Atoms

Atoms

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The Bohr model of the hydrogen atom quantises electron orbits around the nucleus using the principal quantum number n = 1, 2, 3, …. Allowed orbit radii scale as r_n = n² · a₀, where a₀ = 5.29 × 10⁻¹¹ m is the Bohr radius. Total energy falls as E_n = −13.6 / n² eV, so the ground state (n = 1) holds −13.6 eV and ionisation requires exactly 13.6 eV.

• Rydberg formula: 1/λ = R_H (1/n₁² − 1/n₂²), R_H = 1.097 × 10⁷ m⁻¹
• Angular momentum quantisation: m_e v_n r_n = n·h / 2π
• Speed in nth orbit: v_n = 2.187 × 10⁶ / n m/s

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Bohr’s Postulates and Orbit Quantisation

Rutherford’s nuclear model failed to explain atomic stability because a classical orbiting electron would radiate and spiral in within ~10⁻¹¹ s. Niels Bohr rescued stability by imposing two postulates in 1913: the electron moves in a stationary orbit without radiating, and only orbits whose angular momentum equals n·h/2π are allowed. Combining Coulomb’s law with centripetal force gives the orbit radius r_n = n²·h²ε₀ / (π m_e e²), which reduces to r_n = n²·a₀.

Energy Levels and Spectral Series

The total energy E_n = KE + PE = −13.6 / n² eV. Photons are emitted when an electron drops from a higher level n₂ to a lower level n₁; the wavelength obeys the Rydberg formula. Five named series cover different wavelength regions.

Series	Lower level n₁	Region	First line (n₂)	Wavelength
Lyman	1	Ultraviolet	n₂ = 2	121.6 nm
Balmer	2	Visible / near-UV	n₂ = 3	656.3 nm (Hα)
Paschen	3	Infrared	n₂ = 4	1875 nm
Brackett	4	Infrared	n₂ = 5	4051 nm
Pfund	5	Infrared	n₂ = 6	7458 nm

de Broglie Interpretation

Treating the electron as a standing wave around the nucleus gives 2π r_n = n·λ, which recovers Bohr’s quantisation exactly. This explains why orbits are stable: an integer number of wavelengths fits along the circumference, preventing destructive self-interference.

• Ionisation energy = 13.6 eV; first excitation = 10.2 eV (n = 1 → 2).
• Reduced mass correction: R_H uses μ; for precision, use μ = 1836 m_e / 1837.
• Limits of Bohr model: fails for multi-electron atoms, ignores spin, gives no fine structure.

🔴 Extended — Deep Study (3mo+)

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Worked Numerical Example

A hydrogen atom is excited from n = 1 to n = 3. Find (a) the photon wavelength absorbed, (b) the electron’s orbital speed in n = 3, and (c) the longest wavelength the atom can subsequently emit.

1. Absorption: ΔE = 13.6 (1 − 1/9) eV = 13.6 · 8/9 ≈ 12.09 eV, so λ = hc/ΔE = 1240 eV·nm / 12.09 eV ≈ 102.6 nm (Lyman series, UV).
2. Speed: v₃ = 2.187 × 10⁶ / 3 ≈ 7.29 × 10⁵ m/s.
3. Longest emission from n = 3 lands at n = 2 (smallest energy gap): λ = hc / [13.6(1/4 − 1/9)] = 1240 / 1.89 ≈ 656 nm (Balmer Hα, red).

Edge Cases and Exam Traps

Bohr’s model is strictly valid only for hydrogen-like (single-electron) ions such as He⁺, Li²⁺, Be³⁺; for these, replace 13.6 eV by 13.6 · Z² eV and a₀ by a₀/Z. JEE Main asks roughly one question per paper, usually a numerical on wavelength, energy, or orbit radius. JEE Advanced goes deeper into angular momentum graphs, the de Broglie wavelength in a given orbit, and identifying the series from a stated wavelength band. Beware: R_H = 1.097 × 10⁷ m⁻¹ is the experimental hydrogen value; R_∞ = 1.097 × 10⁷ × (1836/1837) m⁻¹ is the infinite-nucleus limit.

Mistake	Correct approach
Setting n = 0	n begins at 1; ground state is n = 1
Writing E = +13.6/n²	Bound states have negative energy: E_n = −13.6/n² eV
Using R_∞ for H	Use R_H = 1.097 × 10⁷ m⁻¹ for hydrogen

Connections and Practice Prompts

Atoms links forward to Nuclei (Bohr’s quantisation reappears in shell models) and to Dual Nature of Matter (de Broglie wavelength). Try these:

1. The ionisation energy of He⁺ is 54.4 eV; verify using Z² scaling and find the wavelength of the n = 2 → 1 transition.
2. Show that the de Broglie wavelength in the nth orbit equals 2π a₀ n / Z, and evaluate for hydrogen n = 2.

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