Module 8: From the Universe to the Atom

NSWPhysicsSyllabus dot point

Inquiry Question 3: How is it known that classical physics cannot explain the properties of the atom?

Investigate the line emission spectra to examine the Balmer-Rydberg equation 1/lambda = R(1/n_f^2 - 1/n_i^2), and assess the limitations of the Bohr model of the hydrogen atom

A focused answer to the HSC Physics Module 8 dot point on the Bohr model of hydrogen. Postulates of stationary orbits and quantised angular momentum, the energy levels E_n = -13.6 eV / n^2, the Balmer-Rydberg formula 1/lambda = R (1/n_f^2 - 1/n_i^2), spectral series (Lyman, Balmer, Paschen), and the limitations of the model.

Generated by Claude OpusReviewed by Better Tuition Academy9 min answer

Have a quick question? Jump to the Q&A page

What this dot point is asking

NESA wants you to state Niels Bohr's three postulates for the hydrogen atom, use the resulting energy levels En=13.6 eV/n2E_n = -13.6 \text{ eV} / n^2 and the Balmer-Rydberg formula 1/λ=R(1/nf21/ni2)1/\lambda = R(1/n_f^2 - 1/n_i^2) to calculate transition wavelengths, identify the named spectral series, and assess the limitations of the model.

The answer

Why Bohr's model was needed

By 1911 Rutherford had established that the atom has a tiny dense nucleus surrounded by electrons. The classical problem: an orbiting electron is accelerating and should radiate electromagnetic waves continuously, losing energy and spiralling into the nucleus in about 101110^{-11} s. Atoms are stable, so classical electromagnetism cannot be the whole story.

A second puzzle was that hot rarefied gases of hydrogen emit a discrete pattern of spectral lines, not a continuous spectrum. Balmer (1885) had found an empirical formula for the visible lines, generalised by Rydberg (1890) for all hydrogen lines:

1λ=R(1nf21ni2),R=1.097×107 m1\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right), \quad R = 1.097 \times 10^7 \text{ m}^{-1}

with ni>nfn_i > n_f for emission. There was no underlying theory for this formula.

Bohr's postulates (1913)

Bohr postulated three rules to fix both problems.

Postulate 1: stationary orbits. The electron in a hydrogen atom occupies certain discrete circular orbits in which it does not radiate. Each such orbit is a stationary state with a well-defined energy.

Postulate 2: quantisation of angular momentum. The allowed orbits are those for which the orbital angular momentum is an integer multiple of =h/(2π)\hbar = h/(2\pi):

mevr=n=nh2π,n=1,2,3,m_e v r = n \hbar = n \frac{h}{2 \pi}, \quad n = 1, 2, 3, \dots

Postulate 3: photon emission. Radiation occurs only when the electron makes a transition between two stationary states. The photon energy equals the energy difference:

hf=EiEfh f = E_i - E_f

The first postulate rejects classical electromagnetism for bound electrons. The second is the new quantum rule. The third converts energy differences into spectral line wavelengths.

Energy levels

Combining quantised angular momentum with the Coulomb-centripetal force balance gives the allowed orbital radii and energies. The result for hydrogen:

rn=n2a0,a0=5.29×1011 m (Bohr radius)r_n = n^2 a_0, \quad a_0 = 5.29 \times 10^{-11} \text{ m (Bohr radius)}

En=13.6 eVn2\boxed{E_n = -\frac{13.6 \text{ eV}}{n^2}}

Properties:

  • The ground state (n=1n = 1) has E1=13.6E_1 = -13.6 eV. This is the ionisation energy: removing the electron to infinity requires 13.6 eV.
  • IMATH_14 as nn \to \infty, the unbound (ionised) limit.
  • Spacing decreases as nn grows (the level "bunching" near zero).

Recovering the Rydberg formula

For a transition ninfn_i \to n_f with ni>nfn_i > n_f:

ΔE=EniEnf=13.6(1nf21ni2) eV\Delta E = E_{n_i} - E_{n_f} = 13.6 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \text{ eV}

Photon wavelength:

1λ=ΔEhc=R(1nf21ni2)\frac{1}{\lambda} = \frac{\Delta E}{h c} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)

with R=13.6 eV/(hc)=1.097×107R = 13.6 \text{ eV} / (h c) = 1.097 \times 10^7 m1^{-1}. This is the Rydberg formula derived, not just postulated.

Named spectral series

Series IMATH_21 Region Examples
Lyman 1 Ultraviolet IMATH_22 : 122, 103, 97 nm
Balmer 2 Visible Hα\alpha 656 nm, Hβ\beta 486 nm, Hγ\gamma 434 nm
Paschen 3 Infrared 1875, 1282, 1094 nm
Brackett 4 Infrared IMATH_26 m

The Balmer series is the visible band first discovered, which is why it has its own name. The full pattern lets astronomers identify hydrogen even from the most distant galaxies.

Worked example: Lyman alpha

Find the wavelength of the photon emitted when an electron drops from n=2n = 2 to n=1n = 1 in hydrogen.

1λ=R(112122)=1.097×107×(10.25)=8.23×106\frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = 1.097 \times 10^7 \times (1 - 0.25) = 8.23 \times 10^6 m1^{-1}.

λ=1.22×107\lambda = 1.22 \times 10^{-7} m = 122 nm (ultraviolet).

