NCERT Solutions for Class 12 Physics Chapter 12: Atoms

These Class 12 Physics Chapter 12 solutions cover Atoms from the NCERT textbook (session 2026–27). Every NCERT Exercise question (12.1–12.9) is reproduced verbatim and solved step by step, with all numericals worked out and verified with correct units. The chapter develops the story of atomic structure from Thomson’s and Rutherford’s models to the Bohr model of the hydrogen atom, the line spectrum of hydrogen, and de Broglie’s explanation of quantisation.

Class: 12 Subject: Physics Chapter: 12 Topic: Atoms Exercises: 12.1 – 12.9 Session: 2026–27
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Class 12 Physics Chapter 12 Atoms – Overview

Chapter 12, Atoms, traces how physicists discovered the internal structure of the atom. Thomson’s plum-pudding model pictured positive charge spread uniformly with electrons embedded in it. The Geiger–Marsden alpha-particle scattering experiment overturned this: most alpha-particles passed straight through gold foil, but a tiny fraction were deflected through large angles, proving that an atom’s positive charge and almost all its mass sit in a tiny, dense nucleus (about 10−100 thousand times smaller than the atom). This led to Rutherford’s nuclear (planetary) model. Because an accelerating orbiting electron must radiate and spiral inward, that classical model could not explain atomic stability or the sharp line spectrum of hydrogen. Niels Bohr resolved this with three postulates — stable non-radiating orbits, quantised angular momentum (L = nh/2π), and photon emission of energy hν = Ei − Ef — giving energy levels En = −13.6/n2 eV. Finally, de Broglie explained the quantisation condition as the orbit holding a whole number of electron matter-waves.

Key Concepts

Alpha-particle scattering: 5.5 MeV alpha-particles fired at a thin gold foil; about 1 in 8000 deflect by more than 90°, showing a small massive positive nucleus.

Distance of closest approach: the centre-to-centre distance at which an alpha-particle’s kinetic energy is fully converted to electric potential energy and it momentarily stops; it sets an upper limit on nuclear size.

Impact parameter (b): perpendicular distance of the alpha-particle’s initial velocity from the nucleus; small b gives large scattering, large b gives small deflection.

Bohr’s postulates: (i) electrons revolve in stable non-radiating stationary orbits; (ii) angular momentum is quantised, L = nh/2π; (iii) a photon of energy hν = Ei − Ef is emitted/absorbed during a transition.

Energy levels: En = −13.6/n2 eV. Ground state (n = 1) is −13.6 eV; ionisation energy of hydrogen is 13.6 eV.

Spectral series: Lyman (transitions to n = 1, ultraviolet), Balmer (to n = 2, visible), Paschen/Brackett/Pfund (to n = 3, 4, 5, infrared).

de Broglie explanation: a stable orbit holds an integral number of electron wavelengths, 2πrn = nλ, which reproduces Bohr’s quantisation L = nh/2π.

Important Formulas

Distance of closest approach: d = 2Ze2 / (4πε0K), where K is the kinetic energy of the alpha-particle.

Quantisation of angular momentum: L = m vn rn = nh/2π

Orbit radius: rn = n2(h/2π)2(4πε0)/(me2) = n2 a0, with Bohr radius a0 = 0.53 × 10−10 m

Orbital speed: vn = e2/(2ε0nh) = (2.18 × 106)/n m/s

Energy of nth level: En = −me4/(8n2ε02h2) = −13.6/n2 eV; with Kn = −En and Un = 2En

Photon emitted: hν = Ei − Ef; wave number 1/λ = R(1/nf2 − 1/ni2), Rydberg constant R = 1.097 × 107 m−1

Useful values: 1/4πε0 = 9.0 × 109 N m2/C2; e = 1.6 × 10−19 C; h = 6.63 × 10−34 J s; me = 9.1 × 10−31 kg; 1 eV = 1.6 × 10−19 J.

