Atoms — NEET notes

Rutherford's Nuclear Model

Rutherford bombarded a thin gold foil with alpha particles and discovered that the atom has a tiny, dense, positively charged nucleus at its centre — demolishing Thomson's 'plum pudding' model. Most alphas sailed through, but a few bounced back, proving the nucleus exists.

b = a 2 2 , a = 1 4 _0 2Ze^2 KE

Thomson's Model

Before Rutherford, J.J. Thomson proposed that the atom is a uniform sphere of positive charge with electrons embedded in it — like plums in a pudding.

J.J. Thomson (1898) proposed the plum pudding model: the atom is a sphere of uniformly distributed positive charge, with electrons embedded in it like plums in a pudding. The total negative charge of the electrons equals the total positive charge, making the atom electrically neutral.

• Also called the 'watermelon model' or 'raisin cake model'
• Positive charge is spread uniformly throughout the sphere — no concentrated nucleus
• Electrons are embedded in equilibrium positions inside this positive sphere
• Thomson's model could explain atomic neutrality and electron emission, but it could NOT explain Rutherford's scattering results

Alpha Scattering Experiment

Geiger and Marsden, under Rutherford's direction, fired alpha particles at a thin gold foil and observed their scattering pattern. The results were shocking — some alphas bounced straight back.

1. A radioactive source ( ^ 214 _ 83 Bi) emits a narrow beam of -particles (He nuclei, charge +2e , mass 4u )
2. The beam hits a thin gold foil ( 2.1 10^ -7 m thick, about 1000 atoms thick)
3. Scattered -particles are detected by a ZnS (zinc sulfide) screen that produces a flash of light on each hit
4. The detector can be rotated around the foil to measure scattering at different angles
5. The number of scattered particles N( ) is counted at each angle

• Gold was chosen because it is very malleable (can be hammered into extremely thin foils) and has a high atomic number ( Z = 79 ), giving large Coulomb repulsion
• The -particle is ^4_2 He ^ 2+ : charge +2e , mass = 4u 6.64 10^ -27 kg
• Typical kinetic energy of -particles: 5.5 MeV
• The experiment was done in vacuum to avoid -particles being stopped by air

Geiger and Marsden did the actual counting (it was extremely tedious — they sat in the dark for hours counting tiny flashes). Rutherford interpreted the results.

Key Observations

Three critical observations from the scattering experiment led Rutherford to propose the nuclear model of the atom.

• Most -particles ( > 99 % ) passed through the foil with little or no deflection → most of the atom is empty space
• A small fraction were deflected through large angles (some even > 90° ) → there must be a massive, positively charged concentration inside
• About 1 in 8000 bounced back ( 180° ) → the positive charge is concentrated in a tiny, very dense region — the nucleus

Rutherford's number distribution formula shows that the number of -particles scattered at angle follows: N( ) 1 ^4( /2) . This means scattering drops off rapidly with angle — most alphas go straight, very few bounce back.

N( ) 1 ^4( /2)

Rutherford's Nuclear Model

Based on his scattering results, Rutherford proposed that the atom has a tiny, dense, positively charged nucleus with electrons orbiting around it, like planets around the Sun.

• All positive charge and nearly all mass is concentrated in a tiny nucleus at the centre
• Nucleus size 10^ -15 m (1 femtometre); atom size 10^ -10 m (1 Angstrom)
• Ratio: R_ nucleus /R_ atom 10^ -5 — the atom is 10^5 times larger than the nucleus
• Electrons orbit the nucleus at large distances (like planets around the Sun)
• The electrostatic attraction between nucleus and electrons provides the centripetal force for orbiting

1 4 _0 Ze 2e r^2 = mv^2 r (Coulomb force = Centripetal force)

The distance of closest approach d_0 gives an upper estimate of nuclear size. An -particle heading directly at the nucleus converts all its kinetic energy to electrostatic potential energy at the turning point.

d_0 = 1 4 _0 2Ze^2 KE_

Impact Parameter

The impact parameter b is the perpendicular distance of the initial velocity vector from the centre of the nucleus. It directly controls the scattering angle.

