Question about Orbitals in an Atom

[CRichard]

For a summer class, I’ve been trying to get the quantum picture of an atom at least in conceptual terms, no math. I understand how, if you think about an atom like a one-dimensional box, an s orbital is like the first fundamental wave, and a p-orbital is like a sine wave, and I understand how the p orbital has higher energy because of greater frequency. But how can you picture 2s and 3s, etc. orbitals? Are they like first fundamental waves with greater amplitudes? I thought that the frequency of a fixed string depends mainly on its length and tension, not amplitude, so if that's the case how does E = hv?

Also, how are they quantized, since can’t you “pluck” that string in a one-dimensional box however far you want (at least as far as it can stretch), and give it an infinite number of amplitudes? In other words, I get how the need to fit a whole number of half wavelengths into the box justifies the quantization of energy into s, p, d, etc orbitals, but what limitation prevents the s orbitals from having a continuous spread in energy?

 

[jtbell]

A string is one-dimensional whereas an atom is three-dimensional. Crudely speaking, the "vibrations" are quantized in each dimension.

Again crudely speaking, the primary quantum number (energy level number) corresponds to quantization in the radial dimension. The s, p, d, f correspond to quantization in the "longitude" dimension (think of going north/south on a globe). Finally, there's a "magnetic" quantum number that corresponds to quantization in the "latitude" dimension (think of going east/west around a globe).

It might help to try to imagine standing waves of sound in a spherical cavity, but I can't find any pictures of those at the moment.

 

[CRichard]

Thanks, that makes sense. The only thing I don't get is how E = hv is not violated for example in the case of a piano string. I remember reading that it doesn't matter how much energy you give to a piano string by striking it hard, the pitch is the same because frequency doesn't change. Shouldn't the added energy and loudness be the result of an increase in frequency?

 

[chogg]

Thanks, that makes sense. The only thing I don't get is how E = hv is not violated for example in the case of a piano string. I remember reading that it doesn't matter how much energy you give to a piano string by striking it hard, the pitch is the same because frequency doesn't change. Shouldn't the added energy and loudness be the result of an increase in frequency?

E=hv is the energy of a single quantum. If you pluck a string harder, you are putting more quanta in it.

If you like, the generalization of the formula is E=nhv, where n is the number of quanta.

 

[Drakkith]

Thanks, that makes sense. The only thing I don't get is how E = hv is not violated for example in the case of a piano string. I remember reading that it doesn't matter how much energy you give to a piano string by striking it hard, the pitch is the same because frequency doesn't change. Shouldn't the added energy and loudness be the result of an increase in frequency?

I don't know for sure, but I think a similar effect happens in a pendulum. Pushing either one harder only causes the amplitude to increase. The way both are built and react, the frequency will not change when you add more energy.

 

[alxm]

The only thing I don't get is how E = hv is not violated for example in the case of a piano string.

E = hv relates the frequency of light to the energy of a photon. It doesn't apply to a piano string.

I remember reading that it doesn't matter how much energy you give to a piano string by striking it hard, the pitch is the same because frequency doesn't change. Shouldn't the added energy and loudness be the result of an increase in frequency?

No, frequency (pitch) and amplitude ('volume', intensity) are independent properties. A flute has a higher pitch than a kettle-drum, but it's not necessarily louder.

 

[chogg]

E = hv relates the frequency of light to the energy of a photon. It doesn't apply to a piano string.

Doesn't it, though? My impression was that for a given mode of vibration, E = hv applies to the phonons in the lattice.

Recently, single phonons were added and removed from a mechanical mode of oscillation:
http://www.nature.com/nature/journal/v464/n7289/abs/nature08967.html

 

[CRichard]

"E = hv relates the frequency of light to the energy of a photon. It doesn't apply to a piano string."

That's what I was thinking too, because you can have many water ripples too with a high frequency but low energy. But then, does E = hv also apply to matter waves like an electron wave?

 

[CRichard]

Sorry to bump this. I can't seem to find a clear answer online, but just to clarify: if you think about the analogy of standing waves on a circular drum, a 2p orbital has electron density farther from the nucleus than a 1s orbital. So, I think it would have a higher energy because of the charge separation. But can you also think about it a different way, and say that the higher energy is due to the higher frequency of the p-orbital by the equation E = hv, (that matter waves are fundamentally quantized) or does this equation only work for electromagnetic energy?