41.1.1 Stationary States of Hydrogen

In one dimension, energy quantization appeared as a consequence of boundary conditions on the wave function. That is, only for certain discrete energies, characterized by the quantum number $n$ , did solutions to the Schrödinger equation satisfy the boundary conditions. In three dimensions, the wave function must satisfy three different boundary conditions. Consequently, solutions to the three-dimensional Schrödinger equation have three quantum numbers and three quantized parameters.

Solutions to the Schrödinger equation for the hydrogen atom potential energy exist only if three conditions are satisfied:

The atom’s energy must be one of the values

$E_{n} = - \frac{1}{n^{2}} (\frac{1}{4 π ϵ_{0}} \frac{e^{2}}{2 a_{B}}) = - \frac{13.60 eV}{n^{2}} n = 1, 2, 3, \dots$ (41.2)

where $a_{B} = 4 π ϵ_{0} ℏ^{2} / m e^{2} = 0.0529 nm$ is the Bohr radius. The integer $n$ is called the principal quantum number. These energies are the same as those predicted by the Bohr model of the hydrogen atom.
The orbital angular momentum $L$ of the electron’s orbit must have magnitude

$L = \sqrt{l (l + 1)} ℏ l = 0, 1, 2, 3, \dots, n - 1$ (41.3)

The integer $l$ is called the orbital quantum number.
The $z$ -component of the angular momentum $L_{z}$ must be one of the values

$L_{z} = m ℏ m = - l, - l + 1, \dots, 0, \dots, l - 1, l$ (41.4)

The integer $m$ is called the magnetic quantum number.

In other words, each stationary state of the hydrogen atom is identified by a triplet of quantum numbers $(n, l, m)$ . Each quantum number is associated with a physical property of the atom.

NOTE The energy of the stationary state depends only on the principal quantum number $n$ , not on $l$ or $m$ .

EXAMPLE 41.1 Listing quantum numbers

List all possible states of a hydrogen atom that have energy $E = - 3.40 eV$ .

SOLVE Energy depends only on the principal quantum number $n$ . States with $E = - 3.40 eV$ have

n = \sqrt{\frac{- 13.60 eV}{- 3.40 eV}} = 2

An atom with principal quantum number $n = 2$ could have either $l = 0$ or $l = 1$ , but $l \geq 2$ is excluded by the rule $l \leq n - 1$ . If $l = 0$ , the only possible value for the magnetic quantum number $m$ is $m = 0$ . If $l = 1$ , then the atom could have $m = - 1, m = 0$ , or $m = + 1$ . Thus the possible quantum numbers are

\begin{array}{r} n & l & m \\ 2 & 0 & 0 \\ 2 & 1 & 1 \\ 2 & 1 & 0 \\ 2 & 1 & - 1 \end{array}

These four states all have the same energy.

Hydrogen turns out to be unique. For all other elements, the allowed energies depend on both $n$ and $l$ (but not $m$ ). Consequently, it is useful to label the stationary states by their values of $n$ and $l$ . The lowercase letters shown in TABLE 41.1 are customarily used to represent the various values of quantum number $l$ . These symbols come from spectroscopic notation used in prequantum-mechanics days, when some spectral lines were classified as sharp, others as principal, and so on.

TABLE 41.1 Symbols used to represent quantum number $l$

$l$	Symbol
0	$s$
1	$p$
2	$d$
3	$f$

Using these symbols, we call the ground state of the hydrogen atom, with $n = 1$ and $l = 0$ , the $1 s$ state. The $3 d$ state has $n = 3, l = 2$ . In Example 41.1, we found one $2 s$ state (with $l = 0$ ) and three $2 p$ states (with $l = 1$ ), all with the same energy.