The impact parameter b is the perpendicular distance between the trajectory of the -particle (extrapolated in a straight line) and the centre of the nucleus. It determines how close the -particle passes to the nucleus, and therefore how much it gets deflected.

b = a 2 2 , where a = 1 4 _0 2Ze^2 KE_

• b large (far from nucleus) → small (barely deflected) — this is the majority of -particles
• b small (near nucleus) → large (strongly deflected)
• b = 0 (head-on collision) → = 180° (complete rebound) — the -particle comes to rest and reverses
• b arrow → arrow 0° (no deflection at all)

Limitations

Rutherford's model, despite correctly predicting the nucleus, was fatally flawed because classical electrodynamics says orbiting electrons must radiate energy and spiral into the nucleus.

• Stability problem: According to classical electrodynamics, an accelerating (orbiting) charge must radiate EM energy continuously. The electron should spiral into the nucleus in 10^ -8 s
• Spectral problem: As the electron spirals inward, its orbital frequency changes continuously → it should emit a continuous spectrum. But atoms emit only discrete line spectra
• No explanation for discrete orbits: Rutherford's model puts no restriction on orbital radius or energy — any orbit should be allowed
• These failures set the stage for Bohr's revolutionary model (1913)

This is a classic NEET 1-marker: 'What was the main limitation of Rutherford's model?' → It could not explain the stability of the atom (orbiting electrons should radiate and spiral in).

Bohr's Model of the Hydrogen Atom

Bohr rescued atomic physics by introducing quantization: electrons can only orbit in specific allowed orbits without radiating. Energy is emitted or absorbed only when electrons jump between these orbits. His formula E_n = -13.6/n^2 eV nailed the hydrogen spectrum perfectly.

E_n = - 13.6 n^2 eV

Bohr's Three Postulates

Bohr proposed three revolutionary postulates that broke from classical physics: quantized orbits, no radiation in stable orbits, and the frequency condition for transitions.

• Postulate 1 (Quantized Orbits): Electrons revolve in certain stable circular orbits called stationary states without radiating energy. These are the only allowed orbits
• Postulate 2 (Angular Momentum Quantization): The angular momentum of the electron in a stationary state is quantized: L = mvr = n h 2 = n , where n = 1, 2, 3, ...
• Postulate 3 (Frequency Condition): When an electron jumps from orbit n_2 to n_1 , a photon is emitted (or absorbed) with frequency = E_ n_2 - E_ n_1 h

L = mvr = n = n h 2 , n = 1, 2, 3, ...

h = E_ n_2 - E_ n_1 (Bohr's frequency condition)

Orbital Radius & Velocity

From Bohr's postulates, the radius grows as n^2/Z and the velocity shrinks as Z/n. The ground state of hydrogen has the famous Bohr radius a_0 = 0.529 Angstrom.

Balancing the Coulomb force with centripetal force AND applying the angular momentum quantization condition mvr = n , we can solve for the orbital radius r_n and velocity v_n .

1. Coulomb force provides centripetal force: 1 4 _0 Ze^2 r^2 = mv^2 r
2. Angular momentum is quantized: mvr = n
3. From equation 2: v = n mr . Substitute into equation 1
4. Solve for r : r_n = n^2 ^2 mZe^2/(4 _0) = n^2 Z a_0
5. Substitute back to get v_n = Z n v_0 , where v_0 = e^2 4 _0 2.18 10^6 m/s

r_n = n^2 Z a_0, a_0 = 0.529 A = 0.529 10^ -10 m

v_n = Z n v_0, v_0 = c 137 2.18 10^6 m/s

• r_n n^2/Z — radius increases as n^2 (orbits spread out fast) and decreases with Z (stronger nuclear pull shrinks orbits)
• v_n Z/n — velocity decreases as n increases (farther orbits = slower electrons)
• The first orbit of hydrogen ( n=1, Z=1 ): r_1 = 0.529 A, called the Bohr radius a_0
• v_0/c = 1/137 — the velocity in the first Bohr orbit is about 1/137 of the speed of light. The number 137 is the famous fine structure constant
• For hydrogen-like ions (He ^+ , Li ^ 2+ ): just replace Z with the nuclear charge

Energy Levels

The total energy of an electron in the nth orbit of a hydrogen-like atom is E_n = -13.6 Z^2/n^2 eV. The negative sign means the electron is bound. This single formula predicts every spectral line of hydrogen.

