Atoms for NEET UG
From Rutherford's gold foil to Bohr's energy levels: how we learnt what is inside an atom, and why atoms give out light only at certain colours.
PhysicsClass 1258 MCQs2 lessons11 flashcards3 memory tricks
What you'll learn
- Rutherford's nuclear model
Explain the alpha-scattering experiment and what it showed about the atom. - Bohr's model and the hydrogen spectrum
Use Bohr's formulas for radius and energy, and find the wavelength of emitted light.
Key formulas
| Distance of closest approach | r₀ = k(2e)(Ze)/K, K = alpha particle's kinetic energy |
|---|---|
| Radius of nth orbit (hydrogen) | rₙ = 0.529 n² Å |
| Energy of nth level (hydrogen) | Eₙ = −13.6/n² eV |
| For hydrogen-like ions (charge Z) | Eₙ = −13.6 Z²/n² eV |
| Rydberg formula | 1/λ = R(1/n_f² − 1/n_i²), R = 1.097 × 10⁷ m⁻¹ |
Memory tricks
Lyman (n = 1, ultraviolet), Balmer (n = 2, visible), Paschen (n = 3, infrared), Brackett (n = 4, infrared), Pfund (n = 5, infrared).
E_n = −13.6/n² eV: −13.6, −3.4, −1.51 eV for n = 1, 2, 3. The ionisation energy from the ground state is 13.6 eV.
Most alpha particles passed straight through, a few were deflected, and about 1 in 8000 bounced back. The atom is mostly empty space, with a tiny, dense, positive nucleus.
Try 3 NEET-style questions
Q1. The shortest wavelength in the Lyman series of hydrogen is about (hc ≈ 1240 eV nm)
- 122 nm
- 656 nm
- 91.2 nm
- 365 nm
Show answer
Answer: (3) 91.2 nm
The series limit corresponds to n = ∞ → 1, energy 13.6 eV. λ = 1240/13.6 ≈ 91.2 nm.
Q2. The energy needed to excite a hydrogen atom from its ground state to the n = 3 state is about
- 1.51 eV
- 10.2 eV
- 13.6 eV
- 12.09 eV
Show answer
Answer: (4) 12.09 eV
ΔE = E₃ − E₁ = (−1.51) − (−13.6) = 12.09 eV.
Q3. Considering all possible transitions, the number of spectral lines emitted when an electron falls from n = 4 to the ground state is
- 3
- 4
- 6
- 10
Show answer
Answer: (3) 6
The number of lines is n(n − 1)/2 = 4 × 3 / 2 = 6 when all transitions are counted.
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