understand that there are discrete electron energy levels in isolated atoms (e.g. atomic hydrogen)

Energy Levels in Atoms and Line Spectra

Learning Objective

Understand that isolated atoms, such as atomic hydrogen, possess discrete electron energy levels, and that transitions between these levels give rise to characteristic line spectra.

Key Concepts

• Quantisation of electron energy in atoms.
• Bohr model of the hydrogen atom.
• Energy of a photon: \$E = h\nu = \dfrac{hc}{\lambda}\$.
• Rydberg formula for hydrogen spectral lines.
• Series of spectral lines (Lyman, Balmer, Paschen, etc.).

Bohr Model Derivation

The Bohr model assumes that an electron moves in a circular orbit around the nucleus with quantised angular momentum:

\$L = m_e v r = n\hbar,\qquad n = 1,2,3,\dots\$

Balancing the centripetal force with the Coulomb attraction gives:

\$\frac{m_e v^{2}}{r} = \frac{k e^{2}}{r^{2}}\$

Eliminating \$v\$ using the angular momentum condition leads to the allowed radii:

\$r{n} = \frac{n^{2}\hbar^{2}}{k m{e} e^{2}} = n^{2}a_{0},\$

where \$a{0} = \dfrac{\hbar^{2}}{k m{e} e^{2}} \approx 5.29\times10^{-11}\,\text{m}\$ is the Bohr radius.

The total energy of the electron in the \$n\$‑th orbit is the sum of kinetic and potential energy:

\$\$E{n} = -\frac{k e^{2}}{2r{n}} = -\frac{k^{2} m_{e} e^{4}}{2\hbar^{2}}\frac{1}{n^{2}}

= -\frac{13.6\ \text{eV}}{n^{2}}.\$\$

Photon Emission and Absorption

When an electron transitions from a higher level \$n{i}\$ to a lower level \$n{f}\$, a photon is emitted with energy equal to the difference between the two levels:

\$\Delta E = E{n{f}} - E{n{i}} = h\nu = \frac{hc}{\lambda}.\$

Conversely, absorption of a photon of the same energy promotes the electron to the higher level.

Rydberg Formula

The wavelength of the emitted or absorbed photon can be expressed using the Rydberg constant \$R_{\infty}\$:

\$\$\frac{1}{\lambda}=R{\infty}\!\left(\frac{1}{n{f}^{2}}-\frac{1}{n_{i}^{2}}\right),\qquad

R_{\infty}=1.097\times10^{7}\ \text{m}^{-1}.\$\$

Hydrogen Spectral Series

Example Calculation

1. Find the wavelength of the Balmer \$\alpha\$ line (\$n{i}=3 \to n{f}=2\$).
2. Use the Rydberg formula:

   \$\frac{1}{\lambda}=R_{\infty}\!\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right).\$

3. Calculate:

   \$\$\frac{1}{\lambda}=1.097\times10^{7}\left(\frac{1}{4}-\frac{1}{9}\right)

   =1.097\times10^{7}\left(\frac{5}{36}\right)

   =1.523\times10^{6}\ \text{m}^{-1}.\$\$

   Thus \$\lambda = \dfrac{1}{1.523\times10^{6}}\approx6.57\times10^{-7}\,\text{m}=657\ \text{nm}.\$

Implications for Atomic Structure

The existence of discrete lines in emission and absorption spectra provides direct evidence that electrons can only occupy certain energy states. This quantisation is a fundamental postulate of quantum mechanics and underlies the stability of atoms.

Common Misconceptions

• Electrons spiral into the nucleus – Incorrect; quantised orbits prevent continuous energy loss.
• All atoms emit the same spectral lines – Incorrect; each element has a unique set of energy levels.
• Energy levels are continuous – Incorrect; they are discrete, as shown by the line spectra.

Summary

1. Isolated atoms have quantised electron energy levels.
2. Transitions between levels produce photons with energies given by \$E = h\nu\$.
3. The Rydberg formula accurately predicts the wavelengths of hydrogen spectral lines.
4. Observation of line spectra confirms the quantum nature of atomic structure.