3.1: The Schroedinger equation for the H atom

The Hamiltonian of H atom

The H atom is the simplest atomic system, consisting of two particles, an electron and a proton, with opposite charges. Since the proton is so much heavier than the electron (about 1800 times) we could treat the proton as fixed in space and located at the center of coordinates, and the electron at a distance \(r \) from that proton. This leads to a small variation in the energy, something that is negligible in the overall understanding of the the H atom for our chemistry purposes.

Note

A better description reduces the system into a single particle with reduced mass \(\mu\), where \[\dfrac{1}{\mu}=\dfrac{1}{m_{electron}} +\dfrac{1}{m_{nucleus}}.\] This particle moves with respect to the center of coordinates.

The Schröedinger equation for the H atom provides the wave equation to understand the probability density for the electron in different states as well as the energy of each of those states. Although we don't need to learn to solve the mathematical equations, seeing their structure help us understand important properties of the H atom.

We start by considering that the energy observable will have its associated operator: the Hamiltonian. This Hamiltonian will have contributions from the Kinetic operator (associated with kinetic energy) and Potential operator (associated with the observable potential energy).

\[\hat{H} = \hat{T}+\hat{V} \label{1}\] where \(\hat{T}\) is the kinetic operator and \(\hat{V}\) is the potential operator. For the H atom, the potential operator \(\textcolor{purple}{\hat{V}}\) is unique and includes the electrostatic interaction between the two particles with opposite charge. According to Coulomb's law \[ V(r) = \dfrac{q_1\cdot q_2}{4\pi \epsilon_o r}\label{2} \nonumber \] were \(q_i\) corresponds to the charge of each particle, \(r \) is the distance between them and \(\epsilon_o\) is the constant of permittivity in vaccuum. Given that the electron and the proton have the same charge \(e=1.60217662 \cdot 10^{-19} C\) with opposite sign, the Coulomb potential becomes \[\hat{V} = -\dfrac{e^2}{4\pi \epsilon_o \hat{r}} \nonumber \]

If we compare the electrostatic interaction between one electron and a nucleus that has a charge \(Z\cdot (-e)\), then the complete potential operator will become
\[\textcolor{purple}{\hat{V} = -\dfrac{Z e^2}{4\pi \epsilon_o \hat{r}}} \label{3}\] .
Equation \ref{3} can be used for the interaction of a single electron at any nucleus with atom of atomic number \(Z\).

The kinetic operator always has the same form, with a contribution from each coordinate leading to
\[ \textcolor{purple}{\hat{T} = -\dfrac{\hbar^2}{2\mu} (\dfrac{\partial^2}{\partial x^2}+\dfrac{\partial^2}{\partial y^2}+\dfrac{\partial^2}{\partial x^2})}\]

where the symbol \(\dfrac{\partial}{\partial x}\) represents the partial derivative with respect to \(x\).
The way it is written, the kinetic component is in Cartesian coordinates, but to combine both components of \(\textcolor{purple}{\hat{H}}\) we need to have them in the same system of coordinates. Since the H-atom has a centro-symmetrical symmetry it is more convenient to use everything in spherical coordinates \(r, \theta, \phi\) where
\begin{align}
x &= r \cdot \cos\phi \cdot \sin \theta  \nonumber \\
y &= r \cdot \sin\phi \cdot \sin \theta  \nonumber\\
z&= r \cdot \cos\theta \label{4}\\
\end{align}

The Schrödinger Equation of H-like Atoms and its Solutions

Once we combine both contributions, the full Schrödinger equation can be written. This equation goes beyond a General Chemistry text, and it is shown here to provide a reference for the work ahead.

\[\textcolor{purple}{\hat{H}} \Psi(r,\theta,\phi) = \textcolor{darkgreen}{E}\cdot \Psi(r,\theta\phi) \]
\[\textcolor{purple}{\dfrac{-\hbar^2}{2 \mu}\left[\dfrac{\partial^2}{\partial r^2}+\dfrac{2}{r}\dfrac{\partial}{\partial r}+\dfrac{1}{\sin\theta} \left(\dfrac{\partial}{\partial \theta}\sin \theta \dfrac{\partial}{\partial \theta}\right)+\dfrac{1}{r^2 \sin\theta}\dfrac{\partial^2}{\partial \phi^2}\right]} \Psi(r,\theta,\phi) = \textcolor{darkgreen}{E}\cdot \Psi(r,\theta\phi) \]

The solution to this equation provides an infinite number of wavefunctions, \[\Psi_{n,l,m_l}(r,\theta,\phi)\label{5}\] each solution is a function of the coordinates \(r, \theta, \phi\) and each solution is characterized by 3 quantum numbers: \(n, l, m_l\). Each wavefunction is called an Orbital, and its probability density \(|Psi_{n,l,m}(r,\theta,\phi)|^2\) provides the probability to find the electron at a particular point in space when it is in the state labeld by \(n, l, m_l\).

