Atoms Class 12 Notes | CBSE Physics Chapter 12 (Free PDF)

Chapter summary

The Atoms chapter traces how the alpha-particle scattering experiment of Geiger and Marsden led Rutherford to the nuclear model, and how Bohr then added quantum postulates to explain the stable orbits and line spectrum of hydrogen. It covers the distance of closest approach, the radius, speed and energy of Bohr orbits, the hydrogen spectral series through the Rydberg formula, and de Broglie’s wave explanation of quantisation. It is a high-yield NEET topic because nearly every year asks numerical questions on Bohr-orbit formulas and the hydrogen spectrum.

Chapter notes

Key Concepts

1. Rutherford’s Nuclear Model

Alpha particle scattering experiment showed that the atom has a tiny, dense, positively charged nucleus at the centre, with electrons orbiting around it.

Limitation: An orbiting electron should continuously radiate energy and spiral into the nucleus - but atoms are stable. Classical physics couldn’t explain this.

2. Bohr’s Model of Hydrogen Atom

Postulates

  1. Electrons revolve in fixed circular orbits (stationary orbits) without radiating energy
  2. Quantisation: Angular momentum is quantised: L = mvr = nh/(2π), n = 1, 2, 3…
  3. Energy is emitted/absorbed only when an electron jumps between orbits: E = hν = E_i − E_f

Key Formulae for Hydrogen (Z = 1)

QuantityFormula
Radius of nth orbitrn = 0.529 × n² Å (= n² × a₀, where a₀ = 0.529 Å)
Velocity in nth orbitvn = 2.18 × 10⁶/n m/s
Energy of nth levelEn = −13.6/n² eV

Ground state (n=1): E = −13.6 eV; First excited (n=2): E = −3.4 eV

3. Hydrogen Spectral Series

1/λ = R(1/n₁² − 1/n₂²) where R = 1.097 × 10⁷ m⁻¹ (Rydberg constant)

SeriesTransition to n₁Region
Lymann₁ = 1Ultraviolet
Balmern₁ = 2Visible
Paschenn₁ = 3Infrared
Brackettn₁ = 4Infrared
Pfundn₁ = 5Infrared

Solved Examples

Example 1

Find the energy of the electron in the 3rd orbit of hydrogen.

Answer: E₃ = −13.6/3² = −13.6/9 = −1.51 eV

Example 2

Find the wavelength of the first line of the Balmer series (n₂ = 3 to n₁ = 2).

Answer: 1/λ = R(1/4 − 1/9) = R(9−4)/36 = 5R/36 = (5 × 1.097 × 10⁷)/36 = 1.524 × 10⁶

λ = 1/(1.524 × 10⁶) = 6.56 × 10⁻⁷ m = 656 nm (red light)


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Quick Revision Points

  • Bohr: quantised orbits, L = nh/(2π), En = −13.6/n² eV
  • rn ∝ n²; vn ∝ 1/n; En ∝ −1/n²
  • Ground state: n=1, E = −13.6 eV; ionisation energy = 13.6 eV
  • Lyman (UV), Balmer (visible), Paschen/Brackett/Pfund (IR)
  • 1/λ = R(1/n₁² − 1/n₂²)

Previous: Ch 11 - Dual Nature
Next: Ch 13 - Nuclei

🃏 Flash Cards: Atoms

Class 12 Physics · Chapter 12 – swipe through all 9 cards to understand the whole chapter.

🥇Start here1/9

Alpha-Scattering & Rutherford’s Nucleus

Geiger and Marsden fired alpha particles at gold foil and a few bounced straight back, revealing a tiny dense nucleus.

N(θ) ∝ 1 / sin4(θ/2)

Nucleus ~10⁻15 m; whole atom ~10⁻10 m (≈105× bigger).

  • Most α pass undeflected → atom is mostly empty space.
  • ≈1 in 8000 rebound >90° → mass + positive charge sit in a tiny nucleus.
  • Flaw: an orbiting electron should radiate, spiral in → classical atom is unstable.
📏Nuclear size2/9

Distance of Closest Approach

In a head-on hit the alpha’s kinetic energy fully converts to electrostatic PE, fixing how close it gets.

r0 = (1/4πε0) · 2Ze2 / K

Upper estimate of nuclear size; r0 ∝ 1/K (faster α → closer).

  • Set K = (1/4πε0)(2e)(Ze)/r0 at the turning point.
  • α charge is +2e — keep the factor 2.
  • Convert MeV → joules (1 MeV = 1.6×10⁻13 J) before plugging in.
🎯Aim & angle3/9

Impact Parameter

How far off-centre the alpha is aimed (b) sets how sharply it scatters.

b = Ze2 cot(θ/2) / (4πε0 K)

b = 0 (head-on) → θ = 180° (perfect rebound).

  • Small b (near head-on) → large scattering angle θ.
  • Large b → small deflection, nearly straight through.
  • The bullseye is a tiny target, so rebounds are rare.
🪜Core model4/9

Bohr’s Three Postulates

Bohr bolted quantum rules onto Rutherford’s atom to make it stable and explain line spectra.

L = mvr = nh/2π ; hν = E_i − E_f

Works for H and one-electron ions (He⁺, Li2⁺); fails for multi-electron atoms.

  • Stationary orbits: electrons revolve without radiating, though accelerating.
  • Angular momentum is quantised in units of h/2π (n = 1,2,3…).
  • Light emitted/absorbed only during a jump: down → emit, up → absorb.
🧲Orbit radius5/9

Radius of the nth Orbit

Coulomb pull as centripetal force plus quantised L gives discrete orbit sizes.

r_n = 0.529 × n2/Z Å

n=1 hydrogen value 0.529 Å is the Bohr radius; r ∝ n2/Z.

  • Radius grows as n2 — outer orbits are far larger.
  • Higher nuclear charge Z pulls the electron in tighter.
  • He⁺ (Z=2) ground radius = 0.529/2 = 0.265 Å.
💨Orbit speed6/9

Speed in the nth Orbit

The orbiting electron slows down as it moves to higher orbits.

v_n = 2.18×106 × Z/n m