Class 12 Physics NCERT Solutions

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Atoms Class 12 Physics NCERT Solutions

The complete NCERT exercise solutions for Chapter 12, Atoms — 9 questions from 12.1 to 12.9, each worked through step by step in the CBSE marking pattern. The Thomson, Rutherford and Bohr models, the hydrogen atom spectrum, energy levels and the de Broglie explanation of atomic stability.

Class:12Subject:PhysicsChapter:12

4 Key Formulas25 Practice MCQs

DWritten byDeep Narayan

UpdatedSeptember 2026

Key Concept Summary

How many questions are in NCERT Class 12 Physics Chapter 12?

Chapter 12 carries 9 exercise questions, numbered 12.1 to 12.9. All of them are solved step by step on this page, along with the chapter's key formulas and exam pointers.

Table of Contents10 sections

Revision & Practice

Chapter Overview

The chapter history goes from Thomson's plum-pudding atom through Rutherford's nuclear atom to Bohr's quantised model, in which the electron orbits at radii r_n = n²r₁ and quantised angular momentum mvr = nh/2π fixes the energies E_n = −13.6/n² eV. Spectral lines come from ΔE = hν between levels, and the ground-state energy splits into kinetic (−E) and potential (2E). Every question below is from the NCERT Class 12 textbook (rationalised edition), solved line by line in the board pattern.

Board pattern

Energy: E_n = −13.6/n² eV for hydrogen, so ΔE = hν = hc/λ for any transition. Bohr speed v_n = (2.18 × 10⁶)/n m/s, radius r_n = n²(5.3 × 10⁻¹¹) m, and period T_n = 2πr_n/v_n ∝ n³. Kinetic energy K = −E_n and potential energy U = 2E_n for a bound level. In Rutherford scattering, scattering falls sharply as the nuclear charge and the 1/r⁴ α-particle dependence reduce deflections.

$Ex 12.1 — Comparing Thomson’s and Rutherford’s models$
$Ex 12.2 — Alpha-particle experiment with a solid hydrogen target$
$Ex 12.3 — Frequency of radiation for a 2.3 eV transition$
$Ex 12.4 — Kinetic and potential energies in the hydrogen ground state$
$Ex 12.5 — Photon absorbed in the n = 1 to n = 4 transition$
$Ex 12.6 — Bohr speeds and orbital periods for n = 1, 2, 3$
$Ex 12.7 — Radii of the n = 2 and n = 3 orbits$
$Ex 12.8 — Series emitted when 12.5 eV electrons bombard hydrogen$
$Ex 12.9 — Quantum number of the Earth’s revolution$

NCERT Exercise 12.1 — Thomson's and Rutherford's Models

1Exercise question

Step-by-step solution

1(a) The atomic size is the same in both models — the correct choice is 'no different from'.
2(b) Thomson's model has a uniform positive charge, so a test electron is in stable equilibrium; in Rutherford's model the orbiting electron continuously radiates and always experiences a net (inward) force. Answer: Thomson's model / Rutherford's model.
3(c) A rapidly accelerating orbiting electron must radiate and spiral in — Rutherford's orbiting model is doomed to collapse classically.
4(d) Thomson's blob spreads mass continuously; Rutherford concentrates it in a tiny nucleus. Answer: Thomson's model / Rutherford's model.
5(e) Almost all mass sits in the positive part in 'both the models'.

Final answer

(a) no different from. (b) Thomson's model / Rutherford's model. (c) Rutherford's model. (d) Thomson's model / Rutherford's model. (e) both the models.

NCERT Exercise 12.2 — Alpha-Particle Scattering from Solid Hydrogen

1Exercise question

Step-by-step solution

1Hydrogen has Z = 1 — a single proton nucleus — against gold's Z = 79, so the Coulomb repulsion an α-particle feels is about 79 times weaker.
2Most α-particles pass through the sheet essentially undeflected.
3The few that do scatter are deflected only through small angles; large-angle (back) scattering is almost absent.
4Also, the proton is only ~1/4 the α-particle's mass, so the energy transfer in a head-on collision is limited and follows the conservation rules for unequal masses.

Final answer

A solid hydrogen target scatters α-particles far less than gold: most pass straight through, deflections are small and large-angle scattering is almost entirely absent.

NCERT Exercise 12.3 — Frequency for a 2.3 eV Transition

1Exercise question

Step-by-step solution

1Energy of the emitted photon: ΔE = hν.
2ν = ΔE/h = (2.3 × 1.6 × 10⁻¹⁹)/(6.63 × 10⁻³⁴) = 5.55 × 10¹⁴ Hz.

Final answer

ν = 5.55 × 10¹⁴ Hz.

NCERT Exercise 12.4 — Kinetic and Potential Energies of the Ground State

1Exercise question

Step-by-step solution

1For a bound Coulomb orbit, total E = −K and U = −2K, with E = K + U.
2K = −E = +13.6 eV = 2.18 × 10⁻¹⁸ J.
3U = 2E = −27.2 eV.

Final answer

K = +13.6 eV (2.18 × 10⁻¹⁸ J); U = −27.2 eV.

NCERT Exercise 12.5 — Photon Absorbed in the n = 1 to n = 4 Transition

1Exercise question

Step-by-step solution

1Level energies: E₁ = −13.6 eV, E₄ = −13.6/16 = −0.85 eV.
2Absorbed energy: ΔE = E₄ − E₁ = −0.85 − (−13.6) = 12.75 eV = 2.04 × 10⁻¹⁸ J.
3Frequency: ν = ΔE/h = (12.75 × 1.6 × 10⁻¹⁹)/(6.63 × 10⁻³⁴) = 3.08 × 10¹⁵ Hz.
4Wavelength: λ = c/ν = (3 × 10⁸)/(3.08 × 10¹⁵) = 9.75 × 10⁻⁸ m = 97.5 nm.

