Question
The first shell may contain up to 2 electrons, the second shell up to 8, the third shell up to 18, and the fourth shell up to 32. Explain this arrangement in terms of quantum numbers.
✓
Answer
For the fust shell $\text{n}=1,\text{l}=0,\text{m}_\text{l}=0,\text{m}_\text{s}=+\frac{1}{2}$ and $-\frac{1}{2}.$ It can have 2 electrons both with opposite spins.
$\text{l}=1,\text{m}_\text{l}=1,0+1,\text{m}_\text{s}=+\frac{1}{2},-\frac{1}{2}.$
Therefore, a total of 2 + 6 = 8 electrons are present.
$\text{l}=1,\text{m}_\text{l}=-1,0+1,\text{m}_\text{s}=\frac{1}{2},-\frac{1}{2}$
$\text{l}=2,\text{m}_\text{l}=-2,-1,0,+1,+2,,\text{m}_\text{s}$
$=+\frac{1}{2},-\frac{1}{2}$
Therefore, a total of 2 + 6 + 10 = 18 electlons aie present.
$\text{l}=1,\text{m}_\text{l}=-1,0+1,\text{m}_\text{s}=+\frac{1}{2},-\frac{1}{2}$
$\text{l}=3,\text{m}_\text{l}=-2,-1,0+1,+2,\text{m}_\text{s}=+\frac{1}{2},-\frac{1}{2}$
$\text{l}=3,\text{m}_\text{l}=-3,-2,-1,0,+1,+2,+3,\text{m}_\text{s}$
$=+\frac{1}{2},-\frac{1}{2}$
Therefore, a total of 2 + 6 + 10 + 14 = 32 electrons are present.
- For $\text{n}=2,$
$\text{l}=1,\text{m}_\text{l}=1,0+1,\text{m}_\text{s}=+\frac{1}{2},-\frac{1}{2}.$
Therefore, a total of 2 + 6 = 8 electrons are present.
- For $\text{n}=3,$ when
$\text{l}=1,\text{m}_\text{l}=-1,0+1,\text{m}_\text{s}=\frac{1}{2},-\frac{1}{2}$
$\text{l}=2,\text{m}_\text{l}=-2,-1,0,+1,+2,,\text{m}_\text{s}$
$=+\frac{1}{2},-\frac{1}{2}$
Therefore, a total of 2 + 6 + 10 = 18 electlons aie present.
- For $\text{n}=4,$ when
$\text{l}=1,\text{m}_\text{l}=-1,0+1,\text{m}_\text{s}=+\frac{1}{2},-\frac{1}{2}$
$\text{l}=3,\text{m}_\text{l}=-2,-1,0+1,+2,\text{m}_\text{s}=+\frac{1}{2},-\frac{1}{2}$
$\text{l}=3,\text{m}_\text{l}=-3,-2,-1,0,+1,+2,+3,\text{m}_\text{s}$
$=+\frac{1}{2},-\frac{1}{2}$
Therefore, a total of 2 + 6 + 10 + 14 = 32 electrons are present.
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