
Atomic Structure: Rutherford & Bohr
Comprehensive study notes on atomic models, alpha-particle scattering, Rutherford’s nucleus, Bohr quantisation, atomic energy levels, hydrogen line spectra, Rydberg formula and atomic excitation, with 20 original MCQs.
What an atomic model is meant to explain
An atom is a microscopic system containing a positively charged nucleus and, in a neutral atom, electrons whose total charge balances the nuclear charge. Earlier ideas of atoms as indivisible hard particles were refined by experiments revealing subatomic charges and matter’s internal structure. A scientific atomic model must explain measurable phenomena such as scattering, stability, line spectra and how chemical properties arise. Different models were introduced when new evidence could no longer be accommodated by the older picture.
For exam preparation, the historical order matters but is not enough by itself. Thomson offered an early distributed-positive-charge model; Rutherford inferred a compact positive nucleus; Bohr introduced quantised stationary states for hydrogen-like atoms; later quantum mechanics replaced classical electron tracks with orbitals and probability amplitudes. Each step addressed particular experimental facts and retained limitations. When a question asks which discovery is associated with a scientist, match the actual experiment to the conclusion rather than attaching every feature of a modern atom to the earliest model.
Discovery of electrons and the Thomson model
Cathode-ray experiments established that rays emitted in evacuated discharge tubes contain negatively charged particles. J. J. Thomson measured the electron charge-to-mass ratio, showing that a common negatively charged constituent existed in different kinds of matter. His atomic model placed electrons within a diffuse positively charged background, often compared with plums embedded in a pudding. It explained overall electrical neutrality in a simple visual way.
However, a widely spread positive background cannot readily produce the rare large-angle deflections seen when energetic alpha particles pass through a thin metal foil. This limitation became important when Geiger and Marsden collected their scattering data under Rutherford’s guidance. The existence of an electron does not by itself determine the location of the atom’s positive charge or explain a line spectrum. Thomson’s picture belongs to the historical development of atomic theory, not to the accepted modern structural model.
The gold-foil experiment and its essential observations
In the alpha-particle scattering experiment, energetic positively charged alpha particles were directed toward a very thin metal foil, typically gold in familiar textbook descriptions. A detecting screen revealed their deflection angles. Most alpha particles passed through with little change in direction, some experienced moderate scattering, and a very small fraction were deflected through large angles, with a few travelling nearly backward.
These observations were striking because a uniformly diffuse positive charge would tend to produce many small deviations rather than occasional strong single encounters. Rutherford’s explanation was that most of the atom’s volume is relatively empty of concentrated positive charge and mass, while a tiny dense nucleus can exert strong Coulomb repulsion during close alpha-particle approaches. The probability of very large deflection is small because a close encounter with the tiny nucleus is uncommon. The experiment does not imply an atom is a vacuum containing literally nothing: electrons and fields still occupy the surrounding region.
Rutherford’s nuclear model and its achievements
Rutherford proposed that the atom contains a small, positively charged, massive central nucleus surrounded by electrons, with most of the volume outside that nucleus. This model explained the highly nonuniform alpha-particle scattering distribution. A compact nuclear charge can strongly repel positive alpha particles that pass nearby while having limited influence on particles passing at large distances. The nucleus contains almost all the atomic mass in the modern view.
The nucleus is much smaller than the atom. Typical atomic dimensions are of order 10⁻¹⁰ m, whereas nuclear radii are of order 10⁻¹⁵ m, varying with element. Thus the radius ratio can be around 10⁵, and a simple volume ratio is enormously larger. Do not confuse these approximate scales with precise universal radii for all atoms and nuclei. Rutherford’s success was locating the positive charge and mass; it did not fully explain why the electron arrangement is stable or why atoms emit discrete wavelengths.
Scattering angles and impact parameter
The impact parameter is the perpendicular distance between the initial straight-line trajectory of an incoming projectile and the nucleus’s centre when the interaction is not yet significant. For Coulomb repulsion, a smaller impact parameter generally leads to a stronger deflection, while a large impact parameter gives a comparatively small change of direction. In the ideal Rutherford point-charge model the scattering-angle relation follows from classical mechanics of the inverse-square repulsive force.