The total energy is the sum of kinetic and potential energy. For an electron in Bohr's orbit: KE = 1 2 mv^2 = Ze^2 8 _0 r and PE = - Ze^2 4 _0 r . Notice that PE = -2 KE , so the total energy is negative (bound state).

E_n = - 13.6 Z^2 n^2 eV = - me^4 Z^2 8 _0^2 h^2 n^2

• Negative energy means the electron is bound to the nucleus. Energy must be supplied to free it
• E_1 = -13.6 eV (ground state of hydrogen) — the most tightly bound state
• E_2 = -3.4 eV, E_3 = -1.51 eV, E_4 = -0.85 eV, ... E_ = 0 eV (free electron)
• KE_n = -E_n = +13.6Z^2/n^2 eV (kinetic energy is always positive)
• PE_n = 2E_n = -27.2Z^2/n^2 eV (potential energy is twice total energy, more negative)
• |PE| = 2 KE always — this is a consequence of the virial theorem

Hydrogen-Like Ions

Bohr's model works perfectly for any atom/ion with only ONE electron. Just replace Z with the nuclear charge. The energy scales as Z^2 and the radius scales as 1/Z.

A hydrogen-like ion is any species with one electron orbiting a nucleus of charge Ze : H ( Z=1 ), He ^+ ( Z=2 ), Li ^ 2+ ( Z=3 ), etc. All Bohr formulas apply directly — just plug in the appropriate Z .

• E_n = -13.6 Z^2/n^2 eV — energy scales as Z^2 (He ^+ ground state = -54.4 eV, Li ^ 2+ ground state = -122.4 eV)
• r_n = 0.529 n^2/Z A — radius scales as n^2/Z (higher Z = more compact orbits)
• v_n = 2.18 10^6 Z/n m/s — velocity scales as Z/n (higher Z = faster electron)
• The Rydberg constant for species with charge Z : R_Z = Z^2 R_H , where R_H = 1.097 10^7 m ^ -1

Quick recall

Which orbit of Li ^ 2+ has the same radius as the first Bohr orbit of hydrogen?

Set n^2/Z = 1^2/1 . For Li ^ 2+ , Z = 3 , so n^2/3 = 1 , giving n^2 = 3 . Since n must be an integer, NO orbit of Li ^ 2+ has exactly the same radius as H's first orbit. But n = 3 gives r = 9/3 = 3a_0 , which equals the 3rd orbit radius of H.

Hydrogen Spectrum & Spectral Series

Hydrogen emits light at very specific wavelengths, organized into named series: Lyman (UV), Balmer (visible), Paschen (IR), Brackett (IR), and Pfund (far IR). Each series corresponds to transitions ending at a particular energy level. The Rydberg formula predicts every single line.

1 = R ( 1 n_1^2 - 1 n_2^2 )

Emission vs Absorption Spectra

When atoms emit light, they produce bright lines on a dark background (emission spectrum). When white light passes through a gas, atoms absorb specific wavelengths, producing dark lines on a bright background (absorption spectrum).

• Emission spectrum: Hot gas at low pressure emits light at specific discrete wavelengths → bright coloured lines on dark background
• Absorption spectrum: White light passing through cool gas → dark lines at the SAME wavelengths as the emission lines, on a continuous bright background
• Continuous spectrum: Hot solid or dense gas emits light at ALL wavelengths (e.g., incandescent bulb filament)
• Every element has a unique set of spectral lines — like a fingerprint. No two elements share the same spectrum
• Kirchhoff's law: a substance absorbs the same wavelengths that it emits

Rydberg Formula

The Rydberg formula gives the wavelength of every spectral line of hydrogen. It was discovered empirically and later explained perfectly by Bohr's model.

For a transition from a higher orbit n_2 to a lower orbit n_1 in a hydrogen-like atom with nuclear charge Z , the wavelength of the emitted photon is given by the Rydberg formula.

