# Why does the energy of an electron inrease with its shell number inside an atom?

According to this:

$$E = \frac{-13.6 Z^2}{n^2}$$

the energy of an electron is, well, higher the farer it is away from the core. I found this confusing as I need to put less energy to release an electron when it is on the 5th shell than when it is on the 1st shell. Am I right? Concluding, one would say "The more energy an electron has the less energy I need to release it". This wouldn't be the case if "they" didn't introduce the minus sign. It probably has a reason and I'm probably neither the first nor the last who asks about that. But, sorry, I still have to ask: Why did they do that? Why is a minus better than a plus?

• Jun 17, 2020 at 14:16

Why did they do that? Why is a minus better than a plus?

Firstly, understand that when the electron is 'infinitely' far from the nucleus its energy is $$0$$ (assuming it is also stationary).

But empirically we also know that when an electron 'falls' from a higher orbital ($$n=n_1$$) to a lower one ($$n=n_2$$ where $$n_2) energy is released in the form of a photon emission, so $$\Delta E <0$$. And indeed:

$$\Delta E=E_2-E_1=-13.6 Z^2\Big(\frac{1}{n_2^2}-\frac{1}{n_1^2}\Big)<0$$

So the choice of sign is correct.

• The higher orbital is $n_1$ and the lower is $n_2$ ?Despite from that I don't see an argument for the minus at that place?
– Ben
Jun 17, 2020 at 13:48
• That IS the argument: put a plus sign in and you'd get $\Delta E >0$ which is contrary to our experience. It's a convention to ensure that for a 'falling' electron, $\Delta E<0$. That's all there is to it.
– Gert
Jun 17, 2020 at 14:05
• Ah, because otherwise, it would gain energy?
– Ben
Jun 17, 2020 at 14:34
• The convention (universal, not just electrons) is that an energy loss is a negative $\Delta$ and a gain is a positive $\Delta$.
– Gert
Jun 17, 2020 at 15:30
• Well, that makes sense, indeed. Confusing but once that is known it makes sense :)
– Ben
Jun 18, 2020 at 5:14