NCERT Solutions – Exercises (12.1 to 12.9)

12.1 Choose the correct alternative from the clues given at the end of the each statement: (a) The size of the atom in Thomson’s model is ………. the atomic size in Rutherford’s model. (much greater than/no different from/much less than.) (b) In the ground state of ………. electrons are in stable equilibrium, while in ………. electrons always experience a net force. (Thomson’s model/ Rutherford’s model.) (c) A classical atom based on ………. is doomed to collapse. (Thomson’s model/ Rutherford’s model.) (d) An atom has a nearly continuous mass distribution in a ………. but has a highly non-uniform mass distribution in ………. (Thomson’s model/ Rutherford’s model.) (e) The positively charged part of the atom possesses most of the mass in ………. (Rutherford’s model/both the models.)

ANSWER (a) No different from. Both models describe an atom of the same order of size (about 10−10 m); they differ only in how charge and mass are arranged inside it. (b) In the ground state of Thomson’s model electrons are in stable equilibrium, while in Rutherford’s model electrons always experience a net force. In Thomson’s uniform charge cloud an electron can sit at a point of zero net force; in Rutherford’s model the orbiting electron always feels the centripetal Coulomb pull of the nucleus. (c) A classical atom based on Rutherford’s model is doomed to collapse, because the accelerating orbiting electron continuously radiates energy and spirals into the nucleus. (d) An atom has a nearly continuous mass distribution in Thomson’s model but a highly non-uniform mass distribution in Rutherford’s model (mass concentrated in the tiny nucleus). (e) The positively charged part possesses most of the mass in both the models — in Thomson’s the positive cloud carries the mass, and in Rutherford’s the positively charged nucleus does.

12.2 Suppose you are given a chance to repeat the alpha-particle scattering experiment using a thin sheet of solid hydrogen in place of the gold foil. (Hydrogen is a solid at temperatures below 14 K.) What results do you expect?

ANSWER The nucleus of a hydrogen atom is a single proton of mass about 1.67 × 10−27 kg, whereas an alpha-particle has mass about 6.64 × 10−27 kg — roughly four times heavier. Because the incident alpha-particle is far more massive than the target proton, it cannot be bounced back even in a head-on collision. (It is like a heavy ball striking a much lighter one at rest: the heavy ball barely changes direction.) Therefore there would be essentially no large-angle scattering. The alpha-particles would pass through almost undeviated, so this experiment would not reveal the nuclear structure as the gold-foil experiment did.

12.3 A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom make a transition from the upper level to the lower level?

ANSWER Given: energy difference E = 2.3 eV = 2.3 × 1.6 × 10−19 J = 3.68 × 10−19 J; h = 6.63 × 10−34 J s. Using E = hν  →  ν = E/h. ν = (3.68 × 10−19 J) / (6.63 × 10−34 J s) ν ≈ 5.6 × 1014 Hz.

12.4 The ground state energy of hydrogen atom is −13.6 eV. What are the kinetic and potential energies of the electron in this state?

ANSWER In the Bohr model the total energy E = −13.6 eV is related to kinetic energy K and potential energy U by K = −E and U = 2E (i.e. U = −2K). Kinetic energy: K = −E = −(−13.6 eV) = +13.6 eV. Potential energy: U = 2E = 2 × (−13.6 eV) = −27.2 eV. Check: K + U = 13.6 + (−27.2) = −13.6 eV = E. ✓

12.5 A hydrogen atom initially in the ground level absorbs a photon, which excites it to the n = 4 level. Determine the wavelength and frequency of photon.

ANSWER Energy absorbed: E = E4 − E1 = (−13.6/42) − (−13.6/12) = (−0.85) − (−13.6) = 12.75 eV. E = 12.75 × 1.6 × 10−19 J = 2.04 × 10−18 J. Wavelength: λ = hc/E = (6.63 × 10−34 × 3 × 108) / (2.04 × 10−18) λ ≈ 9.7 × 10−8 m (≈ 97 nm, in the ultraviolet). Frequency: ν = c/λ = (3 × 108) / (9.7 × 10−8) ≈ 3.1 × 1015 Hz.

12.6 (a) Using the Bohr’s model calculate the speed of the electron in a hydrogen atom in the n = 1, 2, and 3 levels. (b) Calculate the orbital period in each of these levels.

