Important Questions Class 12 Physics Chapter 12: Atoms

Atoms are electrically neutral systems with a dense positive nucleus and electrons arranged in allowed energy states. Hydrogen emits fixed spectral lines because its electron can move only between quantised energy levels.

Atomic structure changed physics by replacing spread-out charge models with a tiny nucleus and quantised electron states. Important Questions Class 12 Physics Chapter 12 help students practise Thomson’s model, Rutherford scattering, Bohr postulates, hydrogen spectrum, energy levels, Bohr radius, ionisation energy, and de Broglie’s explanation. The CBSE 2026 chapter contains direct theory, numerical formulas, model comparisons, and conceptual questions from NCERT Atoms.

Key Takeaways

• Rutherford Scattering: About 1 in 8000 alpha particles deflect by more than 90° in gold foil scattering.
• Atomic Size: An atom has radius about 10^-10 m, while a nucleus has radius about 10^-15 m to 10^-14 m.
• Hydrogen Ground State: The ground state energy of hydrogen is −13.6 eV.
• Bohr Quantisation: Electron angular momentum follows L = nh/(2π).

Important Questions Class 12 Physics Chapter 12 Structure 2026

Important Questions Class 12 Physics Chapter 12 with Answers

The Atoms chapter builds from experimental evidence to quantum restrictions.
Students should separate model facts, scattering logic, and Bohr formulas before solving.
These atoms class 12 important questions follow the NCERT 2026 sequence.

1. What does Important Questions Class 12 Physics Chapter 12 mainly test?

Important Questions Class 12 Physics Chapter 12 mainly test Rutherford model, Bohr model, hydrogen spectrum, energy levels, and de Broglie explanation. The chapter combines theory, numerical formulas, and model comparison.

1. Atomic Models: Thomson, Rutherford, and Bohr.
2. Scattering: Alpha-particle deflection and impact parameter.
3. Spectra: Hydrogen emission and absorption lines.
4. Bohr Formula: En = −13.6/n² eV.
5. Final Result: The chapter tests atomic structure and quantised energy.

2. Why did physicists need atomic models?

Physicists needed atomic models to explain charge distribution inside atoms. Experiments showed that atoms contain electrons and positive charge.

1. Electron Discovery: J. J. Thomson identified electrons.
2. Neutral Atom: Total positive and negative charge balance.
3. Main Question: Positive charge arrangement needed explanation.
4. Final Result: Atomic models explain internal atomic structure.

3. What is the main difference between Thomson and Rutherford atomic models?

Thomson’s model spreads positive charge through the atom, while Rutherford’s model concentrates it in the nucleus. Rutherford’s model also leaves most atomic space empty.

1. Thomson Model: Positive charge fills the atom.
2. Rutherford Model: Positive charge stays in a tiny nucleus.
3. Electron Position: Electrons lie inside Thomson’s sphere but orbit Rutherford’s nucleus.
4. Final Result: Rutherford’s model introduced the nucleus.

Rutherford Atomic Model Class 12 Important Questions

Rutherford’s model came from alpha-particle scattering through thin gold foil.
The experiment showed that mass and positive charge occupy a tiny central region.
These Rutherford atomic model class 12 important questions focus on observations and conclusions.

4. What did Rutherford’s alpha-particle scattering experiment show?

Rutherford’s alpha-particle scattering experiment showed that an atom has a tiny, dense, positive nucleus. Most alpha particles passed through gold foil.

1. Observation 1: Most particles passed through the foil.
2. Observation 2: A few particles scattered by large angles.
3. Observation 3: About 1 in 8000 deflected by more than 90°.
4. Final Result: Positive charge and mass concentrate in the nucleus.

5. Why did most alpha particles pass through the gold foil?

Most alpha particles passed through because most of an atom is empty space. The nucleus occupies a very small volume.

1. Atomic Radius: About 10^-10 m.
2. Nuclear Radius: About 10^-15 m to 10^-14 m.
3. Size Difference: Atom is 10,000 to 100,000 times larger than nucleus.
4. Final Result: Most alpha particles meet empty space.

