de Broglie (1924): the quantisation rule arises because matter waves form standing waves on the orbit (2πr = nλ).
Bohr's model fails for multi-electron atoms, line intensities, fine structure, and Zeeman/Stark effects. Schrödinger's wave mechanics (1926) supersedes it.
Quantity
Symbol
Equation / value
Bohr radius
a₀
5.29×10⁻¹¹ m = 0.529 Å
Radius of n-th orbit
r_n
n² a₀ / Z
Speed of n-th orbit
v_n
(αc)·Z/n where α ≈ 1/137
Energy of n-th level
E_n
−13.6 Z²/n² eV
Ionisation energy of H
I
13.6 eV
Bohr quantisation
L
nℏ = nh/(2π)
Rydberg formula
1/λ
R(1/n_f² − 1/n_i²)
Rydberg constant
R
1.097 × 10⁷ m⁻¹
de Broglie standing-wave
2πr
nλ
Distance of closest approach
r₀
(1/4πε₀)·(2Ze²)/((1/2)mv²)
Photon energy ↔ wavelength
E
1240/λ(nm) eV
NCERT Exercises — Worked Solutions
Exercise 12.1 — Choose correct alternatives
(a) The size of the atom in Thomson's model is …………… that in Rutherford's model. (b) In the ground state of …………… electrons are in stable equilibrium, while in …………… electrons always experience a net force.
(a) About the same as (both ~10⁻¹⁰ m).
(b) Thomson's model electrons are in stable equilibrium (positive jelly balances electron repulsion). In Rutherford's model electrons orbit and continuously experience the central Coulomb force.
Exercise 12.2 — Suppose you are given a chance to repeat the α-scattering experiment using a thin sheet of solid hydrogen in place of the gold foil
(Hydrogen is a solid below 14 K.) What results do you expect?
The hydrogen nucleus has charge +e and mass roughly equal to one-fourth of an α-particle's mass (α has mass ≈ 4 amu). With a much smaller charge (Z=1 vs Z=79), the Coulomb repulsion is greatly reduced, so most α-particles would barely deflect. More importantly, since the α-particle is heavier than the H-nucleus, the H-nucleus would recoil rather than back-scatter the α. There would be very few large-angle deflections — the experiment would not lead to the discovery of the nucleus, and the nuclear-size estimate would fail.
Exercise 12.3 — Three Balmer-series wavelengths
What is the shortest wavelength present in the Paschen series of spectral lines?
Paschen has n_f = 3 and n_i = 4, 5, 6, … The shortest wavelength is the series limit (n_i = ∞):
\(1/\lambda = R/9 = (1.097\times10^{7})/9 = 1.219\times10^{6}\) m⁻¹.
λ = 8.205×10⁻⁷ m = 820.5 nm (near IR).
Exercise 12.4 — Frequency from energy gap
A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom makes a transition from the upper level to the lower level?
The ground state energy of hydrogen atom is −13.6 eV. What are the kinetic and potential energies of the electron in this state?
For a Coulomb orbit: E = −K and U = −2K. So K = −E = +13.6 eV and U = 2E = −27.2 eV.
Exercise 12.6 — Lyman to Balmer
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 the photon.
ΔE = E₄ − E₁ = (−0.85) − (−13.6) = 12.75 eV = 2.04×10⁻¹⁸ J.
ν = ΔE/h = 2.04×10⁻¹⁸/6.626×10⁻³⁴ = 3.08×10¹⁵ Hz.
λ = c/ν = 3×10⁸/3.08×10¹⁵ = 97.4 nm — Lyman-γ, in the far UV.
Exercise 12.7 — Three quantities, three orbits
(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.
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?
r_n = n² a₀.
r₂ = 4 × 5.3×10⁻¹¹ = 2.12×10⁻¹⁰ m.
r₃ = 9 × 5.3×10⁻¹¹ = 4.77×10⁻¹⁰ m.
Exercise 12.9 — Hydrogen at room temperature
A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?
A 12.5 eV electron can excite hydrogen from the ground state (E₁ = −13.6 eV) to a level with energy ≤ −13.6 + 12.5 = −1.1 eV.
