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Which of the following is not a valid set of quantum numbers?

A. [tex]n=2, l=1, m=0[/tex]
B. [tex]n=1, l=0, m=0[/tex]
C. [tex]n=3, l=3, m=3[/tex]

Sagot :

GinnyAnswer
To determine which set of quantum numbers is invalid, we need to review the rules governing the values of quantum numbers in atomic physics:

1. The principal quantum number [tex]\( n \)[/tex]:
- [tex]\( n \)[/tex] must be a positive integer ([tex]\( n > 0 \)[/tex]).

2. The azimuthal (or angular momentum) quantum number [tex]\( l \)[/tex]:
- [tex]\( l \)[/tex] must be an integer such that [tex]\( 0 \le l < n \)[/tex].

3. The magnetic quantum number [tex]\( m \)[/tex]:
- [tex]\( m \)[/tex] must be an integer such that [tex]\( -l \le m \le l \)[/tex].

Let's examine each given set of quantum numbers according to these rules:

### Set 1: [tex]\( n = 2, l = 1, m = 0 \)[/tex]
- [tex]\( n = 2 \)[/tex] (valid, [tex]\( n > 0 \)[/tex]).
- [tex]\( l = 1 \)[/tex], and since [tex]\(0 \le l < 2\)[/tex], [tex]\( l \)[/tex] is valid.
- [tex]\( m = 0 \)[/tex], and since [tex]\( -1 \le 0 \le 1 \)[/tex], [tex]\( m \)[/tex] is valid.
- Conclusion: This set is valid.

### Set 2: [tex]\( n = 1, l = 0, m = 0 \)[/tex]
- [tex]\( n = 1 \)[/tex] (valid, [tex]\( n > 0 \)[/tex]).
- [tex]\( l = 0 \)[/tex], and since [tex]\(0 \le l < 1\)[/tex], [tex]\( l \)[/tex] is valid.
- [tex]\( m = 0 \)[/tex], and since [tex]\( -0 \le 0 \le 0 \)[/tex], [tex]\( m \)[/tex] is valid.
- Conclusion: This set is valid.

### Set 3: [tex]\( n = 3, l = 3, m = 3 \)[/tex]
- [tex]\( n = 3 \)[/tex] (valid, [tex]\( n > 0 \)[/tex]).
- [tex]\( l = 3 \)[/tex], however, [tex]\(0 \le l < 3\)[/tex] must hold, but [tex]\( l = 3 \)[/tex] is not less than [tex]\( n = 3 \)[/tex].
- Conclusion: This set is invalid due to the value of [tex]\( l \)[/tex].

Thus, the set of quantum numbers [tex]\( n=3, l=3, m=3 \)[/tex] is invalid because [tex]\( l \)[/tex] must be less than [tex]\( n \)[/tex]. Therefore, the invalid set of quantum numbers is set 3.
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