6. Why did a few alpha particles scatter through large angles?

A few alpha particles scattered through large angles because they passed close to the positive nucleus. Strong electrostatic repulsion changed their path.

1. Alpha Charge: +2e.
2. Nuclear Charge: +Ze.
3. Force Type: Repulsive Coulomb force.
4. Final Result: Close nuclear encounters produce large deflections.

7. Why is Rutherford credited with discovering the nucleus?

Rutherford is credited with discovering the nucleus because his model explained large alpha-particle deflections. A small dense centre caused strong repulsion.

1. Gold Foil Data: Few particles rebounded.
2. Required Force: Large repulsive force.
3. Conclusion: Positive charge occupies a tiny central volume.
4. Final Result: Rutherford identified the atomic nucleus.

Alpha Particle Scattering Class 12 Questions

Alpha-particle scattering questions test impact parameter, force, and distance of closest approach.
The alpha particle carries charge +2e, while a target nucleus carries charge +Ze.
These alpha particle scattering class 12 questions follow the NCERT experiment and examples.

8. What is impact parameter in alpha-particle scattering?

Impact parameter is the perpendicular distance between the initial velocity line and the nucleus centre. It decides the scattering angle.

1. Small Impact Parameter: Particle passes close to nucleus.
2. Effect: Large scattering angle occurs.
3. Large Impact Parameter: Particle passes far from nucleus.
4. Final Result: Smaller impact parameter gives larger scattering.

9. What force acts between an alpha particle and a gold nucleus?

The Coulomb repulsive force acts between the alpha particle and gold nucleus. Both charges are positive.

1. Alpha Charge: 2e.
2. Gold Nucleus Charge: Ze.
3. Formula Used:
   F = (1/4πε0)(2e)(Ze)/r²
4. Final Result: The force is electrostatic and repulsive.

10. Why does a head-on alpha particle rebound?

A head-on alpha particle rebounds because it loses kinetic energy to electric potential energy. At closest approach, its speed becomes zero.

1. Initial Energy: Kinetic energy K.
2. Closest Approach: Momentary rest point.
3. Energy Form: Electric potential energy.
4. Final Result: The alpha particle reverses before touching the nucleus.

11. Find the distance of closest approach for a 7.7 MeV alpha particle near gold.

The distance of closest approach is 3.0 × 10^-14 m. Use energy conservation for alpha-gold interaction.

1. Given Data:
   K = 7.7 MeV = 1.2 × 10^-12 J
   Z = 79
   e = 1.6 × 10^-19 C
   1/(4πε0) = 9.0 × 10^9 N m²/C²
2. Formula Used:
   d = (1/4πε0)(2Ze²)/K
3. Calculation:
   d = 3.84 × 10^-16 Z m
   d = 3.84 × 10^-16 × 79
   d = 3.0 × 10^-14 m
4. Final Result: d = 30 fm.

12. Why does the alpha particle not touch the gold nucleus in closest approach?

The alpha particle does not touch the nucleus because it reverses before reaching the nuclear surface. Electrostatic repulsion stops it.

1. Closest Approach: d = 30 fm.
2. Actual Gold Nucleus Radius: About 6 fm.
3. Comparison: Closest approach exceeds nuclear radius.
4. Final Result: The alpha particle stops outside the nucleus.

Thomson Model and Rutherford Model Class 12 Physics Questions

Model comparison questions often appear as direct conceptual questions.
Students should compare charge distribution, mass distribution, and stability limits.
This section supports class 12 physics atoms questions with answers through model-based reasoning.

13. What is Thomson’s plum pudding model?

Thomson’s plum pudding model places electrons inside a uniformly positive sphere. It compares electrons to seeds embedded in a watermelon.

1. Positive Charge: Uniformly distributed.
2. Electrons: Embedded inside the positive charge.
3. Neutrality: Positive and negative charges balance.
4. Final Result: Thomson’s model has no nucleus.

14. Why did Rutherford’s experiment reject Thomson’s model?

Rutherford’s experiment rejected Thomson’s model because it found a small dense positive nucleus. Thomson’s spread-out charge could not cause large deflections.

