4.2 · Atomic structure

Energy levels of hydrogen-like atoms

E = −Z²R_H/n² and what it predicts.

By the end you should be able to:

Calculate energies of hydrogen-like atoms and ions, and identify degenerate orbitals

Formula

Energy levels of one-electron species

$E_n = -\frac{Z^2R_H}{n^2}, \qquad R_H = 2.179\times10^{-18}\ \mathrm{J}$

This holds for any species with one electron: H ( $Z = 1$ ), $\ce{He+}$ ( $Z = 2$ ), $\ce{Li^2+}$ ( $Z = 3$ ), $\ce{Be^3+}$ ( $Z = 4$ ).

$E = 0$ at $n = \infty$ , where the electron is free of the nucleus; every bound level is negative.
The levels crowd together as $n$ increases.
A larger nuclear charge pulls the electron in: $\ce{He+}$ levels are 4 times deeper than those of H, and $\ce{Li^2+}$ levels 9 times deeper.

Key idea

Degeneracy in hydrogen

In a one-electron species the energy depends on $n$ alone. All orbitals in a shell are degenerate (equal in energy): 2s = 2p, and 3s = 3p = 3d. Shell $n$ therefore contains $n^2$ degenerate orbitals.

In atoms with more than one electron, shielding and electron–electron repulsion split the subshells (2s < 2p; 3s < 3p < 3d), as module 6 explains. The orbitals within one subshell, such as the three 2p orbitals, remain degenerate in a free atom.

Formula

Transitions and ionization energy

$\Delta E = E_f - E_i = Z^2R_H\left(\frac{1}{n_i^2} - \frac{1}{n_f^2}\right)$

$\Delta E < 0$ : emission ( $n$ decreases); a photon with energy $|\Delta E|$ is released.
$\Delta E > 0$ : absorption ( $n$ increases).

Ionization from level $n$ is the jump to $n = \infty$ :

$\mathrm{IE}_n = \frac{Z^2R_H}{n^2}\ \text{per atom}, \qquad \text{multiply by } N_A \text{ for per mole}$

Method

Solving a hydrogen-like energy problem

Check that the species has exactly one electron, and identify $Z$ .
Compute each level: $E_n = -Z^2R_H/n^2$ .
$\Delta E = E_{\text{final}} - E_{\text{initial}}$ ; the sign tells you emission (−) or absorption (+).
For the photon: $E = |\Delta E|$ and $\lambda = hc/|\Delta E|$ .
For molar values multiply by $N_A = 6.022\times10^{23}$ mol⁻¹, and divide by 1000 for kJ/mol.

Common mistake

Hydrogen-atom traps

Wrong: using $Z$ instead of $Z^2$ , so that $\ce{He+}$ levels come out twice as deep as those of H. Right: the factor is $Z^2$ : $\ce{He+}$ levels are 4 times deeper.

Wrong: applying $E_n = -Z^2R_H/n^2$ to neutral He or Li. Right: it holds only for one-electron species.

Wrong: $\Delta E = E_{\text{initial}} - E_{\text{final}}$ . Right: $\Delta E = E_{\text{final}} - E_{\text{initial}}$ ; the photon energy is its magnitude.

Wrong: "3s is lower in energy than 3p in hydrogen." Right: they are degenerate in every one-electron species.

Worked example

Worked example: ionization energies

From the ground state ( $n = 1$ ), $\mathrm{IE} = Z^2R_H$ :

Species	$Z$	IE per atom	IE per mole
H	1	$2.179\times10^{-18}$ J	1312 kJ/mol
$\ce{He+}$	2	$8.716\times10^{-18}$ J	5249 kJ/mol
$\ce{Li^2+}$	3	$1.961\times10^{-17}$ J	11 810 kJ/mol

From an excited level less energy is needed: an H atom in $n = 2$ needs $\frac{2.179\times10^{-18}}{2^2} = 5.45\times10^{-19}$ J (328 kJ/mol). These predictions match experiment: the measured second ionization energy of He is 5251 kJ/mol and the third of Li is 11 815 kJ/mol.

Worked example

Worked example: a He⁺ transition

$\ce{He+}$ ( $Z = 2$ ) drops from $n = 3$ to $n = 2$ :

$\Delta E = -(2^2)(2.179\times10^{-18})\left(\frac{1}{2^2} - \frac{1}{3^2}\right) = -1.211\times10^{-18}\ \mathrm{J}$

The sign is negative, so a photon is emitted:

$\lambda = \frac{hc}{|\Delta E|} = \frac{(6.626\times10^{-34})(2.998\times10^{8})}{1.211\times10^{-18}} = 1.641\times10^{-7}\ \mathrm{m}$

Answer: 164 nm, in the ultraviolet. Because energies scale as $Z^2/n^2$ , the $\ce{He+}$ transition $6 \to 4$ has exactly the same wavelength as the hydrogen line $3 \to 2$ (656 nm).

Check yourself

Fresh questions every time you visit. Answers count toward your progress.

Keep practising this topic