# De-Broglie's explanation on Bohr's Angluar Momentum quantization?

1. Apr 18, 2013

### easwar2641993

In order to understand about De-Broglie's explanation on Bohr's second postulate,concept of standing waves should be understood.
But condition of λ for a given value of length(string) L is given by L=nλ/s where n =1,2,3 etc.
But for a string whose ends are connected together and its shape is like a ring and let the radius of ring be r. Then standing waves condition is given by
Circumference = nλ/2
2∏r = nλ/2.

But this isn't right.
it should be 2∏r=nλ

I am missing something.Please correct me.

2. Apr 18, 2013

### sophiecentaur

A standing wave can exist if there is one half wavelength around the circumference - in the same way that it can exist in a half wavelength string. It's the equivalent of two waves travelling in opposite directions around the circumference. Only one node is necessary.

3. Apr 18, 2013

### Philip Wood

If there were a rope around the circumference, we'd need a whole number of wavelengths in order for the wave pattern to join up properly, without discontinuities. An odd number of half wavelengths won't do - try drawing it.

Note that this 'rope picture' is nothing like the 'proper' wave function picture, at least not for small values of the principal quantum number, n. Interestingly, for large n, and ell = m = n-1 it gives a remarkably good picture.

4. Sep 9, 2016

### sophiecentaur

I thought this through a bit better and I think you must be right. I was assuming that the nodes at each fixed end were the equivalent to joining them together on a circular string but the phases would be wrong unless you have a full wavelength path around the circle. Cheers!

5. Sep 9, 2016

### DarkBabylon

Yes in fact it should be 2Πr=nλ for circular circumference.
Sommerfeld and Wilson have made a correction to De-Broglie's theory that it should be:
∫pdx = nh for a closed curve, for the general case.
For our specific case assuming p is constant and not a function and λ=h/p:
∫dx=nλ
De-Broglie assumed a standing wave on a non-curved string, which works perfectly as the two waves go back and forth between two ends, but for a curved string this isn't the case, as the wave can continue, and thus destroy itself. Should a particle be a wave it must not destroy itself, as is evident by the fact matter exists for a long enough time, so the solution is to have one wave, that will interfere with itself such that the circumference is exactly an integer multiple of the wavelength, and thus not destroy itself but construct itself, and still be a standing wave.

Last edited: Sep 9, 2016