1. Thomson Prediction: Small deflections only.
2. Observed Result: Some particles scattered backward.
3. Required Cause: Concentrated positive charge.
4. Final Result: Large-angle scattering rejected Thomson’s model.

15. What are the main features of Rutherford’s nuclear model?

Rutherford’s nuclear model places positive charge and most mass in the nucleus. Electrons revolve around it at large distances.

1. Nucleus: Tiny, dense, and positive.
2. Electrons: Move around the nucleus.
3. Space: Most of the atom remains empty.
4. Final Result: Rutherford’s model is the nuclear model.

16. What are the limitations of Rutherford’s atomic model?

Rutherford’s model cannot explain atomic stability or line spectra. Classical physics predicts electron energy loss during circular motion.

1. Stability Problem: Accelerating electrons should radiate energy.
2. Collapse Problem: Electrons should spiral into the nucleus.
3. Spectra Problem: It predicts continuous spectra.
4. Final Result: Rutherford’s model cannot explain stable atoms.

Atomic Spectra Class 12 Physics Important Questions

Atomic spectra show that atoms emit selected wavelengths rather than a continuous band.
The line pattern connects an element’s internal structure with its radiation.
These atomic spectra class 12 physics questions explain emission and absorption lines.

17. What is an emission line spectrum?

An emission line spectrum contains bright lines at specific wavelengths. Excited atoms emit these wavelengths while returning to lower energy states.

1. Source: Excited atomic gas.
2. Appearance: Bright lines on dark background.
3. Meaning: Atoms emit selected frequencies.
4. Final Result: Emission spectra identify elements.

18. What is an absorption spectrum?

An absorption spectrum contains dark lines in a continuous spectrum. Gas atoms absorb selected wavelengths from passing white light.

1. Source: White light passes through gas.
2. Absorption: Atoms absorb matching photon energies.
3. Appearance: Dark lines appear.
4. Final Result: Absorption lines match emission wavelengths.

19. Why does atomic spectrum act like a fingerprint?

Atomic spectrum acts like a fingerprint because each element emits characteristic lines. Different atoms have different energy level gaps.

1. Element Identity: Each element has unique energy levels.
2. Emission Lines: Lines appear at fixed positions.
3. Use: Spectra identify gases.
4. Final Result: Atomic spectra identify elements.

Bohr Model Class 12 Physics Important Questions

Bohr solved hydrogen stability by adding quantum restrictions to Rutherford’s model.
He used stationary orbits, angular momentum quantisation, and photon emission rules.
These Bohr model class 12 physics important questions focus on postulates and formulas.

20. Why did Bohr modify Rutherford’s model?

Bohr modified Rutherford’s model because classical physics predicted atomic collapse. Bohr introduced stable orbits without radiation.

1. Rutherford Issue: Electrons accelerate in circular motion.
2. Classical Prediction: Electrons radiate energy.
3. Bohr Fix: Certain orbits do not radiate.
4. Final Result: Bohr explained hydrogen atom stability.

21. What are Bohr’s three postulates?

Bohr’s postulates describe stationary orbits, angular momentum quantisation, and photon emission during transitions. They explain hydrogen spectra.

1. First Postulate: Electrons move in stable orbits without radiation.
2. Second Postulate: mvr = nh/(2π).
3. Third Postulate: hν = Ei − Ef.
4. Final Result: Bohr’s postulates explain hydrogen line spectra.

22. What is Bohr’s quantisation condition?

Bohr’s quantisation condition states that electron angular momentum equals an integral multiple of h/(2π). Its formula is mvr = nh/(2π).

1. Angular Momentum: L = mvr.
2. Allowed Values: L = nh/(2π).
3. Quantum Number: n = 1, 2, 3, ...
4. Final Result: Electron angular momentum is quantised.

23. What is Bohr’s photon emission condition?

Bohr’s photon emission condition states that photon energy equals the energy difference between two levels. The formula is hν = Ei − Ef.

1. Initial Level: Higher energy Ei.
2. Final Level: Lower energy Ef.
3. Photon Energy: hν = Ei − Ef.
4. Final Result: A downward transition emits one photon.

