B The Mystery of Excited Electrons: Are They Moving Away from the Nucleus?

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Excited electrons in the conduction band are generally considered to be further from the nucleus than valence electrons, as higher-energy electrons tend to occupy higher energy states. However, due to the principles of quantum mechanics, electrons cannot be precisely located, and even high-energy electrons can have a probability of being found near the nucleus. In the conduction band, electrons are not bound to specific atoms, allowing them to move freely over long distances. The discussion highlights that if electrons were strongly localized to a nucleus, electrical conduction would not occur. The nature of the band structure is more about energy states than spatial positioning.
Mustafa Bayram
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when an electron is excited to the conduction band is it move further from the nucleus?
Are free electrons in the conduction band further from valence electrons?
I saw this picture that seems problematic to me. what do you think?

conduction band.png
 
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I am not sure what that picture represents, but (1) it looks like a very classical-minded picture and (2) the band structure is a feature in energy, not in space.

Generally speaking, higher-energy electrons are farther away from the nucleus, and therefore further away from core electrons. But the proper description is quantum mechanical, so you can't assign a precise position to an electron. Even high-energy electrons have a non-zero probability of being found near the nucleus.

In the conduction band, electrons are no longer tethered to a particular atom and are free to move over long distances.
 
DrClaude said:
I am not sure what that picture represents
Me neither, but for some reason I remember that I need to pick up some thing at Target.
DrClaude said:
In the conduction band, electrons are no longer tethered to a particular atom
This is key. Can you write it again?

If all the electrons were strongly localized to a nucleus, you wouldn't have conduction!
 
For the quantum state ##|l,m\rangle= |2,0\rangle## the z-component of angular momentum is zero and ##|L^2|=6 \hbar^2##. According to uncertainty it is impossible to determine the values of ##L_x, L_y, L_z## simultaneously. However, we know that ##L_x## and ## L_y##, like ##L_z##, get the values ##(-2,-1,0,1,2) \hbar##. In other words, for the state ##|2,0\rangle## we have ##\vec{L}=(L_x, L_y,0)## with ##L_x## and ## L_y## one of the values ##(-2,-1,0,1,2) \hbar##. But none of these...

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