ANSWER (a) The speed in the nth orbit is vn = e2/(2ε0nh) = (2.18 × 106)/n m/s. n = 1: v1 = 2.18 × 106 m/s. n = 2: v2 = 2.18 × 106 / 2 = 1.09 × 106 m/s. n = 3: v3 = 2.18 × 106 / 3 = 7.27 × 105 m/s. (b) Orbital period Tn = 2πrn/vn, with rn = n2 a0 and a0 = 5.3 × 10−11 m. n = 1: r1 = 5.3 × 10−11 m; T1 = 2π(5.3 × 10−11)/(2.18 × 106) = 1.52 × 10−16 s. n = 2: r2 = 4 × 5.3 × 10−11 = 2.12 × 10−10 m; T2 = 2π(2.12 × 10−10)/(1.09 × 106) = 1.22 × 10−15 s. n = 3: r3 = 9 × 5.3 × 10−11 = 4.77 × 10−10 m; T3 = 2π(4.77 × 10−10)/(7.27 × 105) = 4.11 × 10−15 s. Note: the period grows as Tn ∝ n3, so T2 = 8T1 and T3 = 27T1, confirming the values above.

12.7 The radius of the innermost electron orbit of a hydrogen atom is 5.3×10−11 m. What are the radii of the n = 2 and n =3 orbits?

ANSWER In the Bohr model rn = n2 r1, where r1 = 5.3 × 10−11 m. n = 2: r2 = 22 × 5.3 × 10−11 = 4 × 5.3 × 10−11 = 2.12 × 10−10 m. n = 3: r3 = 32 × 5.3 × 10−11 = 9 × 5.3 × 10−11 = 4.77 × 10−10 m.

12.8 A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?

ANSWER Energy levels: E1 = −13.6 eV, E2 = −3.40 eV, E3 = −1.51 eV, E4 = −0.85 eV. A ground-state atom can be excited to n = 3 because E3 − E1 = 12.09 eV < 12.5 eV, but not to n = 4 (E4 − E1 = 12.75 eV > 12.5 eV). So the highest level reached is n = 3. From n = 3 the atom can de-excite by the transitions 3→1, 3→2 and 2→1, giving these emitted wavelengths (using 1/λ = R[1/nf2 − 1/ni2], R = 1.097 × 107 m−1): Lyman series (to n = 1): 3→1 gives λ ≈ 103 nm; 2→1 gives λ ≈ 122 nm (ultraviolet). Balmer series (to n = 2): 3→2 gives λ ≈ 656 nm (visible, red). Conclusion: the Lyman series (103 nm and 122 nm) and the first Balmer line (656 nm) will be emitted.

12.9 In accordance with the Bohr’s model, find the quantum number that characterises the earth’s revolution around the sun in an orbit of radius 1.5 × 1011 m with orbital speed 3 × 104 m/s. (Mass of earth = 6.0 × 1024 kg.)

ANSWER Bohr’s quantisation condition: m v r = nh/2π  →  n = 2π m v r / h. n = [2π × (6.0 × 1024 kg) × (3 × 104 m/s) × (1.5 × 1011 m)] / (6.63 × 10−34 J s). Numerator = 2π × 6.0 × 3 × 1.5 × 1024+4+11 = 2π × 27 × 1039 = 1.696 × 1041 kg m2/s. n = (1.696 × 1041) / (6.63 × 10−34) ≈ 2.6 × 1074. Significance: the quantum number is enormous, so the spacing between adjacent levels is utterly negligible and the earth’s orbit appears perfectly continuous — quantisation effects are unobservable for macroscopic bodies (the correspondence principle).

Extra Practice Questions

Short Answer Type Questions

Q1. State the two main drawbacks of Rutherford’s nuclear model of the atom.

ANSWER(i) An orbiting electron is accelerated and must radiate energy continuously, so it should spiral into the nucleus — the atom would be unstable. (ii) As it spirals in, the frequency of revolution changes continuously, so the atom should emit a continuous spectrum, contradicting the observed sharp line spectrum.

Q2. Why is the total energy of an electron in a hydrogen atom negative?

ANSWERThe negative total energy means the electron is bound to the nucleus. Energy must be supplied (positive work) to remove it to infinity, where the energy is taken as zero. If the energy were positive, the electron would be free and would not follow a closed orbit.

Q3. Define the impact parameter and state how it affects the scattering angle.

ANSWERThe impact parameter is the perpendicular distance of the initial velocity vector of the alpha-particle from the centre of the nucleus. A small impact parameter (near head-on) produces a large scattering angle; a large impact parameter produces a small deflection.