1 = RZ^2 ( 1 n_1^2 - 1 n_2^2 ), n_2 > n_1

R = 1.097 10^7 m^ -1 (Rydberg constant)

• R = 1.097 10^7 m ^ -1 — the Rydberg constant (MEMORIZE this value)
• 1/ is called the wave number (unit: m ^ -1 )
• n_1 = lower level (where electron lands), n_2 = upper level (where electron starts)
• For hydrogen, Z = 1 . For He ^+ , Z = 2 . For Li ^ 2+ , Z = 3
• The formula also gives absorption wavelengths — just read the transition in reverse ( n_1 n_2 )

Five Spectral Series

Each spectral series consists of all transitions ending at a specific lower level. Lyman ends at n=1, Balmer at n=2, and so on. Only the Balmer series falls in the visible range.

• Lyman series: n_1 = 1 , transitions from n_2 = 2, 3, 4, ... → Ultraviolet region
• Balmer series: n_1 = 2 , transitions from n_2 = 3, 4, 5, ... → Visible region (the only visible series!)
• Paschen series: n_1 = 3 , transitions from n_2 = 4, 5, 6, ... → Infrared region
• Brackett series: n_1 = 4 , transitions from n_2 = 5, 6, 7, ... → Infrared region
• Pfund series: n_1 = 5 , transitions from n_2 = 6, 7, 8, ... → Far infrared region

Within each series, the first line is the longest wavelength (smallest energy jump: n_2 = n_1 + 1 ) and the series limit is the shortest wavelength (largest energy jump: n_2 = ). For example, in the Balmer series: first line is 3 2 (H , 656 nm, red) and series limit is 2 (365 nm, UV boundary).

Wavelength Calculations

How to calculate the wavelength of any specific spectral line, the series limit, and the ratio of wavelengths between different lines or series.

Use the Rydberg formula 1/ = R(1/n_1^2 - 1/n_2^2) and plug in the values. For the first line of any series: n_2 = n_1 + 1 . For the series limit: n_2 = (so 1/n_2^2 = 0 ).

1. Identify which series → this gives n_1 (Lyman = 1, Balmer = 2, Paschen = 3, etc.)
2. Identify which line → 'first line' means n_2 = n_1 + 1 ; 'second line' means n_2 = n_1 + 2 ; 'series limit' means n_2 =
3. Plug into 1/ = R(1/n_1^2 - 1/n_2^2) with R = 1.097 10^7 m ^ -1
4. Solve for in metres, then convert to nm if needed ( 1 nm = 10^ -9 m)

• Shortest in Lyman (series limit): 1/ = R(1/1 - 0) = R → = 1/R = 91.2 nm
• Longest in Lyman (first line): 1/ = 3R/4 → = 121.6 nm
• Shortest in Balmer (series limit): 1/ = R/4 → = 4/R = 364.7 nm
• Longest in Balmer (first line, H ): 1/ = 5R/36 → = 656.3 nm
• Within any series: _ first line > _ second line > ... > _ series limit

Quick recall

What is the ratio of the longest wavelength of the Lyman series to the shortest wavelength of the Balmer series?

Longest Lyman: _1 = 4/(3R) . Shortest Balmer: _2 = 4/R . Ratio = _1/ _2 = (4/3R)/(4/R) = 1/3 . So the longest Lyman line has 1/3 the wavelength of the Balmer series limit.

Number of Spectral Lines

When electrons in a gas are excited to the nth level, the maximum number of different spectral lines they can emit while returning to ground state is n(n-1)/2.

If hydrogen atoms are excited to the n th energy level, electrons can de-excite through various paths. The total number of possible transitions (and hence spectral lines) is the number of ways to choose 2 levels from n levels.

Maximum number of spectral lines = n(n-1) 2

• n = 2 : only 1 line ( 2 1 )
• n = 3 : 3 lines ( 3 2 , 3 1 , 2 1 )
• n = 4 : 6 lines ( 4 3 , 4 2 , 4 1 , 3 2 , 3 1 , 2 1 )
• n = 5 : 10 lines, n = 6 : 15 lines
• This assumes ALL atoms are excited to level n and ALL possible de-excitation paths are taken

Energy Levels & Transitions

The energy level diagram is your best friend for Atoms questions. Electrons absorb exact photon energies to jump up and emit exact photon energies when they jump down. Excitation energy, ionization energy, and photon energies all come from reading this diagram correctly.