Final answer

ν = 3.08 × 10¹⁵ Hz; λ = 9.75 × 10⁻⁸ m (≈ 97 nm).

NCERT Exercise 12.6 — Bohr Speeds and Orbital Periods

1Exercise question

Step-by-step solution

1Bohr speed: v_n = v₁/n with v₁ = 2.18 × 10⁶ m/s.
2(a) v₁ = 2.18 × 10⁶ m/s; v₂ = 1.09 × 10⁶ m/s; v₃ = 7.27 × 10⁵ m/s.
3Orbit radii: r_n = n²r₁ with r₁ = 5.3 × 10⁻¹¹ m; period T_n = 2πr_n/v_n = n³T₁.
4T₁ = 2π(5.3 × 10⁻¹¹)/(2.18 × 10⁶) = 1.53 × 10⁻¹⁶ s.
5(b) T₁ = 1.53 × 10⁻¹⁶ s; T₂ = 8T₁ = 1.22 × 10⁻¹⁵ s; T₃ = 27T₁ = 4.12 × 10⁻¹⁵ s.

Final answer

(a) v = 2.18 × 10⁶, 1.09 × 10⁶, 7.27 × 10⁵ m/s for n = 1, 2, 3. (b) T = 1.53 × 10⁻¹⁶, 1.22 × 10⁻¹⁵, 4.12 × 10⁻¹⁵ s.

NCERT Exercise 12.7 — Radii of the n = 2 and n = 3 Orbits

1Exercise question

Step-by-step solution

1Bohr radii scale as r_n = n²r₁.
2r₂ = 4 × 5.3 × 10⁻¹¹ = 2.12 × 10⁻¹⁰ m.
3r₃ = 9 × 5.3 × 10⁻¹¹ = 4.77 × 10⁻¹⁰ m.

Final answer

r₂ = 2.12 × 10⁻¹⁰ m; r₃ = 4.77 × 10⁻¹⁰ m.

NCERT Exercise 12.8 — Series Emitted by 12.5 eV Electron Bombardment

1Exercise question

Step-by-step solution

1Excitation energies from the ground state: ΔE₁₂ = −3.4 − (−13.6) = 10.2 eV and ΔE₁₃ = −1.51 − (−13.6) = 12.09 eV.
2ΔE₁₄ = −0.85 − (−13.6) = 12.75 eV > 12.5 eV, so the beam can excite only up to n = 3.
3De-excitations from n = 3 give three lines: 3 → 2, 3 → 1, 2 → 1.
43 → 1: 1/λ = R(1 − 1/9) = (8/9)R ⇒ λ = 102.6 nm (Lyman).
52 → 1: 1/λ = R(1 − 1/4) = (3/4)R ⇒ λ = 121.6 nm (Lyman).
63 → 2: 1/λ = R(1/4 − 1/9) = (5/36)R ⇒ λ = 656.3 nm (Balmer Hα).

Final answer

Three lines are emitted: 102.6 nm and 121.6 nm (Lyman) and 656.3 nm (Balmer Hα).

NCERT Exercise 12.9 — Quantum Number of the Earth's Revolution

1Exercise question

Step-by-step solution

1Angular momentum quantisation: mvr = nh/2π.
2n = 2πmvr/h = 2π(6.0 × 10²⁴)(3 × 10⁴)(1.5 × 10¹¹)/(6.63 × 10⁻³⁴).
3mvr = 2.7 × 10⁴⁰ kg m²/s, so n = (2π × 2.7 × 10⁴⁰)/(6.63 × 10⁻³⁴) = 2.56 × 10⁷⁴.

Final answer

n = 2.56 × 10⁷⁴.

Quick Revision

Key formulas at a glance

Memorise these equations — direct application numericals and derivations in CBSE & JEE frequently hinge on these.

Bohr energy level

E_{n} = - \frac{13.6}{n ^{2}} eV

Bohr orbit radius

r_{n} = 0.529 n^{2} \overset{˚}{A}

Hydrogen spectrum

\frac{1}{λ} = R (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}})

Angular momentum in a Bohr orbit

m v r = \frac{nh}{2 π}

Exam Strategy

How this chapter is asked

High-yield question patterns observed across CBSE boards, JEE Main & Advanced, and NEET.

The energy difference between two levels gives the photon energy exactly, so every spectral question reduces to finding n₁ and n₂ for the series — Lyman to n = 1, Balmer to n = 2.
Bohr's angular momentum quantisation is what keeps the electron from radiating and spiralling into the nucleus; that is the assumption his model rests on.

FAQ

Frequently asked questions

How many questions are in NCERT Class 12 Physics Chapter 12 (Atoms)?

There are 9 exercise questions in this chapter, numbered 12.1 to 12.9. Every one is solved step by step on this page in the official NCERT numbering.

Which formulas come up in Atoms Class 12 Physics?

The formulas this chapter's questions actually turn on are: Bohr energy level, Bohr orbit radius, Hydrogen spectrum, Angular momentum in a Bohr orbit. They are listed with their expressions in the key formulas section below, and the solved questions show where each one is used.

Is Atoms important for JEE Main and NEET?

Moderate — energy-level and spectrum questions are formulaic and worth secure marks, though the chapter is conceptually lighter than the electromagnetic chapters around it.

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Previous: Dual Nature of Radiation and Matter Next: Nuclei Atoms revision notes All Class 12 Physics NCERT Solutions

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