Explain that an isolated atom (e.g. hydrogen) can occupy only certain, discrete electron energy levels and that transitions between these levels produce photons of characteristic wavelengths, giving rise to line spectra.
| Syllabus Requirement | Coverage in These Notes | Action Required |
|---|---|---|
| 1 – 11 (AS core topics) | Only sub‑topic 22.4 (Energy levels & line spectra) is covered. | Create a separate set of lecture notes for the remaining AS topics (e.g. kinematics, dynamics, waves, electricity). Include a mapping table linking each syllabus number to its dedicated notes. |
| 12 – 25 (A‑level extensions) | No A‑level material is present. | Add a “Further Reading / Next Steps” section (see below) that points to the next units: 12 Motion in a circle, 13 Gravitational fields, 22 Quantum physics, 23 Nuclear physics, etc. |
| Depth & Accuracy (AO1) | Correct Bohr model, energy formulae, Rydberg equation and a worked example are provided, but a few technical slips exist. | Correct the Bohr‑radius expression, state explicitly that the 13.6 eV constant applies to hydrogen‑like (single‑electron) atoms, and add a footnote for the Rydberg constant. |
The Bohr model was the first successful attempt to explain why atoms emit discrete spectra. It treats the electron as moving in a circular orbit around the nucleus, but only certain orbits are allowed.
\[
L = m_e v r = n\hbar ,\qquad n = 1,2,3,\dots
\]
\[
\frac{me v^{2}}{r}= \frac{1}{4\pi\varepsilon0}\frac{e^{2}}{r^{2}}
\]
Eliminating the speed \(v\) using the angular‑momentum condition gives the Bohr radius and its higher‑order equivalents:
\[
rn = \frac{n^{2}\hbar^{2}}{k me e^{2}} = n^{2}a_0,
\qquad
a0 = \frac{4\pi\varepsilon0\hbar^{2}}{m_e e^{2}} \approx 5.29\times10^{-11}\,\text{m},
\]
where \(k = \dfrac{1}{4\pi\varepsilon_0}\).
The total energy (kinetic + potential) of an electron in the \(n\)th orbit is
\[
En = -\frac{k e^{2}}{2rn}
= -\frac{k^{2} m_e e^{4}}{2\hbar^{2}}\frac{1}{n^{2}}
= -\frac{13.6\ \text{eV}}{n^{2}} .
\]
Note: The 13.6 eV value is valid for hydrogen‑like (single‑electron) atoms only.
When an electron moves between two allowed levels, a photon is emitted (if the electron drops) or absorbed (if it is raised). The photon energy is given exactly by
\[
hf = Ei - Ef ,
\]
where \(Ei\) (higher) and \(Ef\) (lower) are the energies of the initial and final states. Substituting the Bohr‑hydrogen expression for \(E_n\) yields
\[
hf = 13.6\ \text{eV}\left(\frac{1}{nf^{2}}-\frac{1}{ni^{2}}\right).
\]
In terms of wavelength the relationship becomes
\[
\frac{1}{\lambda}=R{\infty}\!\left(\frac{1}{nf^{2}}-\frac{1}{n_i^{2}}\right),
\qquad
R_{\infty}=1.097\,\times10^{7}\ \text{m}^{-1}\;†.
\]
| Series | Final level \(n_f\) | Typical region of the spectrum |
|---|---|---|
| Lyman | 1 | Ultraviolet |
| Balmer | 2 | Visible |
| Paschen | 3 | Infrared |
| Brackett | 4 | Infrared |
| Pfund | 5 | Infrared |
\[
\frac{1}{\lambda}=R_{\infty}\!\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right).
\]
\[
\frac{1}{\lambda}=1.097\times10^{7}\left(\frac{1}{4}-\frac{1}{9}\right)
=1.097\times10^{7}\left(\frac{5}{36}\right)
=1.523\times10^{6}\ \text{m}^{-1}.
\]
\[
\lambda = \frac{1}{1.523\times10^{6}} \approx 6.57\times10^{-7}\,\text{m}=657\ \text{nm}.
\]
In atoms with more than one electron:
These notes form a self‑contained module on discrete energy levels. To complete the Cambridge syllabus, plan the following follow‑up sessions:
†Rydberg constant \(R{\infty}\) is defined for an infinitely massive nucleus; for real hydrogen the value is slightly lower (\(RH = 1.09678\times10^{7}\ \text{m}^{-1}\)).
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