E_ photon = E_ n_2 - E_ n_1 = 13.6 ( 1 n_1^2 - 1 n_2^2 ) eV

Energy Level Diagram

A visual representation of all allowed energy states of hydrogen. Horizontal lines at heights proportional to E_n show where electrons can exist. Transitions between levels are shown as arrows.

• n = 1 → Ground state ( E_1 = -13.6 eV) — the lowest, most stable energy state
• n = 2 → First excited state ( E_2 = -3.4 eV) — NOT the second level, but the first EXCITED level
• n = 3 → Second excited state ( E_3 = -1.51 eV)
• n = → Free electron ( E_ = 0 eV) — electron is no longer bound
• Energy levels get closer together as n increases — they converge toward 0 eV

Excitation & Ionization Energy

Excitation energy is the energy needed to jump from one level to a higher one. Ionization energy is the energy needed to completely remove the electron from the atom (jump to n = infinity).

Excitation energy = energy needed to take the electron from its current state to a higher state. Ionization energy = excitation energy to n = (remove the electron completely). Both are always positive (energy must be supplied).

Excitation energy (n_1 n_2) = E_ n_2 - E_ n_1 = 13.6 ( 1 n_1^2 - 1 n_2^2 ) eV

Ionization energy from n_1 = E_ - E_ n_1 = 0 - E_ n_1 = 13.6 Z^2 n_1^2 eV

• Ionization energy from ground state ( n = 1 ): 13.6 eV for hydrogen — this is the IE of hydrogen
• First excitation energy ( n = 1 2 ): -3.4 - (-13.6) = 10.2 eV
• Second excitation energy ( n = 1 3 ): -1.51 - (-13.6) = 12.09 eV
• Ionization energy from n = 2 : 0 - (-3.4) = 3.4 eV
• Excitation potential = excitation energy in volts (numerically same in eV if charge = e )

Photon Emission & Absorption

When an electron drops from a higher to a lower energy level, it emits a photon whose energy exactly equals the energy gap. When it absorbs a photon of exactly the right energy, it jumps up.

E_ photon = h = hc = |E_ n_2 - E_ n_1 |

• Emission: Electron drops from higher n_2 to lower n_1 → photon is released with energy E_ n_2 - E_ n_1
• Absorption: Electron jumps from lower n_1 to higher n_2 → photon of EXACTLY E_ n_2 - E_ n_1 must be absorbed
• If the incoming photon has MORE energy than the gap, the excess does NOT get absorbed — the transition simply does NOT happen (unlike the photoelectric effect where excess becomes KE!)
• If the photon has EXACTLY the ionization energy or more, the electron is freed. Excess energy becomes KE of the free electron
• Useful conversion: E (eV) = 13.6(1/n_1^2 - 1/n_2^2) and (nm) = 1240/E (eV)

Excitation by Electron Collision

Unlike photons, free electrons can transfer ANY amount of energy during a collision (not just exact gap energies). The excess energy stays as KE of the free electron.

When a free electron (not a photon) collides with a hydrogen atom, the rules are different. The electron can transfer any amount of energy as long as it is at least equal to the gap. The leftover energy stays as KE of the free electron after the collision.

• A free electron with 11 eV hitting a hydrogen atom CAN excite it to n = 2 (needs 10.2 eV) — leftover 0.8 eV stays as KE of the free electron
• A photon of 11 eV hitting the SAME atom CANNOT excite it — because 11 eV doesn't match any transition exactly
• For electron collisions, the condition is: KE of electron transition energy gap
• For photon absorption, the condition is: photon energy = transition energy gap (exact match required for bound-bound transitions)

Quick recall

A free electron with 12.5 eV collides with a hydrogen atom in the ground state. To which levels can the atom be excited?

