Question

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?

Accepted 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

Science

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?

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