Bohr model and the Balmer-Rydberg formula: HSC Physics Module 8

Rydberg formula:

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

$\frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{4} - \frac{1}{16} \right) = 1.097 \times 10^7 \times \frac{3}{16} = 2.057 \times 10^6$ m$^{-1}$.

$\lambda = 1 / 2.057 \times 10^6 = 4.86 \times 10^{-7}$ m = 486 nm.

The final state is $n_f = 2$, 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, $n_f = 2$ for Balmer, the wavelength in nm, and identification of the series.

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 $h / 2\pi$: $m v r = n h / 2 \pi$ for $n = 1, 2, 3, \dots$.

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

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.

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

Postulate 2. The allowed orbits are those for which the orbital angular momentum is quantised: $m_e v r = n \dfrac{h}{2\pi}$, for $n = 1, 2, 3, \dots$.

Marks: one for each correctly stated postulate (stationary non-radiating orbits; quantised angular momentum).

$E_n = -\dfrac{13.6}{n^2}\ \text{eV}$, so $E_4 = -\dfrac{13.6}{4^2} = -\dfrac{13.6}{16} = -0.850\ \text{eV}$.

The value is negative because the electron is bound to the nucleus; energy must be added to remove it to $n = \infty$ (where $E = 0$), so every bound level lies below zero.

Marks: one for $E_4 = -0.850\ \text{eV}$, one for the correct reasoning that bound-state energies are negative.

$\dfrac{1}{\lambda} = R \left( \dfrac{1}{n_f^2} - \dfrac{1}{n_i^2} \right) = 1.097 \times 10^7 \left( \dfrac{1}{1^2} - \dfrac{1}{3^2} \right) = 1.097 \times 10^7 \times \dfrac{8}{9} = 9.751 \times 10^6\ \text{m}^{-1}$.

$\lambda = \dfrac{1}{9.751 \times 10^6} = 1.026 \times 10^{-7}\ \text{m} = 102.6\ \text{nm}$.

Since $n_f = 1$, this is a Lyman-series line, in the ultraviolet region.

Marks: one for the correct Rydberg substitution, one for $\lambda \approx 102.6\ \text{nm}$ (or $103\ \text{nm}$) with the unit, one for correctly naming the Lyman series and the UV region.

$E_2 = -\dfrac{13.6}{4} = -3.40\ \text{eV}$, $E_5 = -\dfrac{13.6}{25} = -0.544\ \text{eV}$.

Absorbed photon energy: $\Delta E = E_5 - E_2 = -0.544 - (-3.40) = 2.856\ \text{eV} \approx 2.86\ \text{eV}$.

Convert to joules: $\Delta E = 2.856 \times 1.602 \times 10^{-19} = 4.575 \times 10^{-19}\ \text{J}$.

Wavelength from $E = hf = hc/\lambda$: $\lambda = \dfrac{hc}{\Delta E} = \dfrac{(6.626 \times 10^{-34})(3.00 \times 10^8)}{4.575 \times 10^{-19}} = 4.34 \times 10^{-7}\ \text{m} = 434\ \text{nm}$.

Marks: one for both level energies and $\Delta E \approx 2.86\ \text{eV}$, one for converting to joules correctly, one for $\lambda \approx 434\ \text{nm}$ with correct method $\lambda = hc/\Delta E$.

(a) The graph is a straight line through the origin, so $1/\lambda$ is directly proportional to $\left(1/n_f^2 - 1/n_i^2\right)$, exactly as predicted by the Rydberg formula $1/\lambda = R\left(1/n_f^2 - 1/n_i^2\right)$ with $R$ constant.

(b) Gradient $= \dfrac{\Delta y}{\Delta x} = \dfrac{(2.438 - 1.524) \times 10^6}{0.2222 - 0.1389} = \dfrac{0.914 \times 10^6}{0.0833} = 1.097 \times 10^7\ \text{m}^{-1}$.

(c) The gradient equals the Rydberg constant $R$. The calculated value, $1.097 \times 10^7\ \text{m}^{-1}$, matches the accepted value exactly, confirming that the Balmer-series data obey the Rydberg formula.

Marks: one for identifying direct proportionality (straight line through the origin), one for a correctly computed gradient with working shown, one for $\text{gradient} \approx 1.097 \times 10^7\ \text{m}^{-1}$ with the unit, one for identifying the gradient as $R$ and commenting on the match.

$E_n = -13.6/n^2\ \text{eV}$, so the difference between successive levels $E_{n+1} - E_n$ shrinks as $n$ grows because $1/n^2$ decreases ever more slowly for larger $n$ (e.g. $E_2 - E_1 = 10.2\ \text{eV}$, but $E_5 - E_4 \approx 0.31\ \text{eV}$).

As $n \to \infty$, $E_n \to 0$: this is the ionisation limit, where the electron is no longer bound to the nucleus and the energy levels merge into a continuum.

