a) Hund’s first rule says that, consistent with the Pauli principle, the state with the highest total spin (S) will have the lowest energy. What would this predict in the case of the excited states of helium?

(b) Hund’s second rule says that, for a given spin, the state with the highest total orbital angular momentum (L) , consistent with overall antisymmetrization, will have the lowest energy. Why doesn’t carbon have$\mathbf{L}\mathbf{=}\mathbf{2}$? Note that the “top of the ladder”$\left(ML=L\right)$is symmetric.

(c) Hund’s third rule says that if a subshell$\left(n,\mathcal{l}\right)$is no more than half filled,
then the lowest energy level has$\mathit{J}\mathbf{=}\mathit{l}\mathit{L}\mathbf{-}\mathit{S}\mathbf{}\mathit{I}$; if it is more than half filled, then$\mathbf{J}\mathbf{=}\mathbf{L}\mathbf{+}\mathbf{S}$has the lowest energy. Use this to resolve the boron ambiguity inProblem 5.12(b).

(d) Use Hund’s rules, together with the fact that a symmetric spin state must go with an antisymmetric position state (and vice versa) to resolve the carbon and nitrogen ambiguities in Problem 5.12(b). Hint: Always go to the “top of the ladder” to figure out the symmetry of a state.

Orthohelium should have lower energy than parahelium, for corresponding states

(which is true).

Hund’s first rule says S = 1 for the ground state of carbon. But this (the triplet) is symmetric, so the orbital state will have to be antisymmetric. Hund’s second rule favors L = 2, but this is symmetric, as you can see most easily by going to the top of the ladder. $|22>=|11{>}_1|11{>}_2$So the ground state of carbon will be S = 1,L = 1. This leaves three possibilities:${}^{3}P_{2,}{}^{3}P_{1,}and{}^{3}P_{0,}$

For boron there is only one electron in the 2p subshell (which can accommodate a total

of 6), so Hund’s third rule says the ground state will have $J=|L-S|$. We found in

Problem 5.12(b) that$L=1andS=1/2,soJ=1/2$, and the configuration is${}^2{P}_{1/2}$

For carbon we know that S = 1 and L = 1, and there are only two electrons in the outer

subshell, so Hund’s third rule says J = 0, and the ground state configuration must be${}^3{P}_0$.

For nitrogen Hund’s first rule says S = 3/2, which is symmetric (the top of the ladder is

$\overline{)\frac{3}{2}\frac{3}{2}>=\overline{)\frac{1}{2}\frac{1}{2}{>}_1}\overline{)\frac{1}{2}\frac{1}{2}{>}_2}\overline{)\frac{1}{2}\frac{1}{2}{>}_3})}.$.

Hund’s second rule favors L = 3, but this is also symmetric. In fact, the only antisymmetric

orbital configuration here is L = 0. [You can check this directly by working out the

ClebschGordan- co-effcients, but it’s easier to reason as follows: Suppose the three

outer electrons are in the \top of the ladder” spin state, so each one has spin up $\left(\overline{)\frac{1}{2}\frac{1}{2}>}\right)$;

then (since the spin states are all the same) the orbital states have to be different:$|11>,|10>,and|1-1>$ .

In particular, the total z-component of orbital angular momentum has to be zero.

But the only configuration that restricts $LztozeroisL=0.]$The outer subshell is exactly half filled

(three electrons with n = 2, l = 1), so Hund’s third rule

says $J=|L-S|=\left|0-\frac{3}{2}\right|=3/2$Conclusion: The groundstate of nitrogen is${}^4{S}_{3/2}$