Hydrogen:
z = 1
n |
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L | 2 | ||||
K | 1 |
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H |
One electron (1s), the ground state is 2S½
The binding energy (Ionization energy) is 13,6 eV
Helium:
z = 2
n |
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L | 2 | ||||
K | 1 | Ýß | He |
2 electrons (1s) (1s), ground state 1s2
Since both electrons have the same main quantum numbers then the ground state should be 1S0 (total spin S = + ¯ = 0). If we excite an electron into n = 2, i.e. 2s state (1s) (2s) then both electrons have different main quantum numbers n and total spin can be S = 1. Here we show rough scheme of Helium energy levels and possible transitions:
Lithium:
z = 3
n |
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L | 2 | Ý | |||
K | 1 | Ýß | Li |
According to Pauli there is no 1s3 configuration but
1s2 2s º 2S½.
- fulfilled shell with L = 0
and S = 0, J = 0.
If the nucleus has been covered fully by two 1s electrons then the ionization energy would be 13,6 eV/4 = 3,4 eV. Since the wavefunction of 2s electron penetrates deeper into the nucleus therefore we have a bit higher ionization energy: 5,4 eV or Zeff = 1,3.
Since 1s2 shell is fully occupied, that corresponds to the noble
gas therefore it's sometimes written He(2s)
for the Li-ground state configuration. Another way is to use capital letters K, L, M, N.... for
fully occupied shells, or K(1s) for Li.
Beryllium:
z = 4
n |
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L | 2 | Ýß | |||
K | 1 | Ýß | Be |
(1s2) (2s2) ≡1S0
The binding energy would be 24,6 eV/4
= 6,2 eV (24,6 eV ≡ He). Because of
wavefunctions penetration into the nucleus area we have 9,3 eV here. Beryllium
isn't so chemically inert as He.
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Energy states of Beryllium |
Boron:
z = 5
n |
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L | 2 |
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Ý | ||
K | 1 | Ýß | B |
(1s2) (2s2) (2p) ≡2P½
2P3/2 is also principally possible as the ground state. Nevertheless 2P½-state lays a bit deeper energetically (Hunds rule (2)); that can be described as follows: