KAHIBARO
Table of Contents
Why the hydrogen atom matters
The hydrogen atom is the simplest atom, made of one proton and one electron. Because it has only one electron, it is the easiest real atom to study with quantum mechanics. It became one of the great successes of the Schrödinger equation, because the theory predicts discrete energy levels and explains the pattern of hydrogen spectral lines.
In this chapter, the focus is the quantum description of the hydrogen atom itself, its allowed energies, and the structure of its wave functions. Topics such as the Schrödinger equation, quantum numbers in general, and atomic spectra each have their own place in the course, so here they appear only as much as needed to understand hydrogen specifically.
The physical model
In the hydrogen atom, the electron moves under the electric attraction of the proton. The potential energy depends only on the distance $r$ between them, so it is a central potential:
$$
V(r) = -\frac{1}{4\pi\varepsilon_0}\frac{e^2}{r}
$$
This means the force and potential are spherically symmetric. Because of this symmetry, spherical coordinates are the natural choice for solving the problem.
Strictly speaking, both the proton and electron move around their common center of mass. A more accurate treatment replaces the electron mass $m_e$ by the reduced mass $\mu$:
$$
\mu = \frac{m_e m_p}{m_e + m_p}
$$
Since the proton mass $m_p$ is much larger than $m_e$, we often use $\mu \approx m_e$ as a first approximation.
For the hydrogen atom, the electron does not have arbitrary bound-state energies. Quantum mechanics allows only certain discrete energy values.
The Schrödinger equation for hydrogen
The time independent Schrödinger equation is
$$
-\frac{\hbar^2}{2\mu}\nabla^2 \psi + V(r)\psi = E\psi
$$
For hydrogen, this becomes
$$
-\frac{\hbar^2}{2\mu}\nabla^2 \psi - \frac{1}{4\pi\varepsilon_0}\frac{e^2}{r}\psi = E\psi
$$
Because the potential depends only on $r$, the wave function can be separated into a radial part and an angular part:
$$
\psi(r,\theta,\phi) = R(r)Y(\theta,\phi)
$$
More specifically,
$$
\psi_{n\ell m}(r,\theta,\phi) = R_{n\ell}(r)Y_{\ell}^{m}(\theta,\phi)
$$
The functions $Y_{\ell}^{m}$ are the spherical harmonics, which describe the angular dependence, while $R_{n\ell}(r)$ describes how the wave function changes with distance from the nucleus.
Quantization of energy
When the Schrödinger equation is solved for the Coulomb potential, the allowed bound-state energies are
$$
E_n = -\frac{\mu e^4}{2(4\pi\varepsilon_0)^2\hbar^2}\frac{1}{n^2}
$$
where $n = 1, 2, 3, \dots$ is the principal quantum number.
Using the electron mass approximation, this is usually written as
$$
E_n = -\frac{13.6\ \text{eV}}{n^2}
$$
So the first few levels are:
| $n$ | Energy $E_n$ |
|---|---|
| 1 | $-13.6\ \text{eV}$ |
| 2 | $-3.40\ \text{eV}$ |
| 3 | $-1.51\ \text{eV}$ |
| 4 | $-0.85\ \text{eV}$ |
Negative energy means the electron is bound to the proton. As $n$ increases, the energy approaches $0$ from below. The value $E = 0$ corresponds to a free electron infinitely far from the proton.
The hydrogen energy depends only on the principal quantum number $n$:
$$
E_n = -\frac{13.6\ \text{eV}}{n^2}
$$
A larger $n$ means a less tightly bound electron.
Quantum states of hydrogen
A hydrogen atom state is labeled by three quantum numbers:
$$
n, \ell, m
$$
The principal quantum number $n$ determines the energy. For each $n$, the orbital angular momentum quantum number $\ell$ can take values
$$
\ell = 0, 1, 2, \dots, n-1
$$
For each $\ell$, the magnetic quantum number $m$ can take values
$$
m = -\ell, -\ell+1, \dots, \ell
$$
These different combinations describe different wave functions, often called orbitals.
The orbital letters are named as follows:
| $\ell$ | Letter |
|---|---|
| 0 | $s$ |
| 1 | $p$ |
| 2 | $d$ |
| 3 | $f$ |
So the state with $n=1$, $\ell=0$ is the $1s$ state, the lowest energy state. For $n=2$, there are the $2s$ and $2p$ states. In hydrogen, all states with the same $n$ have the same energy in the basic Schrödinger model.
The ground state
The lowest energy state is the ground state, with
$$
n=1,\quad \ell=0,\quad m=0
$$
Its wave function is spherically symmetric. The ground-state wave function is
$$
\psi_{100}(r,\theta,\phi) = \frac{1}{\sqrt{\pi a_0^3}}e^{-r/a_0}
$$
where $a_0$ is the Bohr radius:
$$
a_0 = \frac{4\pi\varepsilon_0\hbar^2}{\mu e^2}
$$
Numerically,
$$
a_0 \approx 5.29\times 10^{-11}\ \text{m}
$$
The Bohr radius sets the characteristic size of the hydrogen atom.
The ground-state wave function of hydrogen is
$$
\psi_{100} = \frac{1}{\sqrt{\pi a_0^3}}e^{-r/a_0}
$$
and the characteristic atomic length scale is the Bohr radius $a_0$.