Bohr Postulates Class 12 Questions

Bohr postulates connect atomic stability with spectral lines.
Students should write each postulate in simple words before applying formulas.
These Bohr postulates class 12 questions match direct theory and derivation answers.

24. What is meant by stationary orbit in Bohr model?

A stationary orbit is an allowed electron orbit where the electron does not radiate energy. Each stationary orbit has fixed energy.

1. Electron Motion: Circular orbit around nucleus.
2. Radiation: No energy emission in allowed orbit.
3. Energy: Definite total energy.
4. Final Result: Stationary orbits keep atoms stable.

25. Why does Bohr model produce discrete spectral lines?

Bohr model produces discrete spectral lines because only certain energy levels exist. Photons carry exact energy differences between levels.

1. Allowed Energies: En has discrete values.
2. Transition Energy: ΔE = Ei − Ef.
3. Photon Frequency: ν = ΔE/h.
4. Final Result: Discrete levels produce discrete frequencies.

26. What is the limitation of Bohr model for multi-electron atoms?

Bohr model fails for multi-electron atoms because electron-electron forces become important. The model handles only hydrogenic atoms.

1. Hydrogenic Atom: One electron only.
2. Multi-electron Atom: Electrons interact with each other.
3. Problem: Bohr model includes only nucleus-electron force.
4. Final Result: Bohr model applies to hydrogenic atoms.

27. Why can Bohr model not explain line intensities?

Bohr model cannot explain line intensities because it does not predict transition probability. Some spectral lines appear stronger than others.

1. Model Strength: Predicts frequencies.
2. Model Weakness: Does not predict relative intensities.
3. Experimental Fact: Some transitions are more favoured.
4. Final Result: Bohr model misses intensity variation.

Energy Levels of Hydrogen Atom Class 12 Questions

Hydrogen energy levels become less negative as the principal quantum number increases.
The ground state has energy −13.6 eV, and the ionised state has energy 0 eV.
These energy levels of hydrogen atom class 12 questions cover excitation and ionisation.

28. What is the energy of nth orbit in hydrogen atom?

The energy of nth orbit is En = −13.6/n² eV. The value becomes less negative for higher n.

1. Formula Used: En = −13.6/n² eV.
2. Ground State: n = 1 gives −13.6 eV.
3. Ionised State: n = ∞ gives 0 eV.
4. Final Result: Hydrogen energy levels are quantised.

29. What is the energy of hydrogen atom in n = 2 level?

The energy at n = 2 is −3.40 eV. Use En = −13.6/n² eV.

1. Given Data: n = 2.
2. Formula Used: En = −13.6/n² eV.
3. Calculation:
   E2 = −13.6/2²
   E2 = −13.6/4
   E2 = −3.40 eV
4. Final Result: E2 = −3.40 eV.

30. What is the energy required to excite hydrogen from n = 1 to n = 2?

The required energy is 10.2 eV. It equals E2 − E1.

1. Given Data:
   E1 = −13.6 eV
   E2 = −3.40 eV
2. Formula Used: ΔE = E2 − E1.
3. Calculation:
   ΔE = −3.40 − (−13.6)
   ΔE = 10.2 eV
4. Final Result: Excitation energy = 10.2 eV.

31. What is ionisation energy of hydrogen atom?

The ionisation energy of hydrogen atom is 13.6 eV from ground state. It removes the electron completely.

1. Ground State Energy: E1 = −13.6 eV.
2. Free Electron Energy: E∞ = 0 eV.
3. Energy Needed: 0 − (−13.6) = 13.6 eV.
4. Final Result: Ionisation energy = 13.6 eV.

Radius of Bohr Orbit Class 12 Questions

Bohr radius gives the size of allowed circular orbits in hydrogen.
For hydrogen, the orbital radius increases as the square of n.
These radius of Bohr orbit class 12 questions focus on radius values and orbit scaling.

32. What is radius of nth Bohr orbit in hydrogen?

The radius of nth Bohr orbit is rn = n²a0 for hydrogen. Here a0 = 5.3 × 10^-11 m.