The classical Rutherford differential scattering cross-section is proportional to 1/sin⁴(θ/2) for a fixed particle energy and nuclear charge in the standard approximation. Very large angles are much rarer than small angles. Such formulas assume sufficiently energetic projectiles, suitable thin targets and interactions dominated by elastic Coulomb scattering rather than strong nuclear effects or multiple collisions. For school-level problems, the physical interpretation of close approaches is more important than memorising the full cross-section expression.
Distance of closest approach for a head-on alpha particle
An alpha particle of charge +2e directed almost head-on toward a nucleus of charge +Ze slows under electrostatic repulsion. Neglecting other interactions, its initial kinetic energy is converted into electric potential energy at its closest approach, where radial speed momentarily vanishes in the head-on idealisation. Taking Coulomb potential zero at infinity, K_initial=(1/(4πε₀))(2Ze²/r_min), giving r_min=(1/(4πε₀))(2Ze²/K_initial).
This is a useful proportionality question: increasing projectile kinetic energy reduces the closest-approach distance, while increasing target nuclear charge increases it under otherwise fixed conditions. The calculation treats both participants through an approximation in which nuclear recoil is negligible and the target nucleus is effectively fixed. For low projectile energies far from nuclear contact, Coulomb repulsion provides the dominant effect. If the impact parameter is nonzero, the particle also has angular momentum and the simple head-on energy condition must be adapted.
Why Rutherford’s model was incomplete
In classical electrodynamics a charged particle accelerating in a curved orbit emits radiation. A classical electron continuously circling a nucleus would accelerate toward the centre and should lose energy, leading to inward spiralling rather than indefinite stability. Rutherford’s model did not offer a satisfactory explanation for why stable atoms persist despite this classical expectation.
It also could not account naturally for the discrete bright spectral lines emitted by excited hydrogen. Classical orbital motion would allow an extended continuum of frequencies and a changing orbit, not the observed specific sets of spectral wavelengths. These two shortcomings motivated Bohr’s postulates. However, the fact that Bohr’s model solved some problems does not mean the electron actually moves along a mechanically observable circular orbit in modern quantum mechanics. The historical model remains valuable as a stepping stone and because its hydrogen energy formula is accurate in its domain.
Bohr’s stationary-orbit postulates
Niels Bohr proposed that an electron in a hydrogen-like atom may occupy only certain allowed stationary states that do not radiate energy merely because it is in that state. He imposed angular momentum quantisation mvr=nħ for a circular orbit, with positive integer quantum number n. The allowed orbits and energies then form a discrete ladder rather than a continuous range.
A transition between allowed states is accompanied by absorption or emission of a photon whose energy equals the difference between the initial and final levels: hf=|E_i−E_f|. A downward transition emits a photon; an upward transition requires energy absorption under appropriate conditions. Bohr combined classical Coulomb force with a new quantisation rule. The quantisation postulate was not derivable from ordinary Newtonian orbital mechanics, which is why the model was a revolutionary but incomplete early quantum theory.
Allowed orbit radii and nuclear charge dependence
For a hydrogen-like ion containing one electron and a nucleus of charge +Ze, the Bohr orbit radius is r_n=a₀ n²/Z in the simplest approximation with an infinitely heavy nucleus, where a₀≈0.0529 nm is the Bohr radius for hydrogen. Thus in hydrogen the second orbit radius is four times the first, while the third is nine times the first. For higher nuclear charge Z, an otherwise corresponding n orbit lies closer to the nucleus.
The derivation sets electrostatic force kZe²/r² equal to centripetal mv²/r and combines it with mvr=nħ. This relation applies to one-electron Coulomb systems, not arbitrary multi-electron atoms whose electrons interact with each other and screen nuclear charge. It also neglects small corrections from reduced mass, fine structure and relativistic effects. In an exam comparing two states of hydrogen, remember that r∝n²; it is not proportional to n alone.
