves doing calculations and performing experiments. The calculations are unambiguous, they don't admit more than one ``interpretations'' and the experiments likewise.That terms like ``quantum'' seem to attract metaphysical speculation has more to do with sociology than with science. While, historically, there is, always, an effort towards trying to understand the correct way of performing calculations and experiments, as the methods are being invented, it is now known how to do them, for quantum mechanics, so there's no point, beyond an interest in the history of the subject, to repeat the mistakes of the past and not learn the correct formulation.

It is well known that the effects of light with dimensions  [L][T-1]can produce an angular frequency with dimensions of [T-1] Equally well angular frequency  [T-1]can produce the effects of velocity [L][T-1]using light