Answer :

To determine which set of quantum numbers is not valid, let’s review the rules for valid quantum numbers:

1. Principal quantum number ([tex]\(n\)[/tex]): This must be a positive integer ([tex]\(n > 0\)[/tex]).
2. Azimuthal quantum number ([tex]\(l\)[/tex]): This must be a non-negative integer such that [tex]\(0 \leq l < n\)[/tex].
3. Magnetic quantum number ([tex]\(m\)[/tex]): This must be an integer such that [tex]\(-l \leq m \leq l\)[/tex].

Now, let's analyze each given set of quantum numbers to check their validity:

1. Set 1: [tex]\(n = 2, l = 1, m = 0\)[/tex]
- [tex]\(n = 2\)[/tex] is a positive integer, so it's valid.
- [tex]\(l = 1\)[/tex] satisfies [tex]\(0 \leq l < n\)[/tex] because [tex]\(0 \leq 1 < 2\)[/tex], so it's valid.
- [tex]\(m = 0\)[/tex] satisfies [tex]\(-1 \leq 0 \leq 1\)[/tex] since [tex]\(l = 1\)[/tex], so it's valid.

Therefore, the set [tex]\(n = 2, l = 1, m = 0\)[/tex] is valid.

2. Set 2: [tex]\(n = 1, l = 0, m = 0\)[/tex]
- [tex]\(n = 1\)[/tex] is a positive integer, so it's valid.
- [tex]\(l = 0\)[/tex] satisfies [tex]\(0 \leq l < n\)[/tex] because [tex]\(0 \leq 0 < 1\)[/tex], so it’s valid.
- [tex]\(m = 0\)[/tex] satisfies [tex]\(-0 \leq 0 \leq 0\)[/tex] since [tex]\(l = 0\)[/tex], so it’s valid.

Therefore, the set [tex]\(n = 1, l = 0, m = 0\)[/tex] is valid.

3. Set 3: [tex]\(n = 3, l = 3, m = 3\)[/tex]
- [tex]\(n = 3\)[/tex] is a positive integer, so it's valid.
- [tex]\(l = 3\)[/tex] does not satisfy [tex]\(0 \leq l < n\)[/tex] because [tex]\(0 \leq 3 < 3\)[/tex] is false.

Therefore, the set [tex]\(n = 3, l = 3, m = 3\)[/tex] is not valid because [tex]\(l\)[/tex] must be less than [tex]\(n\)[/tex] but here [tex]\(l\)[/tex] equals [tex]\(n\)[/tex].

So, the invalid set of quantum numbers is:

[tex]\[ n = 3, l = 3, m = 3 \][/tex]

This is the third set from the list. Hence, the result indicates that the third set is the invalid one.