Electron speeds and orbital angular momentum
In the simple Bohr circular-orbit model, electron speed in a hydrogen-like ion varies approximately as v_n=Zαc/n, where α≈1/137 is the fine-structure constant. Thus a higher principal quantum number n corresponds to smaller orbital speed for fixed Z in this model. Angular momentum L=nħ grows with n, while orbit radius grows as n². These are compatible because mvr is the product of speed and radius.
Hydrogen’s first Bohr-orbit speed is about 2.19×10⁶ m/s, much lower than light speed, supporting the rough nonrelativistic starting approximation. For high-Z ions, relativistic corrections become more significant. Do not confuse Bohr speed with the speed of an electron in a modern probability orbital; precise quantum states do not assign simultaneous classical position and orbital velocity in the same way. For objective numerical ratios, however, these formulae remain a standard teaching tool.
Quantised energy levels in hydrogen
The Bohr energy of a hydrogen-like one-electron atom is E_n=−13.6Z²/n² electronvolts in the infinite nuclear-mass approximation. For hydrogen Z=1, the ground level is −13.6 eV, the n=2 level is −3.4 eV and the n=3 level is approximately −1.51 eV. The negative sign reflects a bound state relative to a free electron at rest infinitely far from the nucleus, chosen to have zero energy.
As n increases, the energy approaches zero from below. Higher n means the electron is less tightly bound and can be removed with less additional energy. An excited atom can emit photons by dropping to a lower level, with photon energy equal to the positive difference E_initial−E_final. A common exam error is to subtract negative energies in the wrong order and report a negative photon energy. Instead calculate the magnitude of the difference and label the process as absorption or emission.
Hydrogen spectral lines and the Rydberg formula
The discrete hydrogen spectrum contains families of lines grouped by the lower principal quantum number of the transition. The Rydberg relation is 1/λ=R_H(1/n_f²−1/n_i²) for emission from n_i>n_f, with R_H approximately 1.097×10⁷ m⁻¹ when appropriate reduced-mass corrections are understood. Each pair of allowed energy levels produces a specific photon wavelength, linking Bohr energy to experimental spectroscopy.
The Lyman series ends at n_f=1 and lies in the ultraviolet. The Balmer series ends at n_f=2 and includes prominent visible lines. The Paschen series ends at n_f=3 and lies in the infrared. Other series such as Brackett and Pfund have still higher final levels and longer wavelengths. These labels are frequently used in matching questions. Always identify the final level first; confusing an initial level n_i=3 with final n_f=3 misclassifies a hydrogen spectral line.
The Balmer series and visible transitions
Transitions ending at n=2 form the Balmer series. The n=3 to n=2 emission produces the well-known H-alpha line near 656 nm, in the red region of visible light. The n=4 to n=2 transition produces H-beta near 486 nm, while higher initial levels yield increasingly shorter wavelengths approaching the Balmer series limit near 365 nm. This pattern follows from the increasing energy difference as n_i rises toward infinity for a fixed final n_f=2.
The series converges because 1/n_i² tends to zero as the initial level rises. At the series limit the photon energy corresponds to ionising an atom initially in the n=2 level; it is not the ground-state ionisation energy of 13.6 eV. A candidate can avoid memorising all visible wavelengths by computing the difference between Bohr energies. For the 3→2 line, ΔE≈13.6(1/4−1/9)≈1.89 eV, yielding wavelength about 1240/1.89≈656 nm.
Excitation energy versus ionisation energy
Excitation raises an atom or electron from one bound level to another without necessarily removing the electron. For hydrogen, exciting ground-state n=1 to n=2 requires energy E₂−E₁=(−3.4)−(−13.6)=10.2 eV. Ground-state ionisation requires 13.6 eV in the ideal Bohr hydrogen model to bring the electron to E=0. The fact that both processes absorb energy should not blur their different final states.
An incident electron beam can excite an atom through collisions if it transfers sufficient energy, while radiation of an appropriate frequency can produce an allowed electronic excitation. Once an atom is excited it can decay by emitting photons, sometimes through a sequence of intermediate transitions. A possible photon energy equals the difference between a pair of levels; not every arbitrary energy value will be emitted as a sharp line in an isolated stationary atom. Selection rules and line broadening become important in more advanced treatments.
