Why is the lowest energy level close to the center?

[LordWhiplash]

TL;DR Summary

This probably is a dumb question, but I need to understand the reason instead of just accepting it.

While studying photon emission, I noticed that I never really understood why the higher energy level is farther from the center of the atom. To me it seems counterintuitive, because usually the forces of attraction are greater at shorter distances, which would imply a higher energy consumption to remove an electron near the nucleus, not the opposite.

Thanks for the help/answer!

 

[PeterDonis]

I never really understood why the higher energy level is farther from the center of the atom

This actually isn't always true. But it's true for the simplest orbitals, the $s$ orbitals (the ones with zero orbital angular momentum), and so it's a reasonable first approximation for understanding the basics.

the forces of attraction are greater at shorter distances, which would imply a higher energy consumption to remove an electron near the nucleus

Yes, and "higher energy consumption to remove the electron from the atom" means "lower total energy of the atom", i.e., lower energy level. Conversely, "higher energy level" means "higher total energy of the atom", meaning "lower energy consumption to remove the electron from the atom".

 

[Dr_Nate]

Summary:: This probably is a dumb question, but I need to understand the reason instead of just accepting it.

While studying photon emission, I noticed that I never really understood why the higher energy level is farther from the center of the atom. To me it seems counterintuitive, because usually the forces of attraction are greater at shorter distances, which would imply a higher energy consumption to remove an electron near the nucleus, not the opposite.

Thanks for the help/answer!

Picture this analogy (you = electron): you are in a hole in the ground. As you climb up the hole you are increasing your gravitational potential energy and yet the gravitational force on you is a tiny bit smaller. If you fall down you convert that energy to kinetic energy (electrons instead emit photons). If you climb out of the hole you are free from the hole (like an ionized electron).

 

[Dr_Nate]

This actually isn't always true. But it's true for the simplest orbitals, the $s$ orbitals (the ones with zero orbital angular momentum), and so it's a reasonable first approximation for understanding the basics.

Are you thinking of the maximum in the probability density?

From the plots I've seen the on average the probability density shifts farther from the nucleus. This must be true because the energy of the state would depend on the weight of the wave function at each point in the potential well.

 

[PeroK]

To me it seems counterintuitive, because usually the forces of attraction are greater at shorter distances, which would imply a higher energy consumption to remove an electron near the nucleus, not the opposite.

That means you have to input energy to get the electron away from the nucleus. The closer the electron is the more energy you have to input to free it. To turn that round, an electron lost more energy getting to the lower orbital in the first place.

This is the same in the classical, gravitational case, where the total energy of an object in a circular orbit is less for a smaller orbit.

 

[PeterDonis]

Are you thinking of the maximum in the probability density?

For the $s$ orbitals, yes, because those are spherically symmetric so the probability density is only a function of distance from the center.

For other orbitals, the probability density is also a function of angle, not just distance from the center, so the concept of "farther from the center" isn't even well defined, strictly speaking. But even if you take an average over all angles, it's not true that higher energy always corresponds to farther from the center, because for orbitals with $l = 0$ the energy is no longer just a function of position in the potential well.