Question

Which one of the following sets of quantum numbers represents an impossible arrangement?

• A. $$\begin{array}{cccc}n& l& m& s\\ 3& 2& -2& \frac{1}{2}\end{array}$$
• B. $$\begin{array}{cccc}n& l& m& s\\ 4& 0& 0& \frac{1}{2}\end{array}$$
• C. $$\begin{array}{cccc}n& l& m& s\\ 3& 2& -3& \frac{1}{2}\end{array}$$
• D. $$\begin{array}{cccc}n& l& m& s\\ 5& 3& 0& -\frac{1}{2}\end{array}$$

Answer 1

The correct option is C $$\begin{array}{cccc}n& l& m& s\\ 3& 2& -3& \frac{1}{2}\end{array}$$

For a given value of $n$ , possible values of $l$ are $0to(n-1)$, possible values of m are $-lto+l$ and possible values of $s=\frac{1}{2}or\frac{-1}{2}$
(a) For $$\begin{array}{cccc}n& l& m& s\\ 3& 2& -2& \frac{1}{2}\end{array}$$
all n, l,m and s values are fine as fir n=3, l=2,m=-2 and s=$\frac{1}{2}$ is possible.
(b) For n=4, l=0,m=0 and and s=$\frac{1}{2}$ is possible.
(c)
$$\begin{array}{cccc}n& l& m& s\\ 3& 2& -3& \frac{1}{2}\end{array}$$
This is the wrong set of quantum number because $|m|$ can not be greater than $l$ as values of m are from $-lto+l$ so, $m$ can't be $-3$.
(d)
$$\begin{array}{cccc}n& l& m& s\\ 5& 3& 0& -\frac{1}{2}\end{array}$$
For, n=3, value of l=3 is possible, for l=3, m=0 is possible and s=$\frac{-1}{2}$