Marks: one for stating the spacing decreases with $n$, one for a numeric or algebraic justification using $1/n^2$, one for correctly identifying $n = \infty$ as $E = 0$, the ionisation limit.

Band-6 plan. (1) State the classical problem (radiating, spiralling electron). (2) State the postulates and show how each solves part of the problem. (3) Link the postulates to the derived formula $1/\lambda = R(1/n_f^2 - 1/n_i^2)$ and its agreement with observed hydrogen spectra. (4) Evaluate: quantitative success for hydrogen versus the named limitations. End with a judgement.

Model answer. Classical electromagnetism predicts that an orbiting (accelerating) electron continuously radiates energy and should spiral into the nucleus within about $10^{-11}\ \text{s}$, yet atoms are observed to be stable and to emit only discrete spectral lines, not a continuous spectrum. Bohr resolved this with three postulates. First, the electron occupies certain stationary orbits in which it does not radiate, directly rejecting the classical prediction and explaining stability. Second, only orbits with quantised angular momentum $m_e v r = nh/2\pi$ are allowed, which restricts the electron to a discrete set of radii and therefore a discrete set of energies $E_n = -13.6\ \text{eV}/n^2$. Third, a photon is emitted only when the electron jumps between two stationary states, with $hf = E_i - E_f$, which explains why the spectrum is a set of sharp lines rather than a continuum.

Combining these postulates gives $1/\lambda = R(1/n_f^2 - 1/n_i^2)$, which reproduces the Lyman, Balmer and Paschen series to high precision and matches the previously empirical Balmer-Rydberg formula, converting a fitted equation into a derived physical result.

However, the model's success is limited to hydrogen and hydrogen-like one-electron ions; it cannot quantitatively predict the spectra of multi-electron atoms, where electron-electron repulsion is not accounted for. It also treats the electron as travelling on a sharp, definite orbit, which the uncertainty principle later showed to be physically untenable, and it has no mechanism for why angular momentum should be quantised in the first place - the rule is postulated, not derived from a deeper theory. It also cannot explain fine structure or line intensities.

Overall, the Bohr model succeeds remarkably as a quantitative theory of the hydrogen atom, correctly explaining both stability and the line spectrum, but it is a transitional, semi-classical model whose ad hoc quantisation rule and single-electron limitation mark it as a bridge to, not a substitute for, full quantum mechanics.

Marker's note: the top band explains BOTH stability and the discrete spectrum by name-checking specific postulates, shows the derived-Rydberg-formula link (not just "it matches experiment"), and closes with a genuine evaluative judgement (transitional bridge model) rather than a bare list of limitations. A response that lists postulates and limitations with no connecting explanation caps in the middle band.

Band-6 plan. Thesis: the equation's journey from empirical curve-fit to derived law is itself the significance. Argue (1) Balmer and Rydberg found the pattern with no underlying theory, (2) Bohr's postulates derived the same formula from first principles, giving it physical meaning and predictive power beyond the fitted data, (3) this predictive power (other series, other elements' hydrogen-like ions) is what made it historically significant. Conclude with a judgement of significance.

Model answer. In 1885 Balmer found a numerical formula that reproduced the four then-known visible hydrogen lines, and Rydberg generalised it to $1/\lambda = R(1/n_f^2 - 1/n_i^2)$ for all hydrogen lines. Both were purely empirical: the formula fitted the data precisely, but nothing in physics at the time explained why hydrogen should obey exactly this pattern, or why $R$ should take the value it did.

Bohr's 1913 model changed this. By postulating stationary non-radiating orbits with quantised angular momentum and photon emission on transitions, Bohr derived energy levels $E_n = -13.6\ \text{eV}/n^2$ and showed that the photon energy released on a transition, converted to a wavelength via $hc/\lambda = E_i - E_f$, reproduces the Balmer-Rydberg formula exactly, with $R$ expressible in terms of fundamental constants ($R = 13.6\ \text{eV}/hc$). An empirical curve-fit became a consequence of a physical model.

This derivation was highly significant. It predicted spectral series beyond the visible Balmer lines before they were all measured (Lyman in the ultraviolet, Paschen in the infrared), and it extended correctly to hydrogen-like ions such as $\text{He}^+$ with a $Z^2$ scaling, giving the model genuine predictive reach beyond the data used to build it. This transformed spectroscopy from a descriptive, empirical science into a quantitative test of atomic structure, and the close numerical agreement between the derived and measured $R$ was some of the strongest early evidence for quantum ideas.

Weighing this, the Balmer-Rydberg equation's significance lies less in the formula itself than in its transformation from an unexplained empirical pattern into a first physical derivation, which is why it is still presented as one of the pivotal confirmations of early quantum theory.

Marker's note: the top band explicitly contrasts "empirical" with "derived," shows the mechanism of the derivation (not just asserts Bohr "explained" it), and gives at least one concrete predictive success (new series or hydrogen-like ions) before closing with an explicit significance judgement. A response that only restates the formula and the postulates without addressing empirical-versus-derived status caps below the top band.