L-24: Structure of Atoms
Physics — Class 12 · NIOS Code 312 · Module 7 · Source: 312_Physics_Eng_Lesson24.pdf
Structure of Atoms
Atomic structure explains why materials differ in their properties. This lesson traces the evolution from Dalton's indivisible atom through Thomson's plum-pudding model, Rutherford's nuclear model, Bohr's quantised hydrogen atom, and the production of X-rays. Module 7: Atoms and Nuclei.
NIOS objectives: Rutherford experiment and model; Bohr radius, velocity, energy; hydrogen spectrum; X-rays, Moseley's and Duane–Hunt laws.
Historical Background
- Dalton (1808): atom as smallest indivisible particle of an element.
- J.J. Thomson (1897): discovery of electron → atom has structure.
- Thomson model: positive sphere with embedded electrons ("plum pudding") — could not explain Rutherford's results.
24.1 Rutherford's α-Particle Scattering
Geiger and Marsden (1911): fine beam of α-particles from source S on thin gold foil T; flashes on rotating ZnS screen + microscope M; evacuated chamber.
Observations: most α-particles suffered small deflections; a few at 90° or more; ~1 in 8000 rebounded at 180°. Thomson model predicted only minor deflections.
24.1.1 Nuclear Model
- Almost entire positive charge and mass confined in tiny nucleus (~10⁻¹⁵ m, ~10⁻⁵ times atom size).
- Electrons revolve around nucleus; atom electrically neutral.
Shortcomings:
- Stability: accelerating electrons should radiate EM waves and spiral into nucleus (Fig 24.5).
- Spectrum: classical model predicts continuous spectrum; atoms emit line spectra at discrete frequencies.
24.2 Bohr's Model of Hydrogen Atom
Bohr (1913) combined classical and quantum ideas to explain hydrogen stability and spectrum.
Four postulates:
- (i) Classical: electron in circular orbit; centripetal force = Coulomb attraction: mv²/r = Ze²/(4πε₀r²).
- (ii) Quantum: only orbits with quantised angular momentum L = mvr = nh/2π (n = 1, 2, 3…).
- (iii) Quantum: no radiation in allowed stationary orbits; energy constant.
- (iv) Quantum: photon emitted/absorbed only on transitions between allowed levels: ΔE = hν.
Radii ratio 1 : 4 : 9 : 16 … for n = 1, 2, 3, 4
Spacing between orbits increases with n
Decreases as n increases
Negative E means electron bound to nucleus
E₁ = −13.6 eV; E∞ = 0 (ionised)
ΔE = En − Em = hν
Rydberg constant R ≈ 1.097 × 10⁷ m⁻¹ (in wavelength form)
24.3 Hydrogen Spectrum
Each series corresponds to electron jumping to a fixed lower orbit m from higher orbits n:
- Lyman (UV): m = 1, n = 2, 3, 4…
- Balmer (visible): m = 2, n = 3, 4, 5… (e.g. Hα: 6563 Å for 3→2)
- Paschen (near IR): m = 3, n = 4, 5, 6…
- Brackett (mid IR): m = 4, n = 5, 6…
- Pfund (far IR): m = 5, n = 6, 7…
Number of spectral lines when electron drops from nth level: ½n(n−1). Bohr predicted series later discovered experimentally.
Fraunhofer lines: dark lines in solar spectrum — cooler chromosphere absorbs specific wavelengths of continuous solar radiation (Kirchhoff's laws).
24.4 X-Rays
Produced when fast electrons from hot cathode strike a heavy-metal target (high Z, high melting point) in an evacuated tube at very low pressure. ~5% energy → X-rays; rest → heat (cooled by circulating water). Intensity ∝ filament current; quality (penetration) ∝ accelerating voltage (10 kV–1 MV).
Properties: affect photographic plate; cause fluorescence; ionise gases; no ordinary reflection/refraction/diffraction (unless crystal techniques); not deflected by E or B fields.
Types of X-Rays
1. Continuous X-rays: all wavelengths down to sharp cutoff λmin; intensity increases with voltage; max emission shifts to shorter λ at higher V. Produced by multiple collisions — photons of all frequencies.
λmin decreases as accelerating voltage V increases
2. Characteristic X-rays: sharp lines superposed on continuous spectrum; positions depend only on target element (not voltage). Explained by electron transitions in target atoms.