Available gaps from n = 1 : to n = 2 (10.2 eV), to n = 3 (12.09 eV). Both are 12.5 eV, so both transitions are possible. The atom CANNOT reach n = 4 (needs 12.75 eV > 12.5 eV). The electron collision excites to the highest possible level: n = 3 , with 12.5 - 12.09 = 0.41 eV leftover.

de Broglie Explanation of Bohr's Quantization

Why is angular momentum quantized? de Broglie gave an elegant answer: the electron is a wave, and only orbits where an integer number of wavelengths fit around the circumference are stable (standing wave condition). This explains Bohr's mysterious n*h/2pi postulate from first principles.

2 r = n where = h mv

Electron as a Standing Wave

de Broglie proposed that electrons have wave nature. For an electron wave to persist in a circular orbit, it must form a standing wave — otherwise it destructively interferes with itself and cancels out.

Louis de Broglie (1924) proposed that every particle has a wave associated with it, with wavelength = h/mv = h/p . For an electron in a Bohr orbit, this wave must close on itself after going around the circumference. This means the circumference must be an integer multiple of the wavelength.

2 r = n , n = 1, 2, 3, ...

• Standing wave condition: the circumference of the orbit must equal a whole number of de Broglie wavelengths
• If 2 r n , the electron wave destructively interferes with itself → that orbit is NOT allowed
• If 2 r = n , the wave reinforces constructively → stable standing wave → allowed orbit
• The integer n is the number of complete wavelengths that fit around the orbit — it is the same quantum number that appears in Bohr's model

de Broglie proposed this in his PhD thesis! His examiner sent the thesis to Einstein, who called it 'a first feeble ray of light on this worst of our physics enigmas.' de Broglie won the Nobel Prize in 1929.

Deriving Bohr's Condition

Substituting the de Broglie wavelength into the standing wave condition directly gives Bohr's angular momentum quantization — no ad hoc assumption needed.

1. Start with the standing wave condition: 2 r = n
2. Substitute de Broglie wavelength: = h/mv
3. This gives: 2 r = n h mv
4. Rearrange: mvr = nh 2 = n
5. This IS Bohr's angular momentum quantization: L = n — derived, not assumed!

2 r = n = n h mv mvr = n h 2 = n

• de Broglie's idea provides the physical reason behind Bohr's quantization — it is not just a mathematical trick
• Bohr postulated L = n in 1913 without knowing why. de Broglie explained it in 1924
• The full quantum mechanical treatment (Schrodinger, 1926) confirmed de Broglie's picture and gave even more detailed results
• n = number of standing wave loops = quantum number = same n in E_n = -13.6/n^2

de Broglie Wavelength in Bohr Orbits

The electron's wavelength in the nth Bohr orbit can be calculated and shows that it equals the circumference divided by n.

From 2 r_n = n _n , the de Broglie wavelength of the electron in the n th Bohr orbit is _n = 2 r_n / n . Since r_n = n^2 a_0 , this gives _n = 2 n a_0 .

_n = 2 r_n n = 2 n a_0 (for hydrogen, Z = 1)

• _n n — the de Broglie wavelength INCREASES with n (electron moves slower, so wavelength is longer)
• For n = 1 : _1 = 2 0.529 A = 3.32 A (comparable to atomic size!)
• For n = 2 : _2 = 2 _1 = 6.64 A
• Ratio of wavelengths: _1 : _2 : _3 = 1 : 2 : 3
• The electron wavelength in atoms is comparable to atomic dimensions — this is why wave effects matter

Quick recall

The de Broglie wavelength of an electron in the 3rd Bohr orbit of hydrogen is how many times the wavelength in the 1st orbit?

Since _n n , the ratio is _3/ _1 = 3/1 = 3 times.

Limitations of Bohr's Model

Bohr's model was a spectacular success for hydrogen but a spectacular failure for everything else. It cannot explain multi-electron atoms, fine structure of spectral lines, Zeeman effect, Stark effect, or the relative intensities of spectral lines. Quantum mechanics was needed to go further.

Bohr works for: H, He^+, Li^ 2+ (single-electron systems ONLY)

Bohr's Successes

Before listing failures, it is worth appreciating what Bohr achieved: perfect prediction of the hydrogen spectrum, derivation of the Rydberg constant from first principles, and correct energy levels for hydrogen-like ions.

