Question

How many possible states are there for an electron in a hydrogen atom when the principle quantum number (a) n= 1 and (b) n =2.?

Answer 1

For a one-electron atom, the number of states is given by the number of quantum states available in all the orbitals that could be occupied. You can actually do this by hand without taking too much time.

Thus, there are `10` states (`2` for `n = 1`, `8` for `n = 2`):

`ul(uarr color(white)(darr))` `+` spin-down possibility
`1s`

`ul(uarr color(white)(darr))` `+` spin-down possibility
`2s`

`ul(uarr color(white)(darr))" "ul(color(white)(uarr darr))" "ul(color(white)(uarr darr))` `xx 3` for three orbitals
`underbrace(" "" "" "" "" "" "" "" ")`
`" "" "" "2p`

`ul(darr color(white)(darr))" "ul(color(white)(uarr darr))" "ul(color(white)(uarr darr))` `xx 3` for three orbitals
`underbrace(" "" "" "" "" "" "" "" ")`
`" "" "" "2p`

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• At `n = 1`, we have only the `1s` orbital, `l = 0`.
• At `n = 2`, we have the `2s` and `2p` orbitals, `l = 0,1`.

The degeneracy `2l+1` of the orbitals then leads to the number of spin-up states possible:

`n = 1, l = 0 -> 2l + 1 = 1`

`n = 2, l = 0 -> 2l + 1 = 1`

`n = 2, l = 1 -> 2l + 1 = 3`

This gives `5` allowed spin-up states. With only one electron, it can take on the other spin with no restrictions. Thus, since `m_s` can be `+1/2` or `-1/2` with no issues, we have `bb10` states possible:

`ul(uarr color(white)(darr))` `+` spin-down possibility
`1s`

`ul(uarr color(white)(darr))` `+` spin-down possibility
`2s`

`ul(uarr color(white)(darr))" "ul(color(white)(uarr darr))" "ul(color(white)(uarr darr))` `xx 3` for three orbitals
`underbrace(" "" "" "" "" "" "" "" ")`
`" "" "" "2p`

`ul(darr color(white)(darr))" "ul(color(white)(uarr darr))" "ul(color(white)(uarr darr))` `xx 3` for three orbitals
`underbrace(" "" "" "" "" "" "" "" ")`
`" "" "" "2p`