Binding energy between an $1s$ electron and its nucleus I've always thought that the binding energy decreases as the electron moves/jumps away from the nucleus.
Then when I see the radial probability distribution for $1s$ electron, there is a probability for finding the electron everywhere. Since the binding energy depends only on $n$, does this mean the binding energy remains constant no matter where the electron is found ?
If that's the case, kindly consider this situation :
1) A $1s$ electron is $100m$ away from the nucleus.
2) It absorbs a photon and jumps into $2s$ orbital. Now can the electron in this orbital stay closer to the nucleus(say at $20m$) ? If yes, isn't this counter intuitive ? (How can the far away $1s$ electron have greater binding energy compared to the closer 2s electron)
 A: Before looking at the atom, consider a classical example of a comet on an elongated elliptical orbit around the Sun. While particles are not anything like planets, still even this rough classical analogy seems to address your concern. The kinetic energy farther from the Sun is lower, the total energy is still the same while the kinetic energy is irrelevant to your question. You can catch one comet with a lower energy farther away and another with a higher energy closer to the Sun. Clearly the second one will be moving faster, but speed also is irrelevant to your question, only the total energy is.
Please keep in mind that electrons are not like comets and don't actually "rotate" around the nucleus, so this example is just a rough analogy to get some helpful intuition.
A: If an electron is 100 meters from the nucleus, it's not in a 1S orbital. Now, that doesn't mean that you can't find it there when it is in a 1S orbital, but once you do, it's now in a superposition of states.
The key is that position eigenstates, $\delta(\vec r-\vec r_0)$ are not energy eigenstates, but they can be expressed as sum over all of the $\psi_{nlm}$.
For clarification: asking where an electron is when its in a stationary state is classical question. It's very much like asking which slit the electron goes through in the double-slit experiment: quantum mechanics doesn't provide a definitive answer, other than "all amplitudes that can contribute to the process contribute to the process". Likewise, all positions that are in the wave function are in the wave function (as amplitudes, and we can't see amplitudes).
