The Real Hydrogen Atom

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The structures of atoms are described by quantum mechanics. This is a complicated theory which developed over time following a great deal of experimental input. Now, it is used routinely to model the behaviour of electrons, atoms and molecules. At its heart lies the Schrodinger equation, which is an equation for an object called the wavefunction, written as $\psi({\bf x}, t)$. The square of the wavefunction is a probability density function such that $|(\psi{\bf x}, t)|^2dV$ is the probability of finding the particle in a small ball of size $dV$ centred on ${\bf x}$ at time $t$. Solving the Schrodinger equation allows chemists to determine the physical form of the electron orbitals of atoms.

The general Schrodinger equation is very complicated, but for the motion of an electron around a single hydrogen nucleus reduces to three relatively simple equations when we use spherical polar coordinates $(r, \theta, \phi)$ (in terms of the globe $\phi$ is the longitude and $\theta$ is the latitude of a point) to describe the location of the electron through three functions $R(r), P(\theta), F(\phi)$

$$
\frac{1}{R}\frac{d}{dr}\left[r^2 \frac{dR}{dr}\right] -\frac{8\pi^2\mu}{h^2}\left(Er^2+\alpha r\right)=l(l+1)\;,
$$
$$
\frac{\sin\theta}{P}\frac{d}{d\theta}\left[\sin\theta \frac{dP}{d\theta}\right] +l(l+1)\sin^2\theta =m^2_l\;,
$$
$$
\frac{1}{F}\frac{d^2F}{d\phi^2}=-m^2_l\;.
$$
These equations involve the effective mass $\mu$ of the electron in terms of the mass of the electron and proton $m_e$ and $m_p$ as $\mu = \frac{m_em_p}{m_e+m_p}$, $\alpha = \frac{e^2}{4\pi\epsilon_0}$, the energy $E$ of the particle and Planck's constant $h =6.626068\times 10^{-34}$ m$^2$ kg s$^{-1}$.

There are three quantum numbers associated with these equations:

The principal quantum number $n=1, 2, 3, \dots$

The orbital quantum numbers $l = 0, 1, \dots, n-1$.

The magnetic quantum number $m= -l, -l+1, \dots, l -1, l$.

The energy $E$ of the electron depends on the principal quantum number as
$$
E=-\frac{\mu e^4}{8 n^2h^2\epsilon^2_0}\;.
$$
Solving this tricky set of equations gives the wave function for the electron to be $\psi({\bf x}, t)=R(r)P(\theta)F(\phi)$, from which the likely positions of the electron can be determined.

The s, p, d and f shells correspond to solutions with $l=0,1,2$ and $3$ respectively.

The goal of the quantum chemist is to attempt to extract solutions from these equations.

Task 0: Familiarise yourself with the structure of these equations. For example, what does the lowest energy ($n=1$) set of equations look like?

Task 1: Show that if $(R_1, P_1, F_1)$ is a solution to the set of equations then $(aR_1, bP_1, cF_1)$ will also be a solution for any non-zero constants $a, b$ and $c$.

Task 2: Start simply: do constant values for either $R, P$ or $F$ satisfy the equations for particular choices of $n, l, m$?

Task 3: Solving the $P$ equation will clearly be likely to involve trigonometric functions. Do $P=\sin\theta, \cos\theta, \tan\theta$ solve the $P$ equation for particular choices of $l$ and $m$? How about other simple combinations of $\sin$ and $\cos$?

Task 4: The $R$ equation contains many constants. Group these together as $a_0 = \frac{\pi e^2\mu}{h^2\epsilon_0}$ and see how the equation simplifies a little.

Task 5: On physical grounds we would expect the chance of the electron being very far away from the nucleus to be very small, tending to zero as $r$ increases. Why would a function $R = f(r)e^{-ar}$ satisfy this condition for any polynomial $f(r)$ and positive constant $a$? Try the simplest case $R=e^{-ar}$. Does this result in a solution? What about the possibilities for $R= (a+br)e^{-cr}$? for constants $a, b, c$?