Question

Using the vector atom model, the possible values of the magnitude of angular momentum of an electron in the \(f\) shell are:

• A. \(\dfrac{3\sqrt{7}\,\hbar}{2},\ \dfrac{\sqrt{35}\,\hbar}{2}\)
• B. \(\dfrac{2\sqrt{7}\,\hbar}{2},\ \dfrac{\sqrt{25}\,\hbar}{2}\)
• C. \(\dfrac{5\sqrt{7}\,\hbar}{2},\ \dfrac{\sqrt{15}\,\hbar}{2}\)
• D. \(\dfrac{\sqrt{7}\,\hbar}{2},\ \dfrac{\sqrt{5}\,\hbar}{2}\)

Hint:

For an \(f\) electron \(l=3\), \(s=\tfrac12\); use \(j = l\pm s\) and \(|J| = \sqrt{j(j+1)}\,\hbar\).

Answer 1

The Correct Option is A

Step 1: In the vector atom model the magnitude of the total angular momentum is \(|\mathbf{J}| = \sqrt{j(j+1)}\,\hbar\), where \(j\) is the total angular momentum quantum number obtained from spin-orbit coupling \(j = l \pm s\).

Step 2: For an \(f\) electron the orbital quantum number is \(l = 3\) and the spin is \(s = \tfrac{1}{2}\). The two possible total values are
\[j = l + s = 3 + \tfrac{1}{2} = \tfrac{7}{2}, \qquad j = l - s = 3 - \tfrac{1}{2} = \tfrac{5}{2}\]
Step 3: For \(j = \tfrac{7}{2}\):
\[|\mathbf{J}| = \sqrt{\tfrac{7}{2}\left(\tfrac{7}{2}+1\right)}\,\hbar = \sqrt{\tfrac{7}{2}\cdot\tfrac{9}{2}}\,\hbar = \sqrt{\tfrac{63}{4}}\,\hbar = \frac{\sqrt{63}}{2}\,\hbar = \frac{3\sqrt{7}}{2}\,\hbar\]
Step 4: For \(j = \tfrac{5}{2}\):
\[|\mathbf{J}| = \sqrt{\tfrac{5}{2}\left(\tfrac{5}{2}+1\right)}\,\hbar = \sqrt{\tfrac{5}{2}\cdot\tfrac{7}{2}}\,\hbar = \sqrt{\tfrac{35}{4}}\,\hbar = \frac{\sqrt{35}}{2}\,\hbar\]
Step 5: The two magnitudes are \(\dfrac{3\sqrt{7}}{2}\hbar\) and \(\dfrac{\sqrt{35}}{2}\hbar\), which is option (A).
\[\boxed{|\mathbf{J}| = \frac{3\sqrt{7}}{2}\hbar \ \text{and}\ \frac{\sqrt{35}}{2}\hbar}\]