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Bohr Atom & Atomic Spectra

MODULE 13 · QUANTUM

QUANTUM STATE

3
n (LEVEL)
−1.51
E (eV)
4.76
r (Å)
7.27
v (×10⁵ m/s)
TRANSITION λ (nm)

CONTROLS

SERIES / PRESETS

TRANSITIONS (n→n')

DATA EXPORT

THE IDEA, STEP BY STEP

Start — atoms have favourite colours

Hold a flame to table salt and it glows orange; heat copper and it burns green; a neon sign shines red-orange. Each element always makes its own exact set of colours — a fingerprint of light. Hydrogen, the simplest atom of all, glows a soft pink-red. The puzzle for a century was: why does an atom emit only a few precise colours instead of a smooth rainbow?

Build — electrons live on an energy staircase

Picture the electron as standing on a staircase, not a ramp. It can rest on step 1, step 2, step 3 — labelled by the principal quantum number $n$ — but never in between. Each step has a fixed energy. For hydrogen the rule is beautifully simple:

$$E_n = -\frac{13.6\ \text{eV}}{n^2}$$

When the electron drops from a high step to a low one, the leftover energy leaves as a single particle of light — a photon — of one exact colour: $\Delta E = hf = hc/\lambda$. Drop from $n=3$ to $n=2$ and the energy released is $E_3-E_2 = (-1.51)-(-3.40) = 1.89$ eV, which comes out as red light at $656$ nm — that pink-red hydrogen glow. A different jump, a different colour. The few allowed jumps are exactly why an atom emits only a few sharp lines.

Deepen — Bohr's quantum leap and the Rydberg formula

Why only certain steps? In 1913 Bohr proposed that an electron's angular momentum can take only whole-number multiples of $\hbar$: $m_e v r = n\hbar$. That single quantum condition forces the orbits — and so the energies — to be discrete. Combining the jumps gives the formula Rydberg had already guessed from data:

$$\frac{1}{\lambda} = R_\infty Z^2\!\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$$

The steps are not evenly spaced: they bunch together as $n$ grows and pile up at $E=0$, where the electron breaks free (ionization). The nuclear charge $Z$ enters squared, so a one-electron ion like He⁺ ($Z=2$) has every energy multiplied by four and every wavelength shrunk by four. In the sim, the n slider picks the upper step, n' the landing step, and Z the nucleus.

Try this in the sim above

Set the lower level $n'=2$ (the Balmer series) and step $n$ up from 3 to 6: watch the emission lines march from red toward violet, crowding together as they approach the $364.6$ nm series limit. Switch $n'=1$ (Lyman) and notice every line jumps into the ultraviolet — invisible to the eye. Finally push $Z$ from 1 to 2: the whole pattern keeps its shape but slides to four-times-higher energy, the signature of hydrogen-like ions.

DERIVATION — BOHR MODEL

Governing Equations

$$E_n = -\frac{Z^2 \cdot 13.6\,\text{eV}}{n^2}, \quad n = 1,2,3,\ldots$$
$$\frac{1}{\lambda} = R_\infty Z^2\!\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right), \quad R_\infty = 1.097\times10^7\,\text{m}^{-1}$$

Symbol Table

SymbolQuantityValue / Unit
nPrincipal quantum number1, 2, 3, …
ZNuclear charge numberdimensionless
E_nEnergy of level neV or J
R_∞Rydberg constant1.097×10⁷ m⁻¹
a₀Bohr radius0.529 Å
λPhoton wavelengthnm
v_nOrbital velocitym/s
r_nOrbital radiusn²a₀/Z (Å)

Step-by-Step Derivation

STEP 1 — COULOMB FORCE = CENTRIPETAL FORCE
$$\frac{Ze^2}{4\pi\varepsilon_0 r^2} = \frac{m_e v^2}{r}$$

The electrostatic attraction between electron and nucleus provides the centripetal force needed for circular orbit.

STEP 2 — BOHR QUANTIZATION POSTULATE
$$m_e v r = n\hbar, \quad n = 1,2,3,\ldots$$

Angular momentum is quantized in integer multiples of ℏ = h/2π.

STEP 3 — ORBITAL RADII
$$r_n = \frac{n^2 \hbar^2 4\pi\varepsilon_0}{Z m_e e^2} = \frac{n^2 a_0}{Z}, \quad a_0 = 0.529\,\text{Å}$$
STEP 4 — TOTAL ENERGY (KE + PE)
$$E_n = \frac{1}{2}m_e v^2 - \frac{Ze^2}{4\pi\varepsilon_0 r_n} = -\frac{Z^2 m_e e^4}{8\varepsilon_0^2 h^2 n^2} = -\frac{Z^2 \cdot 13.6}{n^2}\,\text{eV}$$
STEP 5 — PHOTON EMISSION (RYDBERG FORMULA)
$$\Delta E = hf = \frac{hc}{\lambda} = E_{n_2} - E_{n_1} \Rightarrow \frac{1}{\lambda} = R_\infty Z^2\!\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$$
STEP 6 — SERIES LIMITS

As n₂→∞, λ approaches the series limit. For Lyman (n₁=1): λ_lim = 91.2 nm. For Balmer (n₁=2): λ_lim = 364.6 nm.