Q4. Calculate the ionisation energy of a hydrogen atom in the first excited state (n = 2).

ANSWERIonisation energy = 0 − E2 = 0 − (−13.6/22) eV = 0 − (−3.4 eV) = 3.4 eV. Less energy is needed because the electron is already loosely bound in the excited state.

Q5. Which transition in the hydrogen spectrum gives the longest wavelength line of the Lyman series, and why?

ANSWERThe Lyman series corresponds to transitions ending at n = 1. The longest wavelength (smallest energy gap) is the 2 → 1 transition, since it has the smallest energy difference among Lyman lines; smaller energy means longer wavelength (λ = hc/E).

Long Answer Type Questions

Q1. Describe the Geiger–Marsden alpha-particle scattering experiment and explain how its results led to the nuclear model of the atom.

ANSWERA narrow, collimated beam of 5.5 MeV alpha-particles from a radioactive source was directed at a thin gold foil (about 2.1 × 10−7 m thick) inside a vacuum chamber. A rotatable detector with a zinc-sulphide screen recorded scintillations to count alpha-particles scattered at different angles. Observations: most particles passed straight through with little or no deflection; only about 0.14% scattered by more than 1°, and roughly 1 in 8000 was deflected by more than 90° (some almost straight back). Conclusions: the atom is mostly empty space (hence the undeviated majority); to turn an alpha-particle back, a large repulsive force is needed, which means the positive charge and almost all the mass are concentrated in a tiny central nucleus (size 10−15–10−14 m, about 10,000–100,000 times smaller than the atom). This gave Rutherford’s nuclear (planetary) model, with electrons revolving around the nucleus.

Q2. State Bohr’s postulates and use them to obtain the expression for the energy of the electron in the nth orbit of a hydrogen atom.

ANSWERPostulates: (i) the electron revolves in certain stable, non-radiating stationary orbits; (ii) angular momentum is quantised, L = m v r = nh/2π; (iii) a photon of energy hν = Ei − Ef is emitted or absorbed during a transition between orbits. Derivation: for the orbiting electron the Coulomb force supplies the centripetal force: e2/(4πε0r2) = m v2/r. Combining this with the quantisation condition v = nh/(2πmr) gives the radius rn = n2ε0h2/(πme2). The total energy E = K + U = (½)mv2 − e2/(4πε0r) = −e2/(8πε0r). Substituting rn gives En = −me4/(8n2ε02h2) = −13.6/n2 eV.

Q3. How does de Broglie’s hypothesis explain Bohr’s second postulate (quantisation of angular momentum)?

ANSWERde Broglie proposed that a moving electron behaves as a matter-wave of wavelength λ = h/(mv). For an electron in a circular orbit to persist, it must form a standing wave; otherwise successive waves interfere destructively and die out. A standing wave on the orbit requires the circumference to contain a whole number of wavelengths: 2πrn = nλ = n·h/(mvn). Rearranging gives m vn rn = nh/2π, which is exactly Bohr’s quantisation condition for angular momentum. Thus the quantised orbits and energy levels arise naturally from the wave nature of the electron, and only resonant standing waves survive.

MCQs & Assertion–Reason

1. In the alpha-particle scattering experiment, the large-angle deflection of a few particles shows that the atom has a:

(a) uniform charge distribution    (b) small dense positive nucleus    (c) large negative core    (d) neutral centre

2. The ground-state energy of a hydrogen atom is:

(a) +13.6 eV    (b) −3.4 eV    (c) −13.6 eV    (d) 0 eV

3. According to Bohr’s model, the angular momentum of an electron in the nth orbit equals:

(a) nh    (b) nh/2π    (c) 2πnh    (d) h/2πn

4. The radius of the nth Bohr orbit is proportional to:

(a) n    (b) 1/n    (c) n2    (d) 1/n2

5. The Balmer series of hydrogen lies in the:

(a) ultraviolet region    (b) visible region    (c) infrared region    (d) X-ray region

