# If every electron must have a unique set of four quantum numbers, how many different electrons (sets of four quantum numbers) can there be for each principal quantum number from n = 1 to n = 3?

*print*Print*list*Cite

### 1 Answer

The quantum numbers of an electron are n (determined by electron's energy), l (angular momentum), `m_l` (one component of angular momentum) and `m_s` (spin, or intrinsic magnetic moment), and they have to obey certain restrictions, which are:

`0 <=l<=n-1` (l is an integer)

`-l<=m_l<=l` (`m_l` is an integer)

`m_s` is a positive or negative half-integer.

So for n = 1, the only possible value of l is l=0, so `m_l = 0` as well and `m_s` can be + or - 1/2. So there are two possible electrons for n = 1.

For n = 2, the possible values of l are 0 and 1.

If l = 0, then `m_l = 0` and there are two possible values of `m_s` .

If l = 1, then `m_l` can be -1, 0, or 1, and for each of these there are two possible values of `m_s` .

Thus, for n = 2, there are 2 + 3*2 = 8 possible electrons.

For n = 3, l can be 0, 1, and 2.

As discussed above, there will be 2 electrons with l = 0 and 6 electrons with l = 1.

If l = 2, there will be 5 possible values of `m_l` : -2, -1, 0, 1, 2, and each of these electrons have +1/2 or -1/2 spin (`m_s` .

So for n = 3, there are 2 + 6 + 2*5 = 18 electrons.

To sum up:

**n=1 : 2 electrons**

**n=2: 8 electrons**

**n=3: 18 electrons**

**Sources:**