1. Formula Used: rn = n²a0.
2. Bohr Radius: a0 = 5.3 × 10^-11 m.
3. Scaling: Radius increases as n².
4. Final Result: rn = n²a0.

33. Find radius of n = 2 orbit in hydrogen.

The radius of n = 2 orbit is 2.12 × 10^-10 m. Use rn = n²a0.

1. Given Data:
   n = 2
   a0 = 5.3 × 10^-11 m
2. Formula Used: rn = n²a0.
3. Calculation:
   r2 = 2² × 5.3 × 10^-11
   r2 = 2.12 × 10^-10 m
4. Final Result: r2 = 2.12 × 10^-10 m.

34. Find radius of n = 3 orbit in hydrogen.

The radius of n = 3 orbit is 4.77 × 10^-10 m. Use rn = n²a0.

1. Given Data:
   n = 3
   a0 = 5.3 × 10^-11 m
2. Formula Used: rn = n²a0.
3. Calculation:
   r3 = 3² × 5.3 × 10^-11
   r3 = 4.77 × 10^-10 m
4. Final Result: r3 = 4.77 × 10^-10 m.

35. Why do higher Bohr orbits have larger radius?

Higher Bohr orbits have larger radius because rn is proportional to n². Increasing n increases allowed orbit size.

1. Formula Used: rn = n²a0.
2. For n = 1: r1 = a0.
3. For n = 3: r3 = 9a0.
4. Final Result: Bohr radius grows as n².

Hydrogen Atom Spectrum Class 12 Physics Questions

Hydrogen spectrum questions use exact energy gaps between allowed levels.
A photon appears when an electron jumps from a higher orbit to a lower orbit.
These Hydrogen Atom Spectrum Class 12 Physics Questions connect Bohr’s model with observed spectral lines.

36. How are spectral lines produced in hydrogen atom?

Spectral lines appear when electrons jump between allowed energy levels. Each jump emits or absorbs a photon of exact energy.

1. Emission: Electron falls to lower energy level.
2. Absorption: Electron rises to higher energy level.
3. Formula: hν = Ei − Ef.
4. Final Result: Hydrogen lines come from level transitions.

37. Why does hydrogen show a line spectrum instead of a continuous spectrum?

Hydrogen shows a line spectrum because its electron can occupy only fixed energy levels. Transitions produce fixed photon energies.

1. Allowed Levels: Hydrogen has discrete energy states.
2. Photon Energy: ΔE = Ei − Ef.
3. Observed Result: Fixed wavelengths appear.
4. Final Result: Hydrogen emits discrete spectral lines.

38. What frequency comes from a 2.3 eV energy difference?

The frequency is 5.6 × 10^14 Hz. Use ν = ΔE/h.

1. Given Data:
   ΔE = 2.3 eV = 2.3 × 1.6 × 10^-19 J
   h = 6.6 × 10^-34 J s
2. Formula Used: ν = ΔE/h.
3. Calculation:
   ν = (2.3 × 1.6 × 10^-19)/(6.6 × 10^-34)
   ν = 5.6 × 10^14 Hz
4. Final Result: ν = 5.6 × 10^14 Hz.

39. What photon energy excites hydrogen from n = 1 to n = 4?

The photon energy is 12.75 eV. Use En = −13.6/n² eV.

1. Given Data:
   E1 = −13.6 eV
   E4 = −13.6/16 eV = −0.85 eV
2. Formula Used: ΔE = E4 − E1.
3. Calculation:
   ΔE = −0.85 − (−13.6)
   ΔE = 12.75 eV
4. Final Result: Photon energy = 12.75 eV.

40. What wavelength excites hydrogen from n = 1 to n = 4?

The wavelength is 97.5 nm. Use λ = hc/ΔE.

1. Given Data:
   ΔE = 12.75 eV
   hc = 1240 eV nm
2. Formula Used: λ = hc/ΔE.
3. Calculation:
   λ = 1240/12.75
   λ = 97.25 nm
4. Final Result: λ ≈ 97.5 nm.