WORKED EXAMPLE — H Balmer α Line (n=3→2)

$$\frac{1}{\lambda} = (1.097\times10^7)\left(\frac{1}{4}-\frac{1}{9}\right) = 1.524\times10^6\,\text{m}^{-1}$$ $$\lambda = 656.3\,\text{nm}\quad(\text{red, H}_\alpha\text{ line})$$ $$\Delta E = \frac{hc}{\lambda} = \frac{(6.626\times10^{-34})(3\times10^8)}{656.3\times10^{-9}} = 1.89\,\text{eV}$$

Atkins, P. W. & De Paula, J. (2014). Physical Chemistry, 10th ed., Ch. 9. Oxford University Press.

FREQUENTLY ASKED QUESTIONS

MODELWhy does the Bohr model work for hydrogen but fail for helium?
The Bohr model treats each electron independently in a central Coulomb field. For hydrogen (1 electron), this is exact within the model. For helium (2 electrons), electron–electron repulsion is significant and cannot be handled by the Bohr circular-orbit approach. Multi-electron atoms require quantum mechanics (Schrödinger equation) with exchange and correlation effects.
ENERGYWhy are Bohr energy levels negative?
The zero of energy is defined as the electron at infinite separation from the nucleus (ionized). Any bound state has lower energy than this, hence negative values. The more negative the energy, the more tightly bound the electron. E₁ = −13.6 eV means 13.6 eV of energy is needed to ionize hydrogen from the ground state.
SPECTRUMWhat is the difference between the Lyman, Balmer, and Paschen series?
They differ by the final (lower) energy level. Lyman series (n'=1) falls in the UV; Balmer (n'=2) in visible/near-UV; Paschen (n'=3) in IR. The Balmer series is historically most important as it was observed visually and led to the empirical formula preceding Bohr's model.
QUANTUMHow does changing Z affect the spectrum?
Energy levels scale as Z², so for hydrogen-like ions (He⁺, Li²⁺, etc.), all energies are multiplied by Z². The Rydberg formula becomes 1/λ = R∞Z²(1/n₁² − 1/n₂²). Lines shift to shorter wavelengths (higher energy) as Z increases. He⁺ Lyman-α is at 30.4 nm vs H at 121.6 nm.
PHYSICSWhat is the physical meaning of the Bohr radius a₀?
a₀ = 0.529 Å is the most probable distance of the electron from the nucleus in the ground state of hydrogen. In quantum mechanics, it corresponds to the peak of the radial probability distribution |ψ₁s|²r². It sets the natural length scale for atomic physics.
EMISSIONWhy do atoms emit only specific wavelengths?
Because energy levels are quantized, transitions between them release photons with energies exactly equal to the level difference: ΔE = hf. Since levels are discrete, only specific frequencies (and hence wavelengths) can be emitted. This is the basis of atomic spectroscopy and the "fingerprint" identification of elements.
HISTORYWhat experimental evidence supported the Bohr model?
The model correctly predicted: (1) the Rydberg constant from first principles; (2) all wavelengths of the H spectrum to within 0.1%; (3) the ionization energy of H (13.6 eV); (4) spectra of He⁺, Li²⁺ (hydrogen-like ions). Definitive failure appeared with fine structure and multi-electron spectra.

COMMON MISCONCEPTIONS

✗ MISCONCEPTION: Electrons travel in circular orbits like planets.
This is the Bohr model picture, which is a simplification. Quantum mechanics shows electrons exist as probability clouds (orbitals). The Bohr model works for hydrogen energetically but the idea of a well-defined circular orbit is incorrect — electrons have no definite trajectory.
Taber, K. S. (2005). Learning quanta. Sci. Educ., 89(6), 994–1010.
✗ MISCONCEPTION: Higher energy means higher quantum number n.
True for hydrogen (E_n = −13.6/n²), but this can be misleading: higher n means less negative energy, i.e., closer to zero (ionization). The energy increases (becomes less negative) with n. Students often confuse "higher n = higher orbit = more energy" with the absolute scale where all bound states are negative.
Niaz, M. (2005). Bohr's model in the chemistry curriculum. J. Chem. Educ., 82(11), 1647.
✗ MISCONCEPTION: The Bohr model is wrong and completely useless.
The Bohr model gives exact energies for hydrogen-like species and correctly derives the Rydberg constant from physical constants. It remains pedagogically valuable and conceptually important. Modern quantum mechanics reduces to Bohr results for expectation values of energy in hydrogen.
Petri, J. & Niedderer, H. (1998). Int. J. Sci. Educ., 20(9), 1075–1099.
✗ MISCONCEPTION: Electrons emit light continuously as they orbit.
Classical electromagnetism predicts an accelerating charge radiates continuously — this would cause electrons to spiral into the nucleus. The Bohr model's key postulate is that electrons in stationary orbits do NOT radiate; emission occurs only during quantum jumps between levels.
Stinner, A. (2002). Phys. Educ., 37(1), 28–35.
✗ MISCONCEPTION: Emission and absorption spectra are identical.
Emission spectra show bright lines on dark background (excited atoms releasing photons); absorption spectra show dark lines on continuous background (ground-state atoms absorbing photons). They have the same wavelengths but are complementary — absorption only shows transitions from the ground state (at room temperature).
CERP Review (2015). Chem. Educ. Res. Pract., 16(1), 9–21.
✗ MISCONCEPTION: The Rydberg formula only applies to hydrogen.
The Rydberg formula applies to all hydrogen-like (one-electron) species: H, He⁺, Li²⁺, Be³⁺, etc. For these, the formula 1/λ = R∞Z²(1/n₁² − 1/n₂²) is exact within the Bohr model. For multi-electron atoms, quantum defects modify effective n values.
Atkins & De Paula (2014). Physical Chemistry, 10th ed., p. 326.