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Hydrogen Atom Energy Level Hydrogen Atom Wave Function

I have difficulty to understand this Hydrogen atom's energy level and its wave function. My difficulty is regarding the hydrogen atom itself, not the energy level and the wave function their selves, also not about the energy absorption or photo emission. So does not about the quantum number $n$, the angular momentum $l$, the magnetic quantum number $m_l$, and the magnetic spin $m_s$.

The two pictures above I took from a lecture here, and here. They are the same lecture/video but different time. What is difficult for me to understand about them are how a hydrogen atom with only one electron have energy level $E_n$ and quantum number $n$ more than one, say $n=2$? What does that mean? With $n=2$, then the electron configuration is $1\text s^2\; 2\text s^2\;2\text p^6$. Here, seems like the hydrogen atom is possible to have 10 electrons ($n=2$, $l=1$, $m_l=1$, $m_s=1/2$). Here is my difficulty to digest it. I hope one can explain this in a simple way.

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3 Answers 3

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In the above question, $n=2,3....$ doesn't imply that there is more than $1$ electron.

In the above problem, $n$ represents the principal quantum number. It tells you in which state, is that one electron. In terms of the Bohr model, you can think of this in a simple way: The first electron is in the $K$ shell with $n=1$. However, if you give the electron some energy, it'll get excited and jump to the next level, say $L$, with $n=2$. Does it mean there are 3 electrons now ? No! It just means that the $one$ electron is in an excited state.

In the case of orbitals, we have the numbers $n,l,m_s,m_l$ out of which $n,l$ are needed to determine the orbital. Yes, if you start filling it up from ground level, it will start filling up in a certain order i.e $1s,2s,2p,3s....$

However, we are not doing that here, as in we are not filling up orbitals. We are just giving an electron enough energy to jump to a higher orbital. So, if we say an electron exists in $2p$ orbital, it doesn't mean you have to fill up $1s$ and $2s$ orbitals first. It means that your electron is in an excited state. It has higher energy than your $1s$ electron. There still is only one electron in your system.

Remember the difference between filling up orbitals, and exciting a single electron to a higher energy state. $1s,2s$ fils up before $2p$ doesn't mean that an electron can't exist in $2p$ directly. It just means that if you have an unlimited no. of electrons, and you are filling them up one by one, then these orbitals will fill up first. But a single electron can be excited to any energy level you want. These are different scenarios, don't mix up between the two.

Hope this helps.

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  • $\begingroup$ Interesting. But then, there is a more complicated situation, say that now it is in energy level N (n=4), which the electron configuration that reaches level 4 will be: 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f. With energy level O (n=5), the configuration will be the same complicated. So, how we deal with them? Will we go up-down from energy level N down to M then up again to N, then go up to O, to continue up to P, then down again to N? $\endgroup$ Aug 20, 2021 at 19:58
  • $\begingroup$ @AirCraftLover no, in the non-relativistic treatment of quantum mechanics the energy level depends on $n$, not on the other quantum numbers. Like I said, there is difference between the filling of states, and exciting a single electron. The excited electron can be made to occupy any state, by giving it the appropriate energy. $\endgroup$ Aug 20, 2021 at 21:52
  • $\begingroup$ Quote: Remember the difference between filling up orbitals, and exciting a single electron to a higher energy state. 1s,2s fills up before 2p doesn't mean that an electron can't exist in 2p directly. So, can I say that that single electron can occupy any "room" in orbital s and p, i.e, pz=1 with ms=-1/2 (down)? As the 2nd picture is saying so. $\endgroup$ Aug 21, 2021 at 7:59
  • $\begingroup$ @AirCraftLover Exactly, you're right. But remember, in order to exist in a certain room, the electron needs certain energy. If you give it that energy, it'll exist in that room. It might release that energy and jump to a lower room. $\endgroup$ Aug 21, 2021 at 10:37
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    $\begingroup$ Quote: In order to exist in a certain room, the electron needs certain energy. If you give it that energy, it'll exist in that room. Thanks you. I think this is the answer to my question. $\endgroup$ Aug 23, 2021 at 13:06
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The hydrogen atom has 1 electron. The electron configuration, 1S1, refers to the ground state, $\Psi_{100}$, or $n=1$, $l=0$, $m=0$. Any other quantum numbers indicate an excited state of that one electron.

In the non-relativistic hydrogen atom, the energy only depends on $n$, so anything with $n\ne 0$ is an excited state. With in the context of the non-relativistic Schrödinger equation, excited states are stable...all eigenstates of $\hat H_0$ are stationary. Of course, the presence of the electromagnetic field means that $\hat H_0$ is not exact, so that excited states can decay back to the ground state by emission of photon(s).

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how a hydrogen atom with only one electron have energy level $E_n$ and quantum number $n$ more than one, say $n=2$? What does that mean? With $n=2$, then the electron configuration is $1s^22s^22p^6$.

You got a misunderstanding. The hydrogen atom always has only one electron.

  • When the atom is in a state with $n=1$, then the single electron is in the $1s$ shell. The electron configuration is $1s^1$.

  • When the atom is in a state with $n=2$, then the single electron is in the $2s$ or $2p$ shell. The lower $1s$ shell is empty, there is no electron in it. So the electron configuration is just $2s^1$ or $2p^1$ (not $1s^22s^22p^6$).
    Of course this state is not very stable. After very short time (a few nanoseconds) the electron falls from the $2s$ or $2p$ shell down to the $1s$ shell, while emitting a photon, which carries away the energy difference.

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  • $\begingroup$ From your second point, how an electron to chose whether to "occupy" 2s1 or 2p1? So, in case we have n=4 for the hydrogen energy level, so the electron configuration will be: 4s1 or 3d1 or 4p1 or 5s1 or 4d1 or 5p1 or 6s1 or 4f1? This is truly confusing me. The further question, how then the electron will chose which orbit (subshell) will be occupied? $\endgroup$ Aug 20, 2021 at 19:47
  • $\begingroup$ Quote: The lower 1s shell is empty, there is no electron in it. So the electron configuration is just 2s1 or 2p1 (not 1s2, 2s2, 2p6). I noticed that there is mistake here. As shown by the 2nd picture, there is clearly shown electron in m=0 and m=1, which for sure occupying 2p2 and 2p3 both ms=+1/2 (up) and ms=-1/2 (down). $\endgroup$ Aug 21, 2021 at 7:55

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