41. Which hydrogen series appears when electrons fall to n = 2?

The Balmer series appears when electrons fall to n = 2. Its visible lines made hydrogen spectrum important historically.

1. Final Level: nf = 2.
2. Initial Levels: ni = 3, 4, 5, ...
3. Region: Mostly visible spectrum.
4. Final Result: Transitions to n = 2 form the Balmer series.

De Broglie Explanation of Bohr Model Class 12 Questions

De Broglie explained Bohr’s quantised orbits using electron wave nature.
Only standing waves fit around allowed circular orbits.
This de Broglie explanation of Bohr model class 12 section connects wave-particle duality with angular momentum.

42. How did de Broglie explain Bohr’s quantisation condition?

De Broglie explained Bohr’s quantisation condition by treating the electron as a standing wave. The orbit circumference must contain whole wavelengths.

1. De Broglie Wavelength: λ = h/mv.
2. Standing Wave Condition: 2πr = nλ.
3. Substitution: 2πr = nh/mv.
4. Final Result: mvr = nh/(2π).

43. Why can only certain electron orbits exist in Bohr model?

Only certain electron orbits exist because only those orbits support standing waves. Other wavelengths cancel by destructive interference.

1. Allowed Orbit: 2πr = nλ.
2. Unstable Wave: Non-integral wavelengths cancel.
3. Physical Result: Only resonant orbits persist.
4. Final Result: Standing waves explain allowed orbits.

44. What is de Broglie wavelength of an electron in Bohr orbit?

The de Broglie wavelength is λ = h/mv. It links electron momentum with wave nature.

1. Momentum: p = mv.
2. Formula Used: λ = h/p.
3. For Electron: λ = h/mv.
4. Final Result: Electron wavelength equals h/mv.

Class 12 Physics Chapter 12 Questions and Answers

NCERT problems in Atoms often combine model facts with direct formulas.
Students should identify whether the question needs energy, radius, frequency, or model comparison.
These class 12 physics chapter 12 questions and answers follow the 2026 exercise style.

45. What are kinetic and potential energies in hydrogen ground state?

In hydrogen ground state, kinetic energy is 13.6 eV, and potential energy is −27.2 eV. Total energy equals −13.6 eV.

1. Given Data: E = −13.6 eV.
2. Bohr Relation: K = −E.
3. Potential Energy: U = 2E.
4. Values:
   K = 13.6 eV
   U = −27.2 eV
5. Final Result: K = 13.6 eV and U = −27.2 eV.

46. Why is total energy of electron in hydrogen negative?

The total energy is negative because the electron remains bound to the nucleus. Energy must be supplied to remove it.

1. Bound State: Electron stays in orbit.
2. Zero Energy: Free electron at rest far away.
3. Hydrogen Ground State: E1 = −13.6 eV.
4. Final Result: Negative energy means bound electron.

47. What is the speed of electron in hydrogen ground state?

The electron speed in hydrogen ground state is 2.2 × 10^6 m/s. NCERT derives this from Coulomb force and Bohr radius.

1. Orbit Radius: r = 5.3 × 10^-11 m.
2. Electron Mass: m = 9.1 × 10^-31 kg.
3. Known Result: v = 2.2 × 10^6 m/s.
4. Final Result: Ground-state electron speed = 2.2 × 10^6 m/s.

48. What is the initial classical frequency of light from revolving hydrogen electron?

The initial classical frequency is 6.6 × 10^15 Hz. It comes from orbital frequency v/(2πr).

1. Given Data:
   v = 2.2 × 10^6 m/s
   r = 5.3 × 10^-11 m
2. Formula Used: ν = v/(2πr).
3. Calculation:
   ν = 2.2 × 10^6/(2π × 5.3 × 10^-11)
   ν ≈ 6.6 × 10^15 Hz
4. Final Result: ν = 6.6 × 10^15 Hz.

49. Why does Bohr model still remain useful?

Bohr model remains useful because it explains major features of hydrogen spectrum. It also gives correct energy levels for hydrogenic atoms.

1. Strength 1: Explains atomic stability for hydrogen.
2. Strength 2: Predicts spectral frequencies.
3. Strength 3: Introduces quantised energy states.
4. Final Result: Bohr model works well for hydrogenic atoms.