Spectral absorption, emission and conservation
A cold gas viewed in front of a broadband source can absorb selected wavelengths, creating dark absorption lines in the transmitted spectrum. An excited dilute gas can emit characteristic bright lines as electrons undergo downward transitions. In the ideal simple treatment, the frequencies of absorption and corresponding emission between the same two energy levels match, because the level difference is the same.
The observed line intensities need not be identical in absorption and emission: they depend on level populations, transition probabilities, temperature, excitation process and observation conditions. Spectral lines allow physicists and astronomers to infer chemical composition of luminous sources and surrounding gases. Redshift or blueshift can move observed wavelengths without changing the intrinsic atomic transition energies in a source’s rest frame. Questions about the source of a dark line should distinguish selective absorption from the absence of all light emission by the material.
Bohr model versus modern quantum mechanics
The Bohr model explains hydrogen-like energy levels and provides useful quantitative formulas, but it is not a complete theory for multi-electron atoms, fine structure, selection rules and intensities. Modern quantum mechanics describes bound electrons with wavefunctions and orbitals, labelled by quantum numbers and probability distributions. It does not generally assign each bound electron a definite classical circular path like a planet.
De Broglie’s wave condition gives a useful historical rationale for Bohr quantisation: an allowed orbit can be imagined to support an integer number of standing matter wavelengths, 2πr=nλ. Using λ=h/(mv), the condition 2πr=nλ gives 2πrmv=nh and therefore mvr=nh/(2π)=nħ. The wavefunction approach is more fundamental. The uncertainty principle and electron diffraction support the idea that electron behaviour is not purely classical. Use Bohr formulas within the one-electron model, and do not extend them uncritically to complex atoms.
Atomic spectroscopy, applications and exam technique
Atomic spectra are central to flame tests, astrophysical composition, laboratory spectroscopy and calibration of light sources. Quantised transitions also underlie lasers, though a laser needs population inversion, stimulated emission and an appropriate optical resonator rather than simply ‘atoms emitting coloured light’. The hydrogen spectrum offers a clear bridge between energy conservation, photon properties and atomic structure.
For each numerical problem, identify the element or ion and its nuclear charge Z, the initial and final quantum numbers, and whether the transition emits or absorbs energy. Write E_n=−13.6Z²/n² eV and compute ΔE with a consistent sign. Convert energy to wavelength through hc≈1240 eV·nm. For nuclear size comparisons, use order-of-magnitude estimates rather than treating atom and nucleus as similar sizes. For historical matching, remember Rutherford for compact nucleus, Bohr for quantised hydrogen levels and modern wave mechanics for orbitals.
Formula sheet and assumptions
Bohr energy and radius expressions below apply to the simplest one-electron Coulomb model, with reduced-mass and relativistic corrections neglected unless explicitly stated.
mvr=nħa₀≈0.0529 nmrₙ=a₀n²/ZEₙ=−13.6Z²/n² eVvₙ≈Zαc/nhf=|Eᵢ−E𝒇|1/λ=R(1/n𝒇²−1/nᵢ²)R_H≈1.097×10⁷ m⁻¹λ(nm)≈1240/E(eV)rmin≈k(2Ze²)/K2πr=nλdσ/dΩ∝1/sin⁴(θ/2)Solved numerical exercises
Use each calculation as a model problem. Check whether an energy describes excitation, ionisation or emitted radiation.
Hydrogen r₃/r₁=3²/1²=9.
For hydrogen-like He⁺ (Z=2), n=1 orbit radius is a₀/2≈0.0265 nm.
E₂=−13.6/4=−3.4 eV.
Hydrogen n=1→2 requires (−3.4)−(−13.6)=10.2 eV.
For hydrogen E₁=−13.6 eV and E∞=0: ionisation energy=13.6 eV.
Hydrogen n=3→2 gives ΔE=13.6(1/4−1/9)≈1.89 eV and λ≈656 nm.
Any transition ending at n=3 is in the Paschen series.
If atom radius is 10⁻¹⁰ m and nucleus radius is 10⁻¹⁵ m, atom is about 100,000 times larger by radius.
