Many-electron Atoms and Coupling Schemes Questions and Answers

Answer: D

Auger emission cannot be obtained from hydrogen.

The Auger process needs three electrons: an inner-shell vacancy is created, an electron from a higher level drops into it, and the energy released ejects a third electron (the Auger electron) instead of emerging as an X-ray. An atom must therefore have electrons in at least two shells, at least three in all.

Hydrogen has only one electron (and helium only two), so neither can emit Auger electrons. Carbon, oxygen and nickel have enough electrons; carbon and oxygen Auger lines are in fact standard signals in Auger electron spectroscopy of surfaces.

Answer: A

The ground state of \(Eu^{3+}\) is \({}^{7}F_{0}\).

Apply Hund's rules to the six 4f electrons:

1. Maximum spin: all six spins parallel in different orbitals, so S = 3 and 2S + 1 = 7.
2. Maximum L: they occupy \(m_{l}\) = 3, 2, 1, 0, -1, -2, so L = 3, an F term.
3. The shell is less than half full (six of fourteen), so J = |L - S| = 0.

The term is \({}^{7}F_{0}\). \({}^{8}S_{7/2}\) is the ground state of a half-filled \(4f^{7}\) shell, as in \(Gd^{3+}\) or \(Eu^{2+}\). \({}^{7}F_{6}\) has the J value for a more than half-filled shell, as in \(Tb^{3+}\) (\(4f^{8}\)), and \({}^{6}H_{15/2}\) is the ground state of \(Dy^{3+}\) (\(4f^{9}\)).

Answer: D

The ground term is \({}^{5}D_{0}\).

Apply Hund's rules to \(3d^{4}\):

• Maximum S: four parallel spins, \(S = 2\), multiplicity \(2S + 1 = 5\).
• Maximum L for that S: \(m_{l} = 2, 1, 0, -1\), so \(L = 2\), a D term.
• The shell is less than half full, so \(J = |L - S| = 0\).

The term is \({}^{5}D_{0}\). \({}^{1}D_{2}\) has the wrong spin, and the terms with half-integer J require an odd number of electrons.

Answer: C

There are 6 permitted transitions.

In a weak field, \({}^{2}P_{3/2}\) splits into four levels (\(M_{J} = \pm 3/2, \pm 1/2\), \(g = 4/3\)) and \({}^{2}S_{1/2}\) into two (\(M_{J} = \pm 1/2\), \(g = 2\)). The selection rule \(\Delta M_{J} = 0, \pm 1\) allows:

• from \(+3/2\): to \(+1/2\) only
• from \(+1/2\): to \(+1/2\) and \(-1/2\)
• from \(-1/2\): to \(+1/2\) and \(-1/2\)
• from \(-3/2\): to \(-1/2\) only

That is 6 transitions, the anomalous Zeeman pattern of the \(D_{2}\) line. 4 is the count for \({}^{2}P_{1/2} \to {}^{2}S_{1/2}\), and 2 or 10 do not fit the selection rule.