\[
En = -\frac{Z^{2}R{\!H}}{n^{2}}\qquad\text{with }R_{\!H}=13.6\;\text{eV},
\]
where \(Z\) is the atomic number.
When an electron moves between two levels \(i\) and \(j\) the atom either emits or absorbs a photon whose energy equals the difference between the two levels:
\[
\Delta E = Ej-Ei = h\nu = \frac{hc}{\lambda}=hf
\]
| Spectrum | How it is produced | Appearance on a detector |
|---|---|---|
| Continuous (black‑body) | Thermal radiation from a hot, dense source (filament, stellar surface) | Smooth curve containing all wavelengths |
| Emission line | Low‑density gas of excited atoms relax to lower levels, emitting photons | Bright, narrow lines on a dark background |
| Absorption line | Continuous radiation passes through a cooler gas; photons matching allowed transitions are removed | Dark lines superimposed on a continuous spectrum |
Transitions that share a common lower (or upper) level form a series, identified by the final principal quantum number \(n_f\).
| Series | Lower level \(n_f\) | Region of the spectrum | Typical transition (example) |
|---|---|---|---|
| Lyman | 1 | Ultraviolet (≈ 90–400 nm) | n=2→1 (121.6 nm) |
| Balmer | 2 | Visible (≈ 400–700 nm) | n=3→2 (656.3 nm, Hα) |
| Paschen | 3 | Infra‑red (≈ 800–2500 nm) | n=4→3 (1875 nm) |
| Brackett | 4 | Infra‑red (≈ 1.5–4 µm) | n=5→4 (4051 nm) |
| Pfund | 5 | Infra‑red (≈ 3–7 µm) | n=6→5 (7460 nm) |
| Humphreys | 6 | Far‑infra‑red (≈ 12–30 µm) | n=7→6 (12 µm) |
For a hydrogen‑like ion the wavenumber of a spectral line is
\[
\frac{1}{\lambda}=RZ\!\left(\frac{1}{nf^{2}}-\frac{1}{ni^{2}}\right),\qquad ni>n_f,
\]
where \(RZ = R{\!H}Z^{2}\) and \(R_{\!H}=1.097\,\times10^{7}\;\text{m}^{-1}\).
Derivation from the Bohr model (useful for AO1 22.1):
\[
R{\!H}= \frac{me e^{4}}{8\varepsilon_0^{2}h^{3}c}
\]
This expression shows that the Rydberg constant arises from the fundamental constants that appear in the Bohr model of the hydrogen atom.
| Quantity | Formula | Useful form for calculations |
|---|---|---|
| Energy ↔ Frequency | \(E = hf\) | \(E\,[\text{eV}] = 4.1357\times10^{-15}\,f\,[\text{Hz}]\) |
| Energy ↔ Wavelength | \(E = \dfrac{hc}{\lambda}\) | \(E\,[\text{eV}] = \dfrac{1240}{\lambda\,[\text{nm}]}\) |
| Frequency ↔ Wavelength | \(c = \lambda f\) | \(f\,[\text{Hz}] = \dfrac{c}{\lambda\,[\text{m}]}\) |
\[
Ni \propto gi\,e^{-E_i/kT}
\]
where \(g_i\) is the statistical weight, \(k\) the Boltzmann constant and \(T\) the temperature.
\[
\Delta l = \pm1,\qquad \Delta m_l = 0,\pm1,\qquad \Delta s = 0
\]
Only transitions obeying these rules appear as observable lines.
Question: A photon of wavelength \(410\;\text{nm}\) is observed in the emission spectrum of hydrogen. Identify the transition and calculate the photon energy in electron‑volts.
\[
\frac{1}{\lambda}=R{\!H}\!\left(\frac{1}{2^{2}}-\frac{1}{ni^{2}}\right)
\]
Substituting \(\lambda=410\times10^{-9}\,\text{m}\) gives \(n_i=6\).
Transition: \(6\rightarrow2\) (H\(_\delta\)).
\[
E=\frac{hc}{\lambda}
=\frac{(6.626\times10^{-34})(3.00\times10^{8})}{410\times10^{-9}}
=4.85\times10^{-19}\,\text{J}
\]
\[
E=\frac{4.85\times10^{-19}}{1.602\times10^{-19}}\approx3.03\;\text{eV}
\]
Question: The H\(_\alpha\) line of hydrogen has \(\lambda = 656.3\;\text{nm}\). Determine (a) its frequency and (b) its energy in joules and electron‑volts.
\[
f = \frac{c}{\lambda}
= \frac{3.00\times10^{8}}{656.3\times10^{-9}}
= 4.57\times10^{14}\;\text{Hz}
\]
\[
E = hf = (6.626\times10^{-34})(4.57\times10^{14})
= 3.03\times10^{-19}\;\text{J}
\]
\[
E = \frac{3.03\times10^{-19}}{1.602\times10^{-19}} \approx 1.89\;\text{eV}
\]
Upward arrows denote absorption (labelled with the corresponding wavelength, e.g., 656 nm); downward arrows denote emission.
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