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I have seen in textbooks and videos that an electron must absorb energy (become excited) to enter a farther-away orbital.

The amount of energy that must be gained is equal to the difference in energy between the orbitals. When the electron relaxes and returns to its ground state, it emits that energy.

However, I am not sure I understand why "higher" orbitals are of greater energy, for a couple of reasons.

  1. When I think of the potential energy of the electron, as given by $U = \frac{kQq}{r}$, it seems that $U \propto \frac{1}{r}$, which would indicate less energy as distance increases.

  2. I also know that the electron is attracted to the nucleus due to opposing charge. By this logic, an electron closer to the nucleus would be more stable (have a lower energy state) than an electron farther from the nucleus (which would thus have a higher energy state). This line of thinking is somewhat similar to how a ball that's still high in the air is in a higher energy state than a ball that's closer to the ground.

So #1 is inconsistent with what we are taught about electrons and orbitals, while #2 would support it.

I am attempting to clarify the misunderstanding I must have somewhere (possibly not making a proper analogy) - and better understand why higher orbitals are of greater energy.

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    $\begingroup$ When I think of the potential energy of the electron, as given by $U = \frac{kQq}{r}$, it seems that $U \propto \frac{1}{r}$, which would indicate *less* energy as distance increases. You forget that the nucleus and the electron are oppositely charged, so $Qq$ will be negative. Thus, $U \propto -\frac{1}{r}$. $\endgroup$
    – Philipp
    Sep 12, 2014 at 7:30
  • $\begingroup$ Yes, I did forget that. Realized a little while after posting. Thank you! $\endgroup$
    – Ghalib
    Sep 12, 2014 at 11:45
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    $\begingroup$ @Philipp Could you write this as an answer so we can go ahead and wrap this question up? $\endgroup$
    – LordStryker
    Sep 12, 2014 at 12:32

1 Answer 1

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potential energy is negative between infinity to nucleus

It varies with .Thus, U∝(−1)1/r so it means the father you go from nucleus the value become less negative that is higher in energy than equilibrium energy.

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