Probability density and most likely radius
The wave function itself is not directly observable. The probability density is
$$
|\psi(r,\theta,\phi)|^2
$$
For the $1s$ state, this becomes
$$
|\psi_{100}|^2 = \frac{1}{\pi a_0^3}e^{-2r/a_0}
$$
This gives the probability per unit volume. But in spherical coordinates, the probability of finding the electron between $r$ and $r+dr$ is found from the radial probability distribution:
$$
P(r)\,dr = 4\pi r^2 |\psi_{100}(r)|^2\,dr
$$
For the ground state,
$$
P(r) = \frac{4r^2}{a_0^3}e^{-2r/a_0}
$$
This function is largest at
$$
r = a_0
$$
So in the ground state, the most probable distance of the electron from the proton is the Bohr radius.
This shows an important idea. The wave function is largest at the nucleus, but the radial probability is not, because the spherical shell volume grows like $r^2$.
Shapes of orbitals
The hydrogen orbitals are not little planetary paths. They are standing-wave patterns described by wave functions. Their shapes come from the angular and radial parts together.
The $s$ orbitals are spherically symmetric. The $p$ orbitals have two main lobes with a nodal plane through the nucleus. Higher orbitals such as $d$ and $f$ have more complicated shapes.
A node is a place where the wave function is zero. Hydrogen states can have angular nodes and radial nodes. For hydrogen,
$$
\text{number of radial nodes} = n - \ell - 1
$$
and the number of angular nodes is $\ell$.
| State | $n$ | $\ell$ | Radial nodes | General shape |
|---|---|---|---|---|
| $1s$ | 1 | 0 | 0 | spherical |
| $2s$ | 2 | 0 | 1 | spherical, one radial node |
| $2p$ | 2 | 1 | 0 | two-lobed |
| $3p$ | 3 | 1 | 1 | two-lobed, one radial node |
These drawings are only qualitative. The true orbitals are mathematical probability distributions, not hard-edged objects.
Degeneracy in hydrogen
In the simple Schrödinger theory of hydrogen, all states with the same $n$ have the same energy, even if their $\ell$ and $m$ values differ. This is called degeneracy.
For a given $n$, the number of spatial states is
$$
\sum_{\ell=0}^{n-1}(2\ell+1) = n^2
$$
If electron spin is included, the total number becomes $2n^2$, because each spatial state can be combined with two spin states.
This degeneracy is a special feature of the Coulomb potential and the basic hydrogen model. More refined effects can split these equal energies, but those details belong elsewhere.
Transitions between energy levels
If the electron moves from one allowed energy level to another, the atom absorbs or emits a photon. The photon energy is the difference between the two levels:
$$
hf = |E_i - E_f|
$$
or
$$
\frac{hc}{\lambda} = |E_i - E_f|
$$
For hydrogen, using the energy formula,
$$
\Delta E = 13.6\ \text{eV}\left|\frac{1}{n_f^2} - \frac{1}{n_i^2}\right|
$$
An emission occurs when the electron drops to a lower level. An absorption occurs when the electron rises to a higher level.
A hydrogen atom can emit or absorb only photons whose energies match differences between allowed levels:
$$
\Delta E = hf = \frac{hc}{\lambda}
$$
This is why hydrogen has a line spectrum rather than a continuous spectrum.
Ionization energy
Ionization means removing the electron completely from the atom. For hydrogen in the ground state, the ionization energy is the energy needed to go from
$$
E_1 = -13.6\ \text{eV}
$$
to
$$
E=0
$$
So the ionization energy of ground-state hydrogen is
$$
13.6\ \text{eV}
$$
More generally, from the level $n$ the required ionization energy is
$$
E_{\text{ion}} = \frac{13.6\ \text{eV}}{n^2}
$$
This decreases for higher excited states.
Expectation values in hydrogen
Quantum mechanics often asks for average values rather than exact paths. In hydrogen, the expectation value of the electron's distance from the nucleus in the ground state is
$$
\langle r \rangle = \frac{3}{2}a_0
$$
This is larger than the most probable radius $a_0$. These are different ideas. The most probable radius is where the radial distribution peaks, while the expectation value is the average over many identical measurements.
This is a good example of how quantum particles differ from classical particles. The electron is described by a spread-out probability distribution, not by a single definite orbit.
What hydrogen teaches us
The hydrogen atom shows several central ideas of quantum mechanics in a concrete way. It shows that bound systems have discrete energies. It shows that wave functions determine probabilities. It shows that symmetry helps solve physical problems. It shows that atomic structure is naturally described by quantum numbers and orbitals.
Hydrogen is simple, but its importance is huge. It provides the foundation for understanding more complicated atoms, chemical structure, and atomic radiation.
Key hydrogen atom results:
$$
V(r) = -\frac{1}{4\pi\varepsilon_0}\frac{e^2}{r}
$$
$$
E_n = -\frac{13.6\ \text{eV}}{n^2}
$$
$$
a_0 = \frac{4\pi\varepsilon_0\hbar^2}{\mu e^2}
$$
$$
\psi_{100} = \frac{1}{\sqrt{\pi a_0^3}}e^{-r/a_0}
$$
These formulas summarize the basic quantum structure of the hydrogen atom.