Confirmed atomic number Z as fundamental property
R = Rydberg constant
STRUCTURE OF ATOM — KEY POINTS
=================================
Thomson model : positive sphere + electrons (failed)
Rutherford : small nucleus; α large-angle scatter
Rutherford fail : stability + continuous spectrum
Bohr quantise : L = nh/2π ; stationary orbits
Bohr radius : a₀ = 0.53 Å ; rₙ = n²a₀
Energy (H) : Eₙ = −13.6/n² eV
Ionisation : 13.6 eV (n=1 → ∞)
Series : Lyman, Balmer, Paschen, Brackett, Pfund
X-rays : electrons stopped by heavy target
Duane–Hunt : eV = hc/λ_min
Moseley : ν ∝ (Z−1)² for characteristic lines
Quick Revision
- α-particles ~7000× heavier than electrons — need strong nucleus repulsion for large-angle scatter.
- Only first Bohr postulate is classical; others are quantum.
- En negative → bound electron; E∞ = 0 → free electron.
- Balmer series visible; Lyman UV; rest mostly IR.
- Characteristic X-ray lines identify elements (Moseley); continuous spectrum cutoff set by Duane–Hunt.
Q1. In Rutherford's experiment, the target was bombarded with:
Q2. Large-angle scattering of α-particles indicated:
Q3. Bohr quantised which quantity of the revolving electron?
Q4. The radius of the nth Bohr orbit of hydrogen is proportional to:
Q5. The ground-state energy of hydrogen is:
Q6. The Balmer series of hydrogen lies mainly in the:
Q7. Transitions to n = 4 in hydrogen give which series?
Q8. Duane–Hunt law relates accelerating voltage to:
Q9. Moseley's law shows that characteristic X-ray frequency depends on:
Q10. For hydrogen, ionisation energy (n = 1 → ∞) is:
PYQ — Previous Year Questions
Extracted from NIOS Physics (312) board exam papers in your PDF. Chapter L24 — Structure of Atoms only. Use Model Answer for marking points; Explanation for concept clarity.
Chapter 24 — Structure of Atoms (L24)
12 questions · Sections A & B · Sources: 312/TUS/104A, 312/MAY/204A, 68/ESS/1-312-A
Section A — Objective (1 mark)
PYQ1. In an experiment of scattering of alpha particles, it was shown for the first time that the atom has: (A) electron (B) proton (C) neutron (D) nucleus
Model Answer
(D) nucleus
Rutherford's α-scattering (1911): most α-particles passed through; rare large-angle rebounds implied a tiny, dense, positively charged nuclear core.
Explanation
Geiger–Marsden experiment on gold foil established the nuclear model (L24 §24.1). Electron was known earlier (Thomson); neutron discovered later (1932).
PYQ2. According to Bohr's postulates, electrons revolve around the nucleus in __________ orbits. (A) dynamic (B) stationary (C) lower (D) first
Model Answer
(B) stationary
Allowed orbits are stationary states — no radiation while electron remains in a quantised orbit.
Explanation
Bohr postulate (iii): electron in permitted orbit does not radiate; energy constant until it jumps between levels (L24 §24.2).
PYQ3. Which spectral series of hydrogen lies in the UV region? (A) Paschen (B) Lyman (C) Brackett (D) Balmer
Model Answer
(B) Lyman series — transitions to n = 1 (UV).
Explanation
Lyman: UV; Balmer: visible; Paschen/Brackett/Pfund: infrared (L24 §24.2, hydrogen spectrum).
PYQ4. Hydrogen atoms are excited from ground state to a state with quantum number 4. The maximum number of spectral lines emitted will be: (A) 2 (B) 3 (C) 5 (D) 6
Model Answer
Number of lines = n(n−1)/2 = 4×3/2 = 6 → (D)
Explanation
From n = 4, electron can cascade to n = 3, 2, 1 giving all pairwise transitions (L24 §24.2).
PYQ5. In terms of Bohr radius an, the radius of the third orbit of hydrogen atom will be: (A) 3an (B) 9an (C) 3an (D) an/3
Model Answer
rn = n²a₀; for n = 3: r₃ = 9a₀ → (B)
Explanation
Bohr radii scale as n² (1 : 4 : 9 …). a₀ ≈ 0.53 Å is the Bohr radius (L24 §24.2).
PYQ6. The ionization energy of hydrogen atom is 13.6 eV. The ionization energy of helium atom is: (A) 54.4 eV (B) 27.2 eV (C) 13.6 eV (D) 6.8 eV
Model Answer
E ∝ Z²/n²; for He⁺ (Z = 2, n = 1): E = 13.6 × 4 = 54.4 eV → (A)
Explanation
Helium atom (neutral) has two electrons — board MCQ uses Bohr formula for hydrogen-like He⁺ ion with Z = 2.