• Explained the stability of atoms — electrons in stationary states don't radiate (resolved Rutherford's collapse problem)
• Predicted the hydrogen spectrum perfectly — all wavelengths matched experiment to high precision
• Derived the Rydberg constant R = me^4/(8 _0^2 h^3 c) from fundamental constants — giving a theoretical basis for an empirical formula
• Correctly predicted spectra of hydrogen-like ions (He ^+ , Li ^ 2+ , etc.)
• Explained the Bohr radius ( a_0 = 0.529 A) and gave correct size estimates for atoms
• Introduced the concept of quantized energy levels — which remains central to all of quantum mechanics

Multi-Electron Atoms

Bohr's model works only for single-electron systems. The moment you have two or more electrons, their mutual repulsion makes the problem unsolvable with Bohr's approach.

Bohr's model treats the electron as orbiting a nucleus with a simple Coulomb force F = kZe^2/r^2 . This works when there is only ONE electron. With two or more electrons, each electron repels every other electron AND is attracted to the nucleus — a three-body (or many-body) problem that Bohr's simple circular orbit model cannot handle.

• Helium (2 electrons) — Bohr's model gives WRONG energy levels. The electron-electron repulsion changes everything
• Lithium and beyond — completely impossible with Bohr's approach
• Even for He, the ground state energy predicted by Bohr ( -108.8 eV using Z = 2 ) is significantly off from the experimental value ( -79 eV) because it ignores electron-electron repulsion
• Only hydrogen-like (single-electron) species work: H, He ^+ , Li ^ 2+ , Be ^ 3+ , etc.

Fine Structure

High-resolution spectroscopy reveals that each spectral 'line' of hydrogen is actually a group of closely spaced lines. Bohr's model predicts single lines — it cannot explain this fine structure.

When hydrogen's spectral lines are examined with high-resolution spectrometers, each line is found to be a cluster of very closely spaced lines — called the fine structure. For example, the H line (656 nm) is actually a doublet (two lines very close together).

• Bohr's model predicts one energy for each n → one line per transition
• Experiment shows each n level is actually split into sub-levels characterised by the angular momentum quantum number l
• The fine structure is caused by spin-orbit coupling — the interaction between the electron's orbital motion and its intrinsic spin
• The splitting is very small (hence 'fine') — of the order of ^2 E_n , where 1/137 is the fine structure constant
• Explaining fine structure requires Sommerfeld's extension (elliptical orbits) or full quantum mechanics + relativistic corrections

Sommerfeld (1916) extended Bohr's model with elliptical orbits and relativistic corrections, which partially explained the fine structure. But the full explanation needed Dirac's relativistic quantum mechanics (1928).

Zeeman & Stark Effects

When hydrogen is placed in a magnetic field (Zeeman effect) or electric field (Stark effect), spectral lines split into multiple components. Bohr's model cannot explain this splitting.

• Zeeman Effect: Spectral lines split when atom is placed in an external magnetic field. Discovered by Pieter Zeeman (1896)
• Stark Effect: Spectral lines split when atom is placed in an external electric field. Discovered by Johannes Stark (1913)
• Bohr's model has no mechanism for either effect — it does not include the interaction of the atom with external fields
• The splitting arises because different sub-levels (characterised by magnetic quantum number m_l ) respond differently to external fields
• Quantum mechanics explains both effects using the additional quantum numbers l and m_l that Bohr's model lacks

What Else Bohr Misses

Bohr's model tells you which transitions are possible but cannot predict how bright each spectral line will be, nor does it give selection rules for which transitions are 'allowed' vs 'forbidden'.

• Bohr's model cannot predict the relative intensities of spectral lines — why some lines are bright and others are faint
• It cannot explain selection rules ( l = 1 ) — not all transitions between energy levels are equally probable
• It treats orbits as circular only — Sommerfeld extended this to elliptical orbits to partially explain fine structure
• It assumes electrons have definite positions and velocities — violating the Heisenberg uncertainty principle x p /2
• The concept of well-defined 'orbits' is replaced by orbitals (probability clouds) in quantum mechanics