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Atoms – Class 12 Physics notes, important questions and MCQs

NCERT Class 12 Physics, chapter 12. Notes, key terms, important questions with answers and MCQs.

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Chapter overview

Rutherford's experiment showed a tiny nucleus; Bohr's model explained the hydrogen spectrum with fixed energy levels.

Rutherford and Bohr

Alpha scattering: most particles pass straight; a few deflect strongly → small, dense, positive nucleus.
Bohr's postulates: angular momentum mvr = nh/(2π); electrons do not radiate in stationary orbits.
Hydrogen: rₙ = 0.529 n² Å; Eₙ = −13.6/n² eV.

Spectral series

Emission when an electron jumps down: hν = E_i − E_f.
Lyman (to n = 1, UV), Balmer (to n = 2, visible), Paschen (to n = 3, IR).
1/λ = R(1/n_f² − 1/n_i²), R = 1.097 × 10⁷ m⁻¹.

Key terms

Ground state: lowest energy state of an atom (n = 1)
Ionisation energy: energy to remove the electron completely (13.6 eV for hydrogen)
Distance of closest approach: minimum distance an α-particle reaches from the nucleus in head-on collision, r₀ = 2kZe²/K

Formulas

Eₙ = −13.6/n² eV
rₙ = 0.529 n² Å
1/λ = R(1/n_f² − 1/n_i²)

Important questions with answers

What did Rutherford's α-scattering show about the atom?2 marks

Most of the atom is empty space; all positive charge and almost all mass are concentrated in a tiny nucleus (about 10⁻¹⁵ m), since a few α-particles bounced back.

State Bohr's postulates.3 marks

(1) Electrons revolve in certain stable orbits without radiating energy. (2) Only those orbits are allowed in which angular momentum is an integral multiple of h/2π: mvr = nh/2π. (3) An electron emits or absorbs a photon only when it jumps between orbits: hν = Eᵢ − E_f.

Using Bohr's model, derive the radius and energy of the nth orbit of hydrogen, and find the wavelength of light emitted when an electron jumps from n = 4 to n = 2.5 marks

Centripetal force from Coulomb attraction: mv²/r = ke²/r². With mvr = nh/2π, solving gives rₙ = n²h²ε₀/(πme²) = 0.529n² Å. KE = ke²/2r, PE = −ke²/r, so Eₙ = −ke²/2rₙ = −13.6/n² eV. For 4 → 2: ΔE = 13.6(1/4 − 1/16) = 13.6 × 3/16 = 2.55 eV. λ = 1240/2.55 ≈ 486 nm (blue-green Balmer line).

Quick revision

Bohr radius 0.529 Å.
Hydrogen ground state −13.6 eV.
Lyman series is ultraviolet.

Exam tips

Only the Balmer series lies in the visible region.
Energy of the photon emitted in a transition: hν = Eᵢ − E_f = 13.6(1/n_f² − 1/nᵢ²) eV for hydrogen.

Practice MCQs

Energy of hydrogen's ground state is
1. −3.4 eV
2. 13.6 eV
3. −13.6 eV
4. 0 eV
Show answer
Answer: (c) −13.6 eV. n = 1.
The visible series of hydrogen is
1. Lyman
2. Paschen
3. Balmer
4. Brackett
Show answer
Answer: (c) Balmer. Balmer.
Energy of hydrogen in n = 2 is
1. −13.6 eV
2. −1.51 eV
3. −6.8 eV
4. −3.4 eV
Show answer
Answer: (d) −3.4 eV. −13.6/4.
Radius of the 3rd Bohr orbit of hydrogen is about
1. 1.59 Å
2. 0.529 Å
3. 4.76 Å
4. 2.12 Å
Show answer
Answer: (c) 4.76 Å. 0.529 × 9.
Ionisation energy of hydrogen in the first excited state is
1. 13.6 eV
2. 10.2 eV
3. 3.4 eV
4. 1.51 eV
Show answer
Answer: (c) 3.4 eV. −E₂.
Energy emitted when an electron jumps from n = 3 to n = 2 in hydrogen is
1. 10.2 eV
2. 1.89 eV
3. 12.09 eV
4. 3.4 eV
Show answer
Answer: (b) 1.89 eV. −1.51 − (−3.4).
Rutherford's model could not explain
1. stability of the atom
2. the nucleus
3. positive charge
4. alpha scattering
Show answer
Answer: (a) stability of the atom. Orbiting electrons should radiate and fall in.
The wavelength of the first Balmer line (n = 3 → 2) of hydrogen is about
1. 122 nm
2. 486 nm
3. 656 nm
4. 1875 nm
Show answer
Answer: (c) 656 nm. H-alpha, red.

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Written by Kundali IQ from standard, well-established facts to help with revision. This is not an official NCERT publication; always read your textbook too.