Physics • Year 12 • Module 8 • Lesson 12
Bohr’s Model of the Hydrogen Atom
Apply Bohr’s energy-level formula, the Rydberg equation and photon energy relationships to quantitative problems and real spectral data.
1. Interpret hydrogen spectral line data
The table below lists observed wavelengths for four lines in the Balmer series of hydrogen. 8 marks
| Spectral line | Transition ($n_i \to n_f$) | Observed wavelength (nm) | Calculated $1/\lambda$ (m¹) | Colour / region |
|---|---|---|---|---|
| H$\alpha$ | $3 \to 2$ | 656.3 | ||
| H$\beta$ | $4 \to 2$ | 486.1 | ||
| H$\gamma$ | $5 \to 2$ | 434.0 | ||
| H$\delta$ | $6 \to 2$ | 410.2 |
1.1 Complete the “Calculated $1/\lambda$” column using the Rydberg formula $1/\lambda = R_H(1/n_f^2 - 1/n_i^2)$ with $R_H = 1.097 \times 10^7$ m¹. Show one full calculation for H$\alpha$ in the space below. 4 marks (1 per row)
1.2 Complete the “Colour / region” column. Note that visible light spans approximately 400–700 nm. 2 marks
1.3 As the transition moves from H$\alpha$ to H$\delta$ (i.e. $n_i$ increases), describe and explain the trend in wavelength. 2 marks
2. Interpret an energy-level diagram — hydrogen emission and absorption
The graph below shows energy values (eV) for the first five levels of hydrogen, with four transitions marked. 7 marks
Figure 2. Simplified hydrogen energy-level diagram (not to scale). Downward arrows = emission; upward arrow = absorption.
2.1 For each transition (P, Q, R, S), state the initial and final quantum numbers and calculate the energy (in eV) of the photon emitted or absorbed. 4 marks (1 per transition)
2.2 Which transition (P, Q, R or S) produces the shortest-wavelength photon? Justify your answer without calculating $\lambda$. 2 marks
2.3 Transition R falls in the Lyman series. In which region of the electromagnetic spectrum does it lie? Explain. 1 mark
3. Compare the three main spectral series of hydrogen
Complete the table below for the Lyman, Balmer, and Paschen series. 9 marks (1 per cell)
| Feature | Lyman series | Balmer series | Paschen series |
|---|---|---|---|
| Final level $n_f$ | |||
| Electromagnetic region | |||
| Longest-wavelength transition ($n_i$) |
4. Predict and justify — a hydrogen lamp scenario
A hydrogen discharge tube is cooled until almost all hydrogen atoms are in their ground state ($n = 1$). A beam of light containing photons of energies 1.89 eV, 2.55 eV, 10.2 eV and 15.0 eV is then shone through the tube.
6 marks
4.1 Which photons will be absorbed by the hydrogen gas? For each absorbed photon, identify the electron transition it causes. Show your reasoning using energy-level values. 4 marks
4.2 Predict what happens to the photons that are not absorbed. What would an observer see if a diffraction grating were placed after the tube? 2 marks
Q1.1 — Calculated $1/\lambda$ values
H$\alpha$ ($n_i = 3, n_f = 2$): $1/\lambda = 1.097\times10^7(1/4 - 1/9) = 1.097\times10^7 \times 0.1389 = 1.524\times10^6$ m¹. $\lambda = 656$ nm. [Full calculation shown; award 1 mark.]
H$\beta$ ($n_i = 4, n_f = 2$): $1/\lambda = 1.097\times10^7(1/4 - 1/16) = 1.097\times10^7 \times 0.1875 = 2.057\times10^6$ m¹.
H$\gamma$ ($n_i = 5, n_f = 2$): $1/\lambda = 1.097\times10^7(1/4 - 1/25) = 1.097\times10^7 \times 0.21 = 2.304\times10^6$ m¹.
H$\delta$ ($n_i = 6, n_f = 2$): $1/\lambda = 1.097\times10^7(1/4 - 1/36) = 1.097\times10^7 \times 0.2222 = 2.437\times10^6$ m¹.
