# Quantum Chemistry

Chapter08 - Home
Exercises: 1 - 2 - 3 - 4 - 5 - 6 - 7 - 8 - 9 - 10 - 11 - 12 - 13 - 14 - 15 - 16 - 17 - 18 - 19 - 20 - 21 - 22 - 23 - 24 - 25 - 26 - 27 - 28 - 29 - 30 - 31 - 32 - 33 - 34 - 35 - 36 - 37 - 38 - 39 - 40 - 41 - 42 - 43 - 44

Exercise. This is the solution to exercise 8.1 in the book.

Solution. For the helium ion we have the same wavefunction as for hydrogen atom except $Z=2$. This was shown in exercise 6.39.

${E}_{n}=-\frac{\mu {Z}^{2}{e}^{4}}{8{ϵ}_{0}^{2}{h}^{2}{n}^{2}}\phantom{\rule{1em}{0ex}}n=1,2\dots$

${a}_{0}=\frac{{ϵ}_{0}{h}^{2}}{\pi \mu {e}^{2}}$

$\mu$ in this context is the reduced mass of the nucleus/electron system. The ground state energy is:

${E}_{1}=-\frac{\mu {Z}^{2}{e}^{4}}{8{ϵ}_{0}^{2}{h}^{2}}$

Thus

${E}_{1}=-\frac{{Z}^{2}{e}^{2}}{8{ϵ}_{0}\pi {a}_{0}}=-\frac{{Z}^{2}{e}^{2}}{2{\kappa }_{0}{a}_{0}}$

In atomic units $e=1,{\kappa }_{0}=1,{a}_{0}=1$ and for helium $Z=2$. Therefore:

${E}_{1}=-\frac{{2}^{2}}{2}=-2\phantom{\rule{1em}{0ex}}a.u.$