PYQ7. The maximum number of electrons present in the outermost orbit of any atom is: (A) 2 (B) 8 (C) 32 (D) infinite
Model Answer
(B) 8 (octet rule for main-group elements; Bohr–Bury scheme).
Explanation
Outermost shell holds at most 8 electrons (2 in K, 8 in higher shells for typical atoms) — linked to atomic structure and stability (L24).
Section A — Short Answer (2 marks)
PYQ8. Match Column—I with Column—II: (a) Balmer series — (i) ν = R(1 − 1/n²) (ii) ν = R(1/4 − 1/n²) (iii) ν = R(1/9 − 1/n²) (iv) ν = R(1/16 − 1/n²); (b) Paschen series — same options.
Model Answer
(a) Balmer → (ii) ν = R(1/4 − 1/n²), final state n = 2
(b) Paschen → (iii) ν = R(1/9 − 1/n²), final state n = 3
(i) Lyman n = 1; (iv) Brackett n = 4
Explanation
Rydberg-type formula 1/λ = R(1/n₁² − 1/n₂²) with n₁ = lower (final) level (L24 §24.2).
PYQ9. Match Column—I with Column—II: (i) Visible spectrum — (a) Lyman (b) Balmer (c) Paschen (d) Pfund; (ii) Far infrared spectrum — same options.
Model Answer
(i) Visible spectrum → (b) Balmer series
(ii) Far infrared spectrum → (d) Pfund series (transitions to n = 5)
Explanation
Balmer (n = 2) falls in visible; Pfund (n = 5) in far IR. Paschen/Brackett are nearer infrared (L24 §24.2).
PYQ10. Complete the sentences (options: ultraviolet rays, gamma rays, heat waves, radio waves): (a) The electromagnetic radiations which have wavelengths shorter than X-rays are _____. (b) Infrared rays are also called as _____.
Model Answer
(a) gamma rays — shorter λ, higher frequency than X-rays
(b) heat waves — infrared radiation carries thermal energy
Explanation
EM spectrum order: … X-rays → gamma (shortest). X-rays produced in atomic transitions / target bombardment (L24 §24.3).
PYQ11. Write TRUE for correct statement and FALSE for incorrect: Paschen series of hydrogen atom lies in the UV region.
Model Answer
FALSE
Paschen series (transitions to n = 3) lies in the infrared region, not UV.
Explanation
UV = Lyman (n = 1). Paschen is IR — board T/F from ESS Q27 part (iii) (L24 §24.2).
Section B — Short Answer (2 marks)
PYQ12. Out of X-rays and microwaves, which radiation is more likely to produce photo-emission from a given material? Explain.
Model Answer
X-rays are more likely.
X-rays have much higher frequency/energy (hν) than microwaves → can exceed work function of many materials → eject photoelectrons. Microwaves are low-energy — cannot cause photoemission from metals.
Explanation
X-rays sit high on EM spectrum; produced when fast electrons strike a target (L24 §24.3). Photoelectric effect needs ν above threshold.
Problem Solving — L24 Structure of Atoms
Six problems spanning this chapter’s NIOS syllabus. Every question is built from the notes and formula sheet: solve with equations first, then read the formal textbook-style write-up, the easy explanation, and the topic in depth (formulas, meaning, exam tips). Explanations open by default.
State one limitation of Thomson’s model that Rutherford’s α-scattering addressed.
Solution — step by step with formulas
- Thomson could not explain large-angle α scattering; Rutherford introduced nuclear atom.
Final answer: Nuclear atom from α large-angle scattering
Textbook formal language
Rutherford scattering shows concentrated positive charge and mass in a nucleus.
Working formula set for this problem: (see solution steps). In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.
Easy language (same idea, plain words)
Most α particles pass; few bounce—so atom is mostly empty with a tiny hard core.
Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.
Topic in depth — Atomic models
Rutherford model failed to explain spectral stability (accelerating electrons radiate).
Link to chapter notes (L24 — Atomic models): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: (see solution steps). In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.
Exam tip
Quote the law in one line, then write (see solution steps) before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.
Common mistakes
- Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
- Using the wrong sign convention for work/heat/force direction.
- Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
- Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
Write Bohr’s angular momentum quantisation and ground-state energy of H atom.