Why 3 quantum numbers?

When solving the Schrödinger equation the wavefucntion must obey the Boundary Conditions. These are specific conditions that provide physical meaning to the problem. (For the particle in a 1-D box, we saw that the wavefunction evaluated at the edges of the box had to be zero \(\psi(0)=\psi(l) =0\). This was a requirement to have a smooth wavefunction that outside the box could not exist).

For the H atom, we know that as the radius \(r\) goes to \(infty\), electron and nucleus are so far apart that the potential interaction will \(\rightarrow 0\), which also means that \(\Psi(r_{\rightarrow 0},\theta,\phi) \rightarrow =0\). This boundary condition sets some limits on the wavefunction and as a consequence, the quantum number \(n\) and \( l\) have specific values. An additional boundary condition considers that the value of a wavefunction has to be unique, therefore \[|\Psi(r,\theta, \phi)|^2 = |\Psi(r,\theta + 2\pi, \phi)|^2\] which leads to specific values on the quantum number \(m_l\).

The possible values for these quantum numbers are

\begin{align} &n,\text{ Principal quantum number } 1,2,3...,\infty \nonumber\\
&l,\text{ Angular momentum or Azimuthal quantum number } 0, 1, 2, ...,n-1 \nonumber\\
&m_l, \text{ Magnetic moment quantum number } 0, \pm1, \pm2, ..., \pm l \nonumber
\end{align}

The Energies

Although each orbital is defined by a set of quantum numbers, the energy of the H-like wavefunctions only depends on the principal quantum number \(n\).
\[\textcolor{darkgreen}{E_n = -\dfrac{\mu e^4}{8h^2 \epsilon_o} \dfrac{1}{n^2} =-hcR_H \dfrac{1}{n^2}}\]
where \(R_H\) is the Rydberg constant for the H atom. The equation shows that the energy of the states of H-like atoms is quantized, and the energy required to change between two states is given by
\begin{align}
\Delta E&= E_{final}-E_{initial}\\
&=-hcR_H \left( \dfrac{1}{n^2_{f}}-\dfrac{1}{n^2_i}\right)\\
\end{align}
which resembles the phenomenological equation used to model the line emission spectra of H. If we take into account that \(\Delta E\) of the transition must have the same energy as the emitted photons \(\Delta E = E_{photon} = h\nu\) and that the experimentally known lines correspond to different combination of \(n_{final}\) and \(n_{initial}\) we conclude that starting from the fundamental properties of quantum mechanics we were able to understand an experimental measurement that could not be properly explained by classical mechanics.

It is useful to note that

• The energies are all negative, describing the binding nature of the electron-nucleus interactions
• The energy depends on the inverse of \(n^2\)
• Since the energy does not depend on \(l\) or \(m_l\), there might be multiple states (orbitals) with different combinations of \(l,m_l\) corresponding to the same \(n\) value and therefore having the same energies (degeneracy)
• As the value of n increases, the energy of the orbital gets closer to zero (less negative values).

Example \(\PageIndex{1}\)

Calculate the minimum energy required to extract an electron from the H atom in its ground state.

Solution

To solve this problem we need to identify the initial and final states. The ground state corresponds to the state with lowest energy, which is \(n=1\). To identify the final state we consider that after the electron is removed from an atom there is no more potential interaction between electron and nucleus. The electron is not bound anymore, thus its energy cannot be negative (A free particle always has positive energies). The minimum energy to unbind the electron must be when \(E_{final}\rightarrow 0\) which corresponds to \(n\rightarrow \infty\).

\begin{align}
\Delta E &= -hcR_H \left(\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2}\right) \nonumber\\
&=-hcR_H\left(\dfrac{1}{n(\rightarrow \infty)^2}-\dfrac{1}{1^2}\right)  \nonumber\\
&=-hcR_H \cdot(-1)\\
\end{align}

Using the tabulated values for \(h, c, R_H\) we obtain
\[\Delta E= 2.18 \cdot 10^{-18} Joules = 13.6 eV\]

In the gas phase this energy corresponds to the reaction
\[ H \rightarrow H^+ + e-  \nonumber \]
which is known as the ionization of Hydrogen. If we multiply by Avogadro's number we obtain the ionization energy, IE, per mole of atoms.
\[IE=2.18 \cdot 10^{-18} \frac{J}{atom} \cdot 6.022 \cdot 10^{23} \frac{atoms}{mol} = 1.31 \cdot 10^{-6} \frac{J}{mol}\]