UPSC previous-year-question-style concept analysis
These are original explanations of past-paper-style concepts, not verbatim authenticated UPSC previous-year questions. Consult the official UPSC question-paper archive to confirm original questions, years and answers before attributing them to an examination.
Gold foil and nuclear inference
A classic objective-question theme asks what is inferred from most alpha particles passing through gold foil and a few being deflected backward. Deduce a largely extended atomic electron region containing a tiny dense charged nucleus.
Official UPSC papersSpectral-series matching
Exam-style matching questions may connect Lyman, Balmer and Paschen with ultraviolet, visible and infrared regions. Determine final n as 1, 2 and 3, respectively.
Official UPSC papersExcitation and ionisation distinction
A hydrogen problem may ask whether 10.2 eV is sufficient to ionise a ground-state atom. It is the ideal n=1 to n=2 excitation energy, below the 13.6 eV ground ionisation threshold.
Official UPSC papers20 original MCQs with full explanations
All 20 items are newly authored practice exercises. Attempt the question before expanding its correct answer.
- Thomson
- Rutherford
- Newton
- Faraday
Reveal answer and explanation
Answer B. Rutherford’s model explains strong rare scattering.
- atom has large relatively empty extent
- nucleus is giant
- alpha particles are neutral
- foil has no electrons
Reveal answer and explanation
Answer A. Most trajectories miss the compact nucleus.
- electron charge
- atomic stability and line spectra
- gravity
- alpha charge
Reveal answer and explanation
Answer B. Classical radiating orbital electrons would spiral inward.
- 10⁻¹⁰ m
- 10⁻¹⁵ m
- 10³ m
- 10⁻³ m
Reveal answer and explanation
Answer A. Ordinary atomic size is angstrom scale.
- 10⁻¹⁰ m
- 10⁻¹⁵ m
- 10 m
- 10⁻³ m
Reveal answer and explanation
Answer B. Nucleus is femtometre scale.
- mvr=nħ
- mvr=0
- mv²r=nh
- p=0
Reveal answer and explanation
Answer A. Angular momentum takes nħ values in Bohr model.
- +13.6 eV
- −13.6 eV
- 0 eV
- −3.4 eV
Reveal answer and explanation
Answer B. Ground level n=1 is −13.6 eV.
- −13.6 eV
- −6.8 eV
- −3.4 eV
- 0 eV
Reveal answer and explanation
Answer C. E2=−13.6/4.
- n
- n²
- 1/n
- 1/n²
Reveal answer and explanation
Answer B. r_n=a0n²/Z.
- 2
- 4
- 1/2
- 1/4
Reveal answer and explanation
Answer B. Ratio 2²=4.
- twice
- half
- four times
- same
Reveal answer and explanation
Answer B. r1=a0/Z.
- 3.4 eV
- 10.2 eV
- 13.6 eV
- 1.9 eV
Reveal answer and explanation
Answer C. Energy to continuum from n=1 is 13.6 eV.
- 3.4 eV
- 10.2 eV
- 13.6 eV
- 1.89 eV
Reveal answer and explanation
Answer B. Difference between −3.4 and −13.6.
- absorbs photon
- emits photon
- destroys energy
- creates nuclear charge
Reveal answer and explanation
Answer B. Energy difference leaves as photon.
- n=1
- n=2
- n=3
- n=4
Reveal answer and explanation
Answer A. All Lyman transitions terminate at n=1.
- n=1
- n=2
- n=3
- n=5
Reveal answer and explanation
Answer B. Balmer terminates at n=2.
- ultraviolet
- visible
- infrared
- gamma
Reveal answer and explanation
Answer C. Transitions terminating at n=3 are infrared.
- 3→2
- 2→1
- 4→1
- 1→3
Reveal answer and explanation
Answer A. Balmer H-alpha is n=3 down to n=2.
- −∞
- 0 from below
- +∞
- −13.6
Reveal answer and explanation
Answer B. E_n=−13.6/n² tends to zero.
- definite circular tracks for all electrons
- quantum orbitals and wavefunctions
- only Thomson pudding
- no wave nature
Reveal answer and explanation
Answer B. Quantum mechanics uses probability amplitudes rather than literal planetary orbits.
