Apply the techniques of this Section to the ${\mathbf{H}}^{\mathbf{-}}$and ${\mathbf{Li}}^{\mathbf{+}}$ions (each has two electrons, like helium, but nuclear charges Z=1and Z=3, respectively). Find the effective (partially shielded) nuclear charge, and determine the best upper bound on ${\mathbf{E}}_{\mathbf{gs}}$, for each case. Comment: In the case of ${\mathbf{H}}^{\mathbf{-}}$ you should find that $\left(H\right)\mathbf{>}\mathbf{-}\mathbf{13}\mathbf{.}\mathbf{6}\mathbf{eV}$, which would appear to indicate that there is no bound state at all, since it would be energetically favourable for one electron to fly off, leaving behind a neutral hydrogen atom. This is not entirely surprising, since the electrons are less strongly attracted to the nucleus than they are in helium, and the electron repulsion tends to break the atom apart. However, it turns out to be incorrect. With a more sophisticated trial wave function (see Problem 7.18) it can be shown that ${\mathbf{E}}_{\mathbf{gs}}\mathbf{<}\mathbf{-}\mathbf{13}\mathbf{.}\mathbf{6}\mathbf{eV}$and hence that a bound state does exist. It's only barely bound however, and there are no excited bound states, so${\mathbf{H}}^{\mathbf{-}}$ has no discrete spectrum (all transitions are to and from the continuum). As a result, it is difficult to study in the laboratory, although it exists in great abundance on the surface of the sun.

Step by step solution

01

Definition of electron repulsion.

The idea that electron pairs in a ring around a central atom will want to stay as far apart as feasible. The shape of a molecule or a polyatomic ion is predicted using electron pair repulsion.

02

For H-and Li+effective charge and estimate on energy of ground state.

The only difference from helium atom is that in equation 7.28 , in last parenthesis

instead of 2 , put $Z^{\text{'}}$. $Z^{\text{'}}=1$for H and $Z^{\text{'}}=2$for$L{i}^{+}$ .

Now equation 7.32 becomes:

$$\begin{gathered} \left(H\right)=\left[2Z^2-4Z\left(Z-Z^{\text{'}}\right)-\frac{5}{4}Z\right]E_1 \\ =\left(-2Z^2+4Z{Z}^{\text{'}}\frac{5}{4}Z\right)E_1 \\ =\frac{\partial \left(H\right)}{\partial Z}=0\mathrm{n}-4\mathrm{Z}+4{\mathrm{Z}}^{\text{'}}-\frac{5}{4} \\ =0 \end{gathered}$$

$$\begin{gathered} \mathrm{Z}={\mathrm{Z}}^{\text{'}}-\frac{5}{16}.{\left(\mathrm{H}\right)}_{\min } \\ =\left[2{\left({\mathrm{Z}}^{\text{'}}-\frac{5}{16}\right)}^2-4\left({\mathrm{Z}}^{\text{'}}-\frac{5}{16}\right)\left({\mathrm{Z}}^{\text{'}}-\frac{5}{16}-\mathrm{Z}\right)-\frac{5}{4}\left({\mathrm{Z}}^{\text{'}}-\frac{5}{16}\right)\right]{\mathrm{E}}_1 \\ =\left[2{\mathrm{Z}}^{\text{'}2}-\frac{5}{4}{\mathrm{Z}}^{\text{'}}+\frac{25}{128}+\frac{5}{4}\left({\mathrm{Z}}^{\text{'}}-\frac{5}{16}\right)-{\mathrm{Z}}^{\text{'}}\frac{25}{64}\right]{\mathrm{E}}_1 \\ =\left(2{\mathrm{Z}}^{\text{'}2}-\frac{5}{4}{\mathrm{Z}}^{\text{'}}+\frac{25}{128}\right){\mathrm{E}}_1 \end{gathered}$$

$$\begin{gathered} \mathrm{for}{\mathrm{H}}^{-}{\mathrm{Z}}^{\text{'}}=1\mathrm{effective}\mathrm{charge}\mathrm{and}\mathrm{estimate}\mathrm{on}\mathrm{engry}\mathrm{of}\mathrm{ground}\mathrm{state}, \\ \mathrm{Z}=1-\frac{5}{16} \\ =0.688 \\ {\left(\mathrm{H}\right)}_{\min }=0.945{\mathrm{E}}_1 \\ =-12.85\mathrm{eV} \\ \mathrm{for}{\mathrm{Li}}^{+}{\mathrm{Z}}^{\text{'}}=3\mathrm{effective}\mathrm{charge}\mathrm{and}\mathrm{estimate}\mathrm{on}\mathrm{energy}\mathrm{of}\mathrm{ground}\mathrm{state}, \\ \mathrm{Z}=3-\frac{5}{16} \\ =2.688 \\ {\left(\mathrm{H}\right)}_{\min }=14.445{\mathrm{E}}_1 \\ =196.45\mathrm{eV} \end{gathered}$$

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