What results when sound interacts with an Atom

[Atomic]

Is there any change to the electrical state of the atom, when it's subjected to a frequency or any type of sound wave shape?

I'm not entirely sure what sound is made up of, only that the number of times that a sound wave vibrates in a second is called its frequency and pitch and volume dictate frequency or shape of the wave.
But what is in the wave? How does it interact with Atoms?
I guess more or less I"m wondering if it alters the number of electrons or protons thus changing the electrical charge of the atom.

Excuse this post if it's already been made before, I am a new member here today.

Thanks , and I look forward to your responses.

 

[lightarrow]

Is there any change to the electrical state of the atom, when it's subjected to a frequency or any type of sound wave shape?

I'm not entirely sure what sound is made up of, only that the number of times that a sound wave vibrates in a second is called its frequency and pitch and volume dictate frequency or shape of the wave.
But what is in the wave? How does it interact with Atoms?
I guess more or less I"m wondering if it alters the number of electrons or protons thus changing the electrical charge of the atom.

Excuse this post if it's already been made before, I am a new member here today.

Thanks , and I look forward to your responses.

Hello Atomic!

As first approximation, sound is simply a collective oscillatory movement of atoms or molecules; think about atoms as they were tiny spherical masses connected with tiny "springs" (the interatomic bonds); when you hit a diapason, for example, you move a great number of atoms in the same direction, so the springs transmit that movement to the near atoms with a slight delay, and so on and on, propagatin that initial "disturb" or "perturbation" to the entire piece of metal, making it vibrate macroscopically. The diapason vibration makes the near air vibrate, which propagates away this vibration to our ears.

In air (or a gas) the mechanism is different because air molecules are not bound (no "springs"); when the diapason or the sound box' cone or any other transducer, vibrates, it moves the air molecules, for example pushing them away of itself; after a very short time these molecules collides with other molecules, giving them their speed and the process repeats miriads of times propagating away the initial "push"; when the sound box' cone comes back, it generates a pressure decrease on the near air, so the air molecules are as "pulled" instead of pushed, so creating another region of pressure decrease nearby, "pulling" others air molecules...and so on, so even this perturbation propagates away (and with the same speed).

At the end you have a series of compressed and rarefied regions of air that propagates away from the source, until they hit your ear, putting in vibration the inner diaphragm called "eardrum", which puts in vibration the little "hammer" ... which excites a nerve that transmits the signal to your brain making you perceive the sound.

Instead of asking directly to your question, I gave you the possibility to understand better the concept and so I hope that you could now answer that question by yourself.

 

[per.sundqvist]

The soundwave stretch atoms in equlibrium from each other in a solid. As a result an electric field is induced (polarization), and this field interacts with electrons in the atom. Polarization means that electronic clouds will be redistributed from the nucleus a little bit. This is the basic physical idea behind the so called electron-phonon interaction, even though it requires some knowledge to deal with it!

|<k+q|E|k>|^2 is a transition (per time unit) from an atomic state k to k+q (where q=the momentum of phonon) and it could be determined by using ab initio (DFT) calculations.