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Physics 3% exam weight

Atoms

Part of the NEET UG study roadmap. Physics topic phy-026 of Physics.

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Atoms

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

  • Thomson → Rutherford → Bohr: Thomson’s plum-pudding failed Rutherford’s α-scattering (large-angle deflection proved a tiny, dense, positive nucleus); Bohr then quantized electron orbits in that nucleus to fix the collapse predicted by classical EM.
  • Bohr postulates: (i) electrons orbit in stationary orbits without radiating; (ii) angular momentum is quantized, L = mvr = nh/(2π); (iii) a photon of energy hν = |E₂ − E₁| is emitted/absorbed on transition.
  • Bohr radius a₀ = 0.529 Å; orbit radius rₙ = n² a₀/Z, energy Eₙ = −13.6 Z²/n² eV.
  • Rydberg formula: 1/λ = RZ²(1/n₁² − 1/n₂²); series landing on n = 1,2,3,4,5 = Lyman, Balmer, Paschen, Brackett, Pfund.
  • Energy signs for n=1: KE = +13.6 Z² eV, PE = −27.2 Z² eV, TE = −13.6 Z² eV (so |PE| = 2|TE| = 2 KE).

🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Rutherford’s Nuclear Atom

Geiger–Marsden fired α-particles at thin gold foil. Most passed straight through, but a tiny fraction rebounded at angles > 90°. Rutherford’s conclusion: a concentrated positive charge (the nucleus) sits at the centre of an atom that is mostly empty space. This ruled out Thomson’s model but left a stability problem — accelerating electrons in circular orbits should radiate continuously and spiral into the nucleus in ~10⁻¹¹ s.

Bohr’s Quantization Conditions

Bohr kept Rutherford’s nucleus but imposed three postulates: (a) only certain orbits (radii rₙ) are permitted; (b) angular momentum is quantized, mvr = nh/(2π), with n = 1, 2, 3…; (c) transitions between orbits emit/absort photons with hν = E₂ − E₁.

Radii, Velocities and Energies

Balancing Coulomb force with centripetal force gives v = Ze²/(4πε₀mr) and rₙ = n²h²ε₀/(πmZe²) = n²a₀/Z, with a₀ ≈ 0.529 Å. Substituting back:

  • Velocity: vₙ = (Ze²)/(2ε₀nh)
  • Kinetic energy: KE = +13.6 Z²/n² eV
  • Potential energy: PE = −27.2 Z²/n² eV
  • Total energy: Eₙ = −13.6 Z²/n² eV

Hence |PE| = 2 |TE| = 2 KE — a useful identity for NEET numericals.

Hydrogen Spectrum and Spectral Series

The energy released when an electron drops from n₂ to n₁ < n₂ becomes a photon:

1/λ = RZ² (1/n₁² − 1/n₂²), with R = 1.097 × 10⁷ m⁻¹.

de Broglie Connection

For allowed Bohr orbits, the electron’s matter wave forms a standing wave: 2πrₙ = nλ, giving λₙ = h/mvₙ = 2πrₙ/n, consistent with the angular-momentum rule.

Typical NEET Patterns

Expect 3–5 questions worth 12–20 marks: matching a series to its wavelength range, computing ionization energy or photon wavelength from a given transition, applying vₙ ∝ Z/n, and one assertion-reason on de Broglie/Bohr equivalence.

🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Edge Cases and Boundary Behaviour

For hydrogen-like ions (He⁺, Li²⁺…) the single electron sees an effective nuclear charge Z, so every energy, radius, velocity and Rydberg term picks up a factor of Z, Z² or 1/Z respectively. The ionization energy equals −E₁ = 13.6 Z² eV; for He⁺ that is 54.4 eV, not 13.6 eV. Watch the sign: rₙ scales as n²/Z (so rₙ decreases with Z), while vₙ ∝ Z/n (velocity rises with Z).

Limitations of the Bohr Model

Bohr’s picture is a semi-classical hybrid: it treats electrons as particles obeying Newtonian mechanics but imposes an ad-hoc quantum rule. It fails to explain fine structure (splitting of spectral lines), the Zeeman (magnetic) and Stark (electric) effects, multi-electron spectra, and the relative intensities of spectral lines. Wave mechanics (Schrödinger) and spin replace it, yet NEET still tests Bohr because it explains the gross structure of one-electron systems cleanly.

Common Mistakes

  • Writing L = nh instead of nh/(2π) — a frequent sign error trap.
  • Forgetting the Z² factor when shifting from H to He⁺/Li²⁺.
  • Using n in place of n² for radius scaling.
  • Mixing up transition directions: emission needs n₂ > n₁; the bracket in the Rydberg formula is positive.

Worked Example

Find the wavelength of the photon emitted when an electron in He⁺ falls from n = 3 to n = 2.

1/λ = RZ² (1/2² − 1/3²) = (1.097 × 10⁷)(4)(1/4 − 1/9) = (1.097 × 10⁷)(4)(5/36) ≈ 6.09 × 10⁶ m⁻¹ λ ≈ 164 nm — the first line of the Balmer series of He⁺, sitting in the ultraviolet.

Practice Prompts

  1. Compute the de Broglie wavelength of the electron in the ground state of hydrogen and show it equals 2πa₀.
  2. For a Li²⁺ ion, calculate the photon energy (in eV) for the transition n = 4 → n = 2, and identify the series.

Exam Strategy

Atoms consistently contributes 3–5 NEET questions (~12–20 marks). Prioritize: Rydberg numericals, energy-sign identities, spectral-series matching, de Broglie/Bohr equivalence, and hydrogen-like ion scaling.

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