Q1.2 — Colour / region
H$\alpha$ (656 nm): red (visible). H$\beta$ (486 nm): blue-green (visible). H$\gamma$ (434 nm): violet (visible, near UV limit). H$\delta$ (410 nm): violet / near-UV (visible but very close to the UV boundary; accept “deep violet” or “near-UV”).
Q1.3 — Trend as $n_i$ increases
As $n_i$ increases, the wavelength decreases (the lines shift toward shorter wavelengths / higher frequency). Higher $n_i$ levels have energy closer to zero, so the gap $|E_{n_i} - E_2|$ increases with $n_i$. A larger energy gap means the emitted photon has more energy, hence higher frequency and shorter wavelength. The lines converge toward the series limit at $\lambda \approx 365$ nm (corresponding to $n_i \to \infty$).
Q2.1 — Transition energies
P ($n=3 \to n=2$, emission): $\Delta E = |-1.51 - (-3.40)| = 1.89$ eV.
Q ($n=4 \to n=2$, emission): $\Delta E = |-0.85 - (-3.40)| = 2.55$ eV.
R ($n=3 \to n=1$, emission): $\Delta E = |-1.51 - (-13.6)| = 12.09$ eV.
S ($n=1 \to n=3$, absorption): $\Delta E = |-13.6 - (-1.51)| = 12.09$ eV.
Note: R and S involve the same energy difference but R is emission and S is absorption. Award 1 mark per correct energy value with correct emission/absorption designation.
Q2.2 — Shortest-wavelength transition
Transition R ($n=3 \to n=1$) produces the shortest-wavelength photon [1] because it has the largest energy difference (12.09 eV compared to 1.89 eV for P and 2.55 eV for Q). Since $E = hc/\lambda$, a larger photon energy means a shorter wavelength [1].
Q2.3 — Region of transition R
Ultraviolet. Transitions to $n_f = 1$ (Lyman series) have energies of 10.2 eV or more, corresponding to wavelengths of 122 nm or less — well below the visible range (<400 nm), placing them firmly in the UV region.
Q3 — Spectral series comparison
Final level $n_f$: Lyman = 1; Balmer = 2; Paschen = 3. Electromagnetic region: Lyman = ultraviolet; Balmer = visible; Paschen = infrared. Longest-wavelength transition: Lyman: $n_i = 2$ ($\lambda \approx 122$ nm); Balmer: $n_i = 3$ ($\lambda \approx 656$ nm, H$\alpha$); Paschen: $n_i = 4$ ($\lambda \approx 1875$ nm).
Q4.1 — Which photons are absorbed
Only photons whose energy exactly matches a transition energy are absorbed by ground-state atoms.
- $\Delta E_{2\to1} = -3.40 - (-13.6) = 10.2$ eV → the 10.2 eV photon is absorbed; transition $n=1 \to n=2$.
- $\Delta E_{3\to1} = -1.51 - (-13.6) = 12.09$ eV → not matched by any photon in the list.
- 1.89 eV: matches $n=3 \to n=2$, but atoms are in ground state ($n=1$), so this transition is not available. Not absorbed.
- 2.55 eV: matches $n=4 \to n=2$; not available from ground state. Not absorbed.
- 15.0 eV: exceeds ionisation energy (13.6 eV). This photon would ionise the atom rather than cause a discrete transition. Accept “ionised” or “not absorbed as a bound-state transition.”
Award 1 mark for correctly identifying 10.2 eV absorbed with transition stated; 1 mark for correctly rejecting 1.89 eV and 2.55 eV with reasoning; 1 mark for 15.0 eV discussion; 1 mark for correct use of energy-level values throughout.
Q4.2 — Unabsorbed photons and diffraction grating
The photons that are not absorbed (1.89 eV, 2.55 eV, and 15.0 eV if ionisation is excluded) pass through the gas without interacting [1]. Through a diffraction grating, an observer would see a continuous spectrum from the original light source but with dark absorption lines at the wavelengths corresponding to the absorbed 10.2 eV photon (Lyman-$\alpha$, $\approx$ 122 nm — in the UV, so not visible to the naked eye) [1].