recall and use hf = E1 – E2

Energy Levels in Atoms and Line Spectra (Cambridge AS/A‑Level 22.4)

Syllabus Outcome 22.4

“Recall and use the relationship \(hf = E{1}-E{2}\) to analyse atomic line spectra, calculate frequencies, wavelengths and photon energies, and interpret the results in terms of the hydrogen‑atom energy‑level structure.”

The sections below are explicitly mapped to this outcome (see the coloured icons in the margin of the teacher’s guide).

Learning Objective

Students will be able to:

  • State the equation \(hf = E{\text{higher}}-E{\text{lower}}\) and explain its meaning.

  • Apply the Bohr‑model energy expression and the Rydberg formula to calculate \(\lambda\), \(f\) and \(E\) for hydrogen transitions.

  • Identify the spectral series (Lyman, Balmer, Paschen, …) and the corresponding region of the electromagnetic spectrum.

  • Use the electric‑dipole selection rules to decide whether a transition is allowed.

Key Physical Constants

SymbolQuantityValue (SI)
hPlanck’s constant6.626 × 10⁻³⁴ J s
cSpeed of light in vacuum2.998 × 10⁸ m s⁻¹
eElementary charge1.602 × 10⁻¹⁹ C
R∞Rydberg constant (in m⁻¹)1.097 373 × 10⁷ m⁻¹

Key Formulae (Cambridge Syllabus)

FormulaPurposeNotes
\(En = -\dfrac{hcR{\infty}}{n^{2}} = -\dfrac{13.6\;\text{eV}}{n^{2}}\)Energy of level \(n\) in hydrogen (Bohr model)n = 1, 2, 3,…
\(hf = E{\text{higher}}-E{\text{lower}}\)Photon energy for emission or absorptionPositive \(\Delta E\) = photon emitted (drop) or absorbed (rise)
\(\displaystyle\frac{1}{\lambda}=R{\infty}\!\left(\frac{1}{nf^{2}}-\frac{1}{ni^{2}}\right)\)Rydberg formula (hydrogen)\(ni>nf\) for emission; \(nf\) determines the series
\(c = \lambda f\)Link between wavelength and frequencyUse after finding \(\lambda\) or \(f\)
\(E = hc/\lambda = hf\)Convert wavelength/frequency to photon energyUseful for part‑c of exam questions

Fundamental Concepts

  • Quantised energy levels – Electrons can only occupy discrete energies \(E_n\). The Bohr model gives a simple analytic expression for hydrogen (see formula above).

  • Photon emission/absorption – When an electron moves between two levels,

    \[

    hf = \Delta E = E{\text{higher}}-E{\text{lower}} .

    \]

    • Emission: electron drops → a photon of energy \(hf\) leaves the atom.

    • Absorption: electron climbs → a photon of the same energy must be supplied.

  • From energy to wavelength – Combine \(hf\) with \(c=\lambda f\) to obtain

    \[

    \lambda = \frac{hc}{\Delta E}.

    \]

    For hydrogen the Rydberg formula is faster and is the method required by the syllabus.

  • Quantum numbers & selection rules (electric‑dipole)

    Quantum numberSymbolAllowed values
    Principaln1, 2, 3,…
    Azimuthal (orbital)ℓ0 ≤ ℓ ≤ n‑1
    Magneticmℓ-ℓ ≤ mℓ ≤ ℓ
    Spinm_s±½

    Selection rules:

    • Δℓ = ±1

    • Δm_ℓ = 0, ±1

    • Δn can be any integer (subject to the ℓ‑rule)

    Only transitions satisfying these rules produce observable spectral lines.

  • Spectral series (hydrogen)

    SeriesFinal level \(n_f\)Region of spectrum
    Lyman1Ultraviolet
    Balmer2Visible
    Paschen3Infrared
    Brackett4Infrared
    Pfund5Infrared

  • Fine‑structure (optional) – Small splittings due to relativistic and spin–orbit effects. Not required for the core syllabus but useful for extension questions.

Derivation of the Working Equation

For a transition from a higher level \(Ei\) to a lower level \(Ef\):

\[

Ei = Ef + hf \;\;\Longrightarrow\;\; hf = Ei - Ef .

\]

The same algebra holds for absorption because the photon supplies the energy needed to raise the electron.

Step‑by‑Step Procedure (Using the Rydberg Formula)

  1. Identify the initial (\(ni\)) and final (\(nf\)) principal quantum numbers.

    • For emission: \(ni > nf\).
    • For absorption: \(ni < nf\).