NCERT Class 12 Physics Chapter 12 Questions

NCERT Class 12 Physics Chapter 12 questions often test exact model comparison and hydrogen formulas.
The exercise includes Thomson model, Rutherford model, energy gaps, Bohr radii, and energy levels.
These NCERT Class 12 Physics Chapter 12 questions use values given in the 2026 textbook.

50. Is atom size different in Thomson and Rutherford models?

The atom size is no different in Thomson and Rutherford models. Both models describe the same atomic scale.

1. Thomson Model: Atom has distributed positive charge.
2. Rutherford Model: Atom has central nucleus.
3. Size Comparison: Overall atomic size remains similar.
4. Final Result: The atom size is no different.

51. Which model gives stable equilibrium for electrons?

Thomson’s model gives stable equilibrium for electrons. Rutherford’s model gives electrons a net centripetal force.

1. Thomson Model: Electrons can stay in equilibrium.
2. Rutherford Model: Electrons revolve under attraction.
3. Net Force: Rutherford electrons always experience force.
4. Final Result: Thomson’s model gives stable equilibrium.

52. Why does a classical atom based on Rutherford model collapse?

A classical Rutherford atom collapses because accelerating electrons radiate energy. The electron then spirals into the nucleus.

1. Electron Motion: Circular motion causes acceleration.
2. Classical Rule: Accelerating charge radiates.
3. Energy Loss: Orbit radius decreases.
4. Final Result: Rutherford’s classical atom collapses.

53. What happens if alpha scattering uses solid hydrogen instead of gold foil?

Alpha particles would suffer much smaller deflections with solid hydrogen. Hydrogen nuclei have much smaller charge than gold nuclei.

1. Gold Nucleus: Z = 79.
2. Hydrogen Nucleus: Z = 1.
3. Repulsive Force: Proportional to Z.
4. Final Result: Large-angle scattering becomes rare.

54. What series emits when hydrogen reaches n = 4 and falls to lower levels?

Hydrogen can emit Lyman, Balmer, and Paschen series lines after excitation to n = 4. Different final levels create different series.

1. To n = 1: Lyman series.
2. To n = 2: Balmer series.
3. To n = 3: Paschen series.
4. Final Result: Transitions from n = 4 can produce multiple series.

Class 12 Physics Chapter 12 Derivations

Derivations in Atoms focus on force balance, angular momentum quantisation, and photon emission.
Students should write assumptions before formulas because the chapter uses model-based reasoning.
These Class 12 Physics Chapter 12 derivations cover Bohr radius, energy, and quantisation.

55. How do you derive Bohr radius formula for hydrogen?

Bohr radius formula comes from Coulomb force and angular momentum quantisation. The nth radius varies as n².

1. Coulomb Force Provides Centripetal Force:
   e²/(4πε0r²) = mv²/r
2. Bohr Quantisation:
   mvr = nh/(2π)
3. Eliminate v:
   rn = n²h²ε0/(πme²)
4. Hydrogen Form:
   rn = n²a0
5. Final Result: rn = n²a0.

56. How do you derive total energy of electron in Bohr orbit?

Total energy comes from kinetic energy plus electrostatic potential energy. The final energy becomes negative.

1. Kinetic Energy:
   K = e²/(8πε0r)
2. Potential Energy:
   U = −e²/(4πε0r)
3. Total Energy:
   E = K + U
   E = −e²/(8πε0r)
4. With Quantised Radius:
   En = −13.6/n² eV
5. Final Result: En = −13.6/n² eV.

57. How do you derive Bohr quantisation from de Broglie wavelength?

Bohr quantisation follows from the standing wave condition around the orbit. A whole number of wavelengths must fit.

1. Standing Wave Condition:
   2πr = nλ
2. De Broglie Relation:
   λ = h/mv
3. Substitution:
   2πr = nh/mv
4. Rearrange:
   mvr = nh/(2π)
5. Final Result: L = nh/(2π).

Class 12 Physics Chapter-Wise Important Questions