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Topic: De Broglie and the Standing Wave  (Read 4986 times)

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Offline Il Divo

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De Broglie and the Standing Wave
« on: June 29, 2012, 05:11:27 AM »
So conceptually, I'm having some trouble understanding certain aspects of De Broglie's hypothesis that electron orbitals function as standing waves, which do not travel along a string, meaning that nodes are always stationary. The macroscopic comparison I've seen used is to a guitar string.

The equation of a guitar string is L = n (λ/2) where n is the energy level, L the length of the string, and lambda the wavelength. The fundamental frequency will be dependent on the length of the string. And since the string's length can be held constant, the values of the frequency/wavelength at higher levels are restricted according to the integer value of n, hence quantization of energy.

My questions are as follows:

1) I've seen this applied to the orbit of an electron. If it behaves as a standing wave (similar to the guitar string), it is then necessary for the wavelength to fit the circumference of the orbit exactly, according to the equation: 2(pi)r = nλ where r is the radius. According to my textbook the value of r remains restricted, due to the fact that n must be an integer.

Where I'm having some confusion is that unlike the guitar string, my textbook says that according to the Bohr model the radius is also dependent on the value of n, so unlike the guitar string it is not held constant as we increase the value of n. I was hoping someone could clearly explain to me this equation, the relation between wavelength and radius, and how the electron functions as a standing wave in a circular orbit.

2) Regarding De Broglie's equation that λ = h/p. This indicates that anything which has mass also possesses a wavelength and vice versa. Ergo, light also has mass. I did some searches on this topic and found something about relativistic mass and was wondering if this is what the mass term for an object's momentum refers to, because I've also heard that light has no rest mass.

Thanks all, in advance. Also, let me know if I've made any big mistakes in the guitar string comparison.

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Re: De Broglie and the Standing Wave
« Reply #1 on: June 29, 2012, 05:24:52 AM »
Where I'm having some confusion is that unlike the guitar string, my textbook says that according to the Bohr model the radius is also dependent on the value of n, so unlike the guitar string it is not held constant as we increase the value of n. I was hoping someone could clearly explain to me this equation, the relation between wavelength and radius, and how the electron functions as a standing wave in a circular orbit.

λ is constant.

You need an integer number of waves on the orbit, otherwise you will not observe a standing wave.

That means [itex]2\pi r=n\lambda[/itex], or [itex]r= \frac{n\lambda} {2\pi}[/itex]. Radius is not constant, but it can take only discrete values. In the case of the string it was the other way around - length was fixed, so it was λ that was taking only discrete values.
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Offline Il Divo

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Re: De Broglie and the Standing Wave
« Reply #2 on: June 29, 2012, 05:35:26 AM »
Where I'm having some confusion is that unlike the guitar string, my textbook says that according to the Bohr model the radius is also dependent on the value of n, so unlike the guitar string it is not held constant as we increase the value of n. I was hoping someone could clearly explain to me this equation, the relation between wavelength and radius, and how the electron functions as a standing wave in a circular orbit.

λ is constant.

You need an integer number of waves on the orbit, otherwise you will not observe a standing wave.

That means [itex]2\pi r=n\lambda[/itex], or [itex]r= \frac{n\lambda} {2\pi}[/itex]. Radius is not constant, but it can take only discrete values. In the case of the string it was the other way around - length was fixed, so it was λ that was taking only discrete values.

And suddenly, it all becomes much clearer. Thank you very much for this.

One additional question: what effects the value of lambda? Similar to the example of the guitar string, we can vary length, but for a given set of vibrations, it will remain constant. Is the wavelength for any particular electron dependent on the atom we are examining? Ex: Comparing the radius of a hydrogen atom vs. a helium atom.

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Re: De Broglie and the Standing Wave
« Reply #3 on: June 29, 2012, 05:57:59 AM »
λ is assumed to be constant characteristic for the particle, electron in this case, so it will be the same for every atom.

Please note this is only a simplified model, getting too deep will not bring any enlightenment, I would rather expect more confusion ;)
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