  2. Insert the numbers into the Rydberg formula

    \[

    \frac{1}{\lambda}=R{\infty}\!\left(\frac{1}{nf^{2}}-\frac{1}{n_i^{2}}\right).

    \]

  3. Calculate \(\lambda\) (metres). If the question asks for frequency, use \(f=c/\lambda\). If photon energy is required, use \(E=hc/\lambda\) or \(E=hf\).

  4. Convert units as required (e.g. m ↔ nm, J ↔ eV). Keep at least three significant figures for constants until the final step.

Worked Example – Emission (Balmer line)

Question: Find the wavelength of the photon emitted when an electron falls from \(n=5\) to \(n=2\).

  1. Identify levels: \(ni=5,\; nf=2\).

  2. Rydberg calculation:

    \[

    \frac{1}{\lambda}=R_{\infty}\!\left(\frac{1}{2^{2}}-\frac{1}{5^{2}}\right)

    =1.097373\times10^{7}\!\left(0.25-0.04\right)

    =2.31\times10^{6}\;\text{m}^{-1}.

    \]

  3. Invert: \(\displaystyle\lambda=\frac{1}{2.31\times10^{6}}=4.33\times10^{-7}\,\text{m}=433\;\text{nm}.\)

  4. Interpretation: 433 nm lies in the violet part of the visible spectrum, confirming the transition belongs to the Balmer series.

Worked Example – Absorption (Lyman series)

Question: Determine the frequency of the photon required to excite a hydrogen electron from \(n=1\) to \(n=3\).

  1. Energy of the two levels (using \(E_n=-13.6\;\text{eV}/n^{2}\)):

    \[

    E1=-13.6\;\text{eV},\qquad E3=-\frac{13.6}{9}=-1.51\;\text{eV}.

    \]

  2. Energy needed: \(\Delta E = E3-E1 = 12.09\;\text{eV}\).

  3. Convert to joules: \(\Delta E =12.09\times1.602\times10^{-19}=1.94\times10^{-18}\;\text{J}\).

  4. Frequency: \(\displaystyle f=\frac{\Delta E}{h}= \frac{1.94\times10^{-18}}{6.626\times10^{-34}}=2.93\times10^{15}\;\text{Hz}.\)

Common Pitfalls & How to Avoid Them

  • Reversing the energies – Always write \(E{\text{higher}}-E{\text{lower}}\). A negative result signals the order is wrong.

  • Unit mismatches – Rydberg constant is in m⁻¹, so \(\lambda\) comes out in metres. Convert to nm (1 nm = 10⁻⁹ m) before interpreting the colour.

  • Choosing the wrong series – First decide the final level \(n_f\); this fixes the series (Lyman = 1, Balmer = 2, …).

  • Ignoring selection rules – Some mathematically possible transitions are forbidden (e.g., Δℓ = 0). Mention the rule even if the Bohr model does not show it.

  • Early rounding – Keep constants to at least three significant figures until the final answer to avoid cumulative rounding errors.

Practice Questions

  1. Calculate the wavelength of the photon emitted when an electron falls from \(n=4\) to \(n=2\). State the series and the region of the spectrum.

  2. What frequency of radiation is required to promote a hydrogen electron from \(n=2\) to \(n=6\)?

  3. Explain why the Balmer series is observed in the visible region whereas the Lyman series lies in the ultraviolet.

  4. Using the selection rules, decide whether the transition \(3p\;(n=3,\;\ell=1)\rightarrow1s\;(n=1,\;\ell=0)\) is allowed, and identify the series it belongs to.

Suggested Diagram

Energy‑level diagram for hydrogen showing n=1–5 and the Lyman, Balmer and Paschen transitions

Energy‑level diagram for hydrogen with the main series (Lyman, Balmer, Paschen) indicated.

Summary

  • Atoms possess discrete energy levels; a transition between two levels releases or absorbs a photon whose energy is given by \(hf = E{\text{higher}}-E{\text{lower}}\).

  • The Bohr‑model expression for hydrogen (\(En = -hcR{\infty}/n^{2}\)) together with the Rydberg formula provides a quick route to \(\lambda\), \(f\) and \(E\).

  • Identifying the final level \(n_f\) tells you the spectral series (Lyman, Balmer, Paschen, …) and the region of the electromagnetic spectrum in which the line appears.

  • Electric‑dipole selection rules (Δℓ = ±1, Δm_ℓ = 0, ±1) decide whether a transition is observable.

  • Mastery of these ideas enables accurate prediction and interpretation of atomic line spectra – a cornerstone of the Cambridge AS/A‑Level Physics syllabus.