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3d Electron in hydrogen atom.

In order to identify the total angular momentum states $\left(\text{j},{\text{m}}_j\right)$ of a 3d hydrogen electron, the ordering rules for j and $m_j$will be required.

Total angular momentum quantum number j would be restricted to values-

$j=|l-s|,|l-s|+1,鈥�.,l+s-1,l+s$

The z projection of the total angular momentum quantum number $m_j$has the

values:$m_j:-j,-j+1,鈥�鈥�,j-1,j$

Here j is the total angular momentum quantum number.

Ignore the coupling between the L and S, the possibilities for the total angular momentum J are when L and S are parallel, or anti parallel.

Consequently, the values for the j for the two states:

$$\begin{gathered} j_{\min }=l-s \\ {j}_{\min }=l+s \end{gathered}$$

Since the d shell has an l of 2 , and that the electron has a spin s of$1/2$that becomes:

$$\begin{gathered} j_{\min }=l-s,j_{\min }=l+s \\ {j}_{\min }=\left(2\right)-\left(\frac{1}{2}\right),j_{\min }=\left(2\right)+\left(\frac{1}{2}\right) \\ {j}_{\min }=\frac{3}{2},j_{\min }=\frac{5}{2} \end{gathered}$$

So, then the two sets of$m_j$will be:

$$\begin{gathered} m_{j\min }=-\frac{3}{2},-\frac{1}{2},+\frac{3}{2} \\ {m}_{j\max }=-\frac{5}{2},-\frac{3}{2},-\frac{1}{2},+\frac{1}{2},+\frac{3}{2},+\frac{5}{2} \end{gathered}$$

Conclusion:

The total angular momentum state$\left(\text{j},{\text{m}}_j\right)$will be:

$$\begin{gathered} \left(\frac{3}{2},-\frac{3}{2}\right),\left(\frac{3}{2},-\frac{1}{2}\right),\left(\frac{3}{2},+\frac{1}{2}\right),\left(\frac{3}{2},+\frac{3}{2}\right), \\ \left(\frac{5}{2},-\frac{5}{2}\right),\left(\frac{5}{2},-\frac{3}{2}\right),\left(\frac{5}{2},-\frac{1}{2}\right),\left(\frac{5}{2},+\frac{1}{2}\right),\left(\frac{5}{2},+\frac{3}{2}\right),\left(\frac{5}{2},+\frac{5}{2}\right) \end{gathered}$$