6. The kinetic energy of the electron in the ground state of hydrogen is:

(a) −13.6 eV    (b) +13.6 eV    (c) −27.2 eV    (d) +27.2 eV

7. The speed of the electron in the nth Bohr orbit varies as:

(a) n    (b) n2    (c) 1/n    (d) 1/n2

8. The ionisation energy of a hydrogen atom in its ground state is:

(a) 3.4 eV    (b) 10.2 eV    (c) 12.09 eV    (d) 13.6 eV

9. de Broglie’s condition for a stable Bohr orbit is:

(a) 2πr = nλ    (b) πr = nλ    (c) r = nλ    (d) 2πr = λ/n

10. When a hydrogen atom is excited from n = 1 to n = 4, the energy absorbed is:

(a) 10.2 eV    (b) 12.09 eV    (c) 12.75 eV    (d) 13.6 eV

Answer key: 1-(b), 2-(c), 3-(b), 4-(c), 5-(b), 6-(b), 7-(c), 8-(d), 9-(a), 10-(c).

For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.

A-R 1. Assertion: Most alpha-particles pass straight through a thin gold foil.

Reason: An atom is largely empty space, with its mass and positive charge concentrated in a tiny nucleus.

A-R 2. Assertion: The total energy of an electron in a hydrogen atom is negative.

Reason: The electron is bound to the nucleus, so energy must be supplied to free it.

A-R 3. Assertion: Rutherford’s classical atom cannot be stable.

Reason: An accelerating orbiting electron radiates energy and spirals into the nucleus.

A-R 4. Assertion: Bohr’s model can fully explain the spectrum of the helium atom.

Reason: Bohr’s model applies only to hydrogenic (single-electron) atoms.

A-R 5. Assertion: Quantisation effects are not observed in the earth’s motion around the sun.

Reason: The quantum number for the earth’s orbit is extremely large, making energy-level spacing negligible.

Answer key: 1-(A), 2-(A), 3-(A), 4-(D), 5-(A).

Common Mistakes to Avoid

Watch out for these

  • Forgetting to convert eV to joules (multiply by 1.6 × 10−19) before using E = hν or λ = hc/E.
  • Mixing up the sign relations: kinetic energy K = −E (positive), potential energy U = 2E (negative), total energy E = −13.6/n2 eV.
  • Using rn ∝ n instead of rn ∝ n2, or vn ∝ n instead of vn ∝ 1/n.
  • Assuming a 12.5 eV beam can reach n = 4; it can only reach n = 3 (12.09 eV < 12.5 eV < 12.75 eV).
  • Thinking the frequency of revolution equals the frequency of the emitted spectral line — the line frequency is (Ei − Ef)/h.
  • Applying Bohr’s model to multi-electron atoms; it works only for hydrogenic (single-electron) systems.

Exam tips for Atoms

Memorise the four ready-to-use Bohr results — En = −13.6/n2 eV, rn = 0.53n2 Å, vn = 2.18 × 106/n m/s and Tn ∝ n3 — so numericals can be solved in one step. For spectral-line problems use 1/λ = R(1/nf2 − 1/ni2) and remember the series: Lyman (nf = 1, UV), Balmer (nf = 2, visible), Paschen (nf = 3, IR). Always write the formula, substitute with units, and state the final answer with the correct power of ten. The two drawbacks of Rutherford’s model and the three Bohr postulates are frequent 2–3 mark theory questions.

Frequently Asked Questions

What is Class 12 Physics Chapter 12 Atoms about?

Chapter 12, Atoms, explains the structure of the atom — Thomson’s and Rutherford’s models, the alpha-particle scattering experiment, the Bohr model of the hydrogen atom, energy levels and the line spectrum of hydrogen, and de Broglie’s explanation of Bohr’s quantisation condition.

How many exercises are there in Class 12 Physics Chapter 12?

The NCERT textbook has nine exercise questions, numbered 12.1 to 12.9. All of them are reproduced verbatim and solved step by step on this page, with numericals verified with units.

What is the ground state energy of a hydrogen atom?

The ground state (n = 1) energy is −13.6 eV. In this state the kinetic energy of the electron is +13.6 eV and the potential energy is −27.2 eV, and 13.6 eV is the ionisation energy of hydrogen.

Are these Class 12 Physics Chapter 12 solutions free?

Yes. All solutions are free and follow the official NCERT Physics textbook for session 2026–27.

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