This is the Lyman alpha line, a powerful tracer of cold neutral hydrogen in the universe.

Try it: Rydberg spectrum calculator to compute wavelengths for arbitrary ninfn_i \to n_f transitions in hydrogen and hydrogen-like ions.

Limitations of the Bohr model

The Bohr model works astonishingly well for hydrogen (and for hydrogen-like ions such as He+^+ and Li2+^{2+}, with Z2Z^2 corrections), but it has clear limitations:

  • Multi-electron atoms. The energy levels in helium, lithium and beyond are not predicted accurately. The model has no way to handle electron-electron repulsion.
  • Definite orbits. The model puts the electron on a sharp circular path with a definite position and velocity, contrary to the uncertainty principle (which we now know to be exact).
  • Fine structure. Spectral lines split into closely spaced sub-lines (fine structure) that the Bohr model does not predict. Relativistic and spin-orbit effects are needed.
  • Zeeman and Stark effects. Splitting in external magnetic or electric fields is unexplained.
  • Intensities. The model gives line positions but not their intensities. Selection rules and transition probabilities require the full quantum-mechanical treatment.
  • No mechanism for quantisation. The angular-momentum rule is postulated, not derived. De Broglie later showed it can be motivated as a standing-wave condition; Schrödinger's wave equation gave it a proper derivation.

The Bohr model is best seen as a transitional model: not the final theory, but the bridge between the Rutherford atom and quantum mechanics.

Common traps

Using ni<nfn_i < n_f in the Rydberg formula. For emission, ni>nfn_i > n_f, and the formula 1/λ=R(1/nf21/ni2)1/\lambda = R(1/n_f^2 - 1/n_i^2) is positive. For absorption, swap them or take the magnitude.

Reporting the energy as positive. Bound-state energies are negative (En<0E_n < 0). The transition energy ΔE=EiEf\Delta E = E_i - E_f is positive for emission because the initial state (higher nn) is closer to zero.

Forgetting the units of RR. R=1.097×107R = 1.097 \times 10^7 m1^{-1}, so 1/λ1/\lambda is in m1^{-1} and λ\lambda in m. Convert to nm at the end.

Applying the Bohr energy formula directly to helium. Bohr applies to hydrogen-like one-electron systems. For neutral helium use full quantum mechanics; for He+^+ scale by Z2=4Z^2 = 4.

Treating the model as fundamentally correct. It is a useful semi-classical picture, valuable for understanding spectra and ionisation energies, but the quantum-mechanical orbital picture (Schrödinger) replaces it for any serious calculation.

In one sentence

Bohr postulated stationary orbits, quantised angular momentum and photon transitions, deriving the hydrogen energy levels En=13.6 eV/n2E_n = -13.6 \text{ eV}/n^2 and the Rydberg formula 1/λ=R(1/nf21/ni2)1/\lambda = R(1/n_f^2 - 1/n_i^2) which exactly explains the hydrogen spectrum, while limitations (multi-electron atoms, fine structure, definite orbits) point to the full quantum-mechanical treatment that follows.

Past exam questions, worked

Real questions from past NESA papers on this dot point, with our answer explainer.

2022 HSC5 marksAn electron in a hydrogen atom transitions from the n = 4 level to the n = 2 level. Calculate the wavelength of the emitted photon and identify the spectral series. (R = 1.097 x 10^7 m^-1.)
Show worked answer →

Rydberg formula:

1λ=R(1nf21ni2)\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)

1λ=1.097×107(14116)=1.097×107×316=2.057×106\frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{4} - \frac{1}{16} \right) = 1.097 \times 10^7 \times \frac{3}{16} = 2.057 \times 10^6 m1^{-1}.

λ=1/2.057×106=4.86×107\lambda = 1 / 2.057 \times 10^6 = 4.86 \times 10^{-7} m = 486 nm.

The final state is nf=2n_f = 2, so the transition is part of the Balmer series. The 486 nm line is in the visible region (blue-green) and corresponds to H-beta.

Markers reward correct Rydberg formula, nf=2n_f = 2 for Balmer, the wavelength in nm, and identification of the series.

2019 HSC4 marksState the postulates of the Bohr model of the hydrogen atom and identify two limitations of the model.
Show worked answer →

Postulates:

  1. The electron in a hydrogen atom occupies certain discrete stationary orbits in which it does not radiate energy, contrary to classical electromagnetism.

  2. The allowed orbits are those for which the angular momentum is quantised in units of h/2πh / 2\pi: mvr=nh/2πm v r = n h / 2 \pi for n=1,2,3,n = 1, 2, 3, \dots.

  3. The electron radiates only when making a transition between two stationary states, emitting a photon of energy equal to the energy difference: hf=EiEfh f = E_i - E_f.

Limitations (any two of):

  • Applies only to hydrogen (or hydrogen-like one-electron ions). Cannot quantitatively predict the spectra of multi-electron atoms.
  • Treats the electron as a particle in a definite orbit. Quantum mechanics shows that the electron has a probability distribution (orbital), not a sharp orbit.
  • Cannot explain the fine structure of spectral lines or their splitting in magnetic fields (Zeeman effect).
  • Provides no mechanism for why angular momentum is quantised, beyond postulating it.
  • Does not explain the intensities of spectral lines.

Markers reward all three postulates and any two distinct limitations.

Related dot points