Solution — step by step with formulas
- mvr = nh/(2π).
- E₁ = −13.6 eV.
Final answer: L = nh/2π; E₁ = −13.6 eV
Formulas used in this problem
Textbook formal language
Bohr quantised angular momentum and assumed non-radiating stationary orbits.
Working formula set for this problem: mvr = nh/2π; E_n = −13.6/n² eV. In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.
Easy language (same idea, plain words)
Only certain orbits allowed; lowest hydrogen energy is −13.6 eV.
Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.
Topic in depth — Bohr postulates
Explains hydrogen spectrum lines via ΔE = hf.
Link to chapter notes (L24 — Bohr postulates): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: mvr = nh/2π; E_n = −13.6/n² eV. In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.
Exam tip
Quote the law in one line, then write mvr = nh/2π; E_n = −13.6/n² eV before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.
Common mistakes
- Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
- Using the wrong sign convention for work/heat/force direction.
- Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
- Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
Name the series for transitions ending at n=2 and give its spectral region.
Solution — step by step with formulas
- Balmer series; visible (and near UV).
Final answer: Balmer; mostly visible
Formulas used in this problem
Textbook formal language
Line spectra arise from discrete energy level differences.
Working formula set for this problem: 1/λ = R(1/n₁² − 1/n₂²). In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.
Easy language (same idea, plain words)
Electrons jumping down to level 2 make the familiar coloured H lines.
Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.
Topic in depth — Hydrogen spectrum
Lyman (UV) to n=1; Paschen (IR) to n=3.
Link to chapter notes (L24 — Hydrogen spectrum): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: 1/λ = R(1/n₁² − 1/n₂²). In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.
Exam tip
Quote the law in one line, then write 1/λ = R(1/n₁² − 1/n₂²) before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.
Common mistakes
- Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
- Using the wrong sign convention for work/heat/force direction.
- Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
- Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
Minimum energy to remove electron from ground-state H atom?
Solution — step by step with formulas
- 13.6 eV.
Final answer: 13.6 eV
Formulas used in this problem
Textbook formal language
Ionisation energy is energy to take electron to continuum E=0 from bound state.
Working formula set for this problem: Ionisation energy = 13.6 eV for H. In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.
Easy language (same idea, plain words)
Need 13.6 eV to free the electron completely from hydrogen’s ground state.
Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.
Topic in depth — Excitation/ionisation
Photon absorption requires hf ≥ ΔE for that transition.
Link to chapter notes (L24 — Excitation/ionisation): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: Ionisation energy = 13.6 eV for H. In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.
Exam tip
Quote the law in one line, then write Ionisation energy = 13.6 eV for H before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.
Common mistakes
- Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
- Using the wrong sign convention for work/heat/force direction.
- Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
- Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
How did de Broglie waves support Bohr’s orbits?
Solution — step by step with formulas
- Standing electron waves: circumference = nλ ⇒ mvr = nh/2π.
Final answer: Integer wavelengths fit around orbit
Formulas used in this problem
Textbook formal language
Quantisation appears as constructive standing-wave condition on the orbit.
Working formula set for this problem: 2πr = nλ. In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.
Easy language (same idea, plain words)
Only orbits where an electron wave closes on itself smoothly are allowed.
Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.
Topic in depth — Wave link
Bridge between Bohr and quantum mechanics.
Link to chapter notes (L24 — Wave link): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: 2πr = nλ. In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.
Exam tip
Quote the law in one line, then write 2πr = nλ before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.
Common mistakes
- Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
- Using the wrong sign convention for work/heat/force direction.
- Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
- Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
State the origin of continuous X-ray spectrum (Bremsstrahlung).
Solution — step by step with formulas
- Deceleration of fast electrons in target converts KE to photon continuum up to eV.
Final answer: Electron braking radiation; hf_max = eV
Formulas used in this problem
Textbook formal language
Sudden acceleration of charge radiates; maximum photon energy equals electron KE.
Working formula set for this problem: eV = hf_max (Bremsstrahlung). In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.
Easy language (same idea, plain words)
Electrons slam into metal and radiate a smear of X-ray energies.
Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.
Topic in depth — X-ray idea
Characteristic lines come from atomic shell transitions in the target.
Link to chapter notes (L24 — X-ray idea): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: eV = hf_max (Bremsstrahlung). In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.
Exam tip
Quote the law in one line, then write eV = hf_max (Bremsstrahlung) before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.
Common mistakes
- Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
- Using the wrong sign convention for work/heat/force direction.
- Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
- Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).