Possible levels: E₃ = −1.51 eV (allowed, since 12.09 eV is enough); E₄ = −0.85 eV (NOT allowed, 12.75 eV needed).
So the highest level reached is n = 3. Possible transitions: 3 → 1 (Lyman), 3 → 2 (Balmer-α), 2 → 1 (Lyman-α).
Wavelengths: 102.6 nm, 656.3 nm, 121.5 nm respectively. Lyman series (UV) and Balmer-α (visible red) will be emitted.
Exercise 12.10 — Frequency of revolution in Bohr orbit
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.)
n = 2π m v r / h = (2π × 6×10²⁴ × 3×10⁴ × 1.5×10¹¹) / (6.626×10⁻³⁴) = 2.57 × 10⁷⁴.
For such a colossal n, the levels are spaced by ΔE/E ~ 1/n² = 10⁻¹⁵⁰ — utterly indistinguishable. The orbit looks classical and continuous, exactly as we observe.
Additional 1 — Distance of closest approach
An α-particle of kinetic energy 7.7 MeV approaches a gold nucleus head-on. Calculate the distance of closest approach. (Z = 79; e = 1.6×10⁻¹⁹ C.)
At r₀: K = (2)(Ze²)/(4πε₀ r₀).
\(r_0 = (1/4\pi\varepsilon_0)\,(2Ze^2/K) = (8.99\times10^{9})(2 \times 79 \times (1.6\times10^{-19})^{2})/(7.7\times10^{6}\times1.6\times10^{-19})\)
= (8.99×10⁹)(4.05×10⁻³⁶)/(1.232×10⁻¹²) = 2.95 × 10⁻¹⁴ m.
Additional 2 — Wavelength corresponding to 13.6 eV
The ground-state binding energy of hydrogen is 13.6 eV. What wavelength of light is just sufficient to ionise a ground-state hydrogen atom?
λ = hc/E = 1240 nm·eV / 13.6 eV = 91.2 nm — the Lyman-series limit, in the far UV.
Additional 3 — He⁺ Lyman-α
Calculate the wavelength of the photon emitted when a singly-ionised helium ion (He⁺, Z = 2) makes a transition from n = 2 to n = 1.
For He⁺: 1/λ = RZ²(1/1² − 1/2²) = R × 4 × (3/4) = 3R = 3.29×10⁷ m⁻¹.
λ = 1/3.29×10⁷ = 30.4 nm. (Compare hydrogen Lyman-α at 121.6 nm — He⁺ photon is 4× more energetic.)
Activity — Build the H Atom from Scratch
Use the data below to compute every key quantity for an electron in the n = 5 Bohr orbit of hydrogen.
Q3. (Short Answer) State two limitations of Bohr's model.
(1) It works only for hydrogen-like one-electron systems; it cannot handle helium or larger atoms because of inter-electron repulsion. (2) It cannot explain the relative intensities of the spectral lines, the fine structure, or the Zeeman/Stark splitting in external fields.
Q4. (Fill in the blank) The Rydberg constant has the value ______ m⁻¹.
1.097 × 10⁷ m⁻¹.
Q5. (HOT) An electron in the n = 4 state of hydrogen jumps down to n = 1 either directly or via intermediate states. How many distinct photons are possible? List them.
From n = 4 the electron can take any cascade route. Possible transitions: 4→3, 4→2, 4→1, 3→2, 3→1, 2→1 — that's 6 distinct photons. In general, ⁿC₂ = n(n−1)/2 lines for an n-th excited state.
Assertion–Reason — Mixed Revision
Options: (A) Both true, R correct explanation of A. (B) Both true, R not the correct explanation. (C) A true, R false. (D) A false, R true.
Assertion: The energy of an electron in a Bohr orbit is the same as for the corresponding free-particle orbit at the same radius.
Reason: Bohr's quantisation does not affect the kinetic and potential energies for given r.
(A) Both correct. The orbit at any given r has K = e²/(8πε₀r) regardless of how it was selected; quantisation merely picks out which r values are allowed.
Assertion: The ionisation energy of hydrogen is exactly +13.6 eV.
Reason: The energy of the ground state is E₁ = −13.6 eV.
(A) Both correct and the reason explains the assertion. Removing the electron means moving from E₁ to E∞ = 0, requiring 13.6 eV.
Assertion: de Broglie's hypothesis successfully explains why angular momentum is quantised in Bohr's model.
Reason: The matter wave of an electron must form a standing wave on the orbit.
(A) Both correct and the reason explains the assertion: 2πr = nλ ⇔ mvr = nh/(2π).
Frequently Asked Questions - NCERT Exercises and Solutions: Atoms
What are the key NCERT exercise types in Chapter 12 Atoms?
NCERT Class 12 Physics Chapter 12 Atoms exercises cover conceptual MCQs, short numerical problems (2-3 marks), long numerical derivations (5 marks), and assertion-reason questions. The MyAiSchool solution set provides step-by-step worked solutions for every NCERT exercise question, aligned with the CBSE board exam pattern. Students should focus on dimensional analysis, formula application, and unit consistency to score full marks.
How should students approach numerical problems in Atoms?
For numerical problems in NCERT Class 12 Physics Chapter 12 Atoms: (1) list all given data with units, (2) write the relevant formula(e), (3) substitute values carefully, (4) calculate with proper significant figures, (5) state the final answer with the correct unit. Always draw a free-body or schematic diagram where relevant. The MyAiSchool solutions follow this 5-step CBSE-aligned format consistently.
What are the most-asked CBSE board questions from Chapter 12?
From NCERT Class 12 Physics Chapter 12 (Atoms), the most-asked CBSE board questions test conceptual understanding of fundamental principles, application of derived formulas to standard scenarios, and proof-style derivations. 5-mark questions usually combine derivation + application. The MyAiSchool exercise set tags each question by board frequency so students can prioritize high-yield problems before exams.
How do I check the dimensional correctness of my answer?
Dimensional correctness in NCERT Class 12 Physics is verified by ensuring the LHS and RHS of any equation have the same dimensional formula. For Chapter 12 Atoms problems, write the dimensions of each quantity, substitute, and simplify. If the dimensions match, the equation is dimensionally valid (necessary but not sufficient). The MyAiSchool solutions include dimensional checks in every numerical answer.
What are common mistakes students make in Chapter 12 exercises?
Common mistakes in NCERT Class 12 Physics Chapter 12 Atoms exercises include: (1) unit conversion errors (CGS vs SI), (2) sign convention mistakes for charges/currents/vectors, (3) forgetting to consider all field/force contributions, (4) algebra errors during derivations, (5) misreading the problem. The MyAiSchool solutions highlight these traps with red-flag annotations so students learn to avoid them.
How does the MyAiSchool solution differ from other NCERT solution sets?
MyAiSchool Class 12 Physics Chapter 12 Atoms solutions use NEP 2024-aligned pedagogy with Bloom Taxonomy tagged questions (L1 Remember to L6 Create), step-by-step working with reasoning, dimensional checks, interactive simulations, and Competency-Based Questions (CBQs) for board exam practice. Each solution is verified against NCERT and CBSE marking schemes for accuracy.
🤖
AI Tutor
Physics Class 12 Part II – NCERT (2025-26)
Ready
🤖
Hi! 👋 I'm Gaura, your AI Tutor for NCERT Exercises and Solutions: Atoms. Take your time studying the lesson — whenever you have a doubt, just ask me! I'm here to help.
Part A - Physics (Section A) · Part A - Physics (Section B) · Part B - Chemistry (Section A) · Part B - Chemistry (Section B) · Part C - Mathematics (Section A) · Part C - Mathematics (Section B)
75Questions
180mDuration
+300Max Marks
Class 11Class 12
🎯 Practise Physics
Sit a full paper on what you have been studying, marked question by question.
Part A - Physics (Section A) · Part A - Physics (Section B) · Part B - Chemistry (Section A) · Part B - Chemistry (Section B) · Part C - Mathematics (Section A) · Part C - Mathematics (Section B)