91影视

Orthohelium should have lower energy than parahelium, for corresponding states

(which is true).

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

For nitrogen Hund鈥檚 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鈥檚 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鈥檚 easier to reason as follows: Suppose the three

outer electrons are in the \top of the ladder鈥� 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鈥檚 third rule

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