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From what i've read, electrons in a lower energy level have more kinetic enery and momentum to stop "falling" into the positive nucleus. However, when an electron falls from a higher energy level to a lower one, the loss of energy is emitted as a photon. So how does it gain kinetic energy? Thanks in advance

Emile Roxs
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  • On various websites, sorry i dont remember a specific one – Emile Roxs Nov 08 '16 at 21:04
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    Take note that in the classical system (gravitation) orbiting bodies in close orbits have higher velocities than those in distant orbits. So it is not a surprise that $\langle T \rangle$ (possibly the closest analogy to "the kinetic energy of an atomic orbital") increases for lower lying orbitals in the quantum system. You might find it easier to reason the problem through in the classical system, however. – dmckee --- ex-moderator kitten Nov 08 '16 at 23:48
  • @CountTo10 please see my answer. Are you sure that your comment is correct? – Prof. Legolasov Nov 09 '16 at 00:43
  • My question is ...............the energy of electron is considered to be negative so when the energy is supplied though photons the energy of electron increase(tends toward positive direction) and they get away from nucleus so does it means that it behave like Proton as it gets away from nucleus (as proton has a positive) or its attraction becomes less and why applying ENERGY it does not become attractive to nucleus. ........plz answer – TPL Oct 17 '21 at 18:06

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I think the confusion comes from mixing up two terms: kinetic energy and energy.

Energy is a sum of kinetic energy and potential energy. Energy states, as is well-known, depend on the quantum number $n$ in ascending order: electrons farther away from the nucleus have greater energy. It is why when photons are emitted the electron transits to a closer-to-the-nucleus orbit.

But kinetic energy indeed is higher for the small-$n$ orbits. This is because they have huge negative potential energy, as they are close to the attracting nucleus, which is "compensated" by huge positive kinetic energy.

Btw, I think that the comment by @CountTo10 is incorrect as (s)he also confuses these terms (no offense!). Please correct me if I got something wrong.

Prof. Legolasov
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    Hi, no offence at all, I am glued to the TV for some reason at the moment........:) I have deleted the comment, I should have distinguished between them, it's a he, btw, regards –  Nov 09 '16 at 01:01
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    The stationary states (orbitals) of an atom are not eigenstates of kinetic energy, so it is not really proper to talk of them having a kinetic energy. However, you can workout the expectation value for kinetic energy $\langle T \rangle$ and use that as a kind of stand in as long as you understand the distinction. So it is not true that small $n$ orbitals (not orbits: orbitals!) have high kinetic energy, but their probability distribution for kinetic energy has a higher average than those of higher $n$ orbitals. – dmckee --- ex-moderator kitten Nov 09 '16 at 02:00
  • @dmckee I agree – Prof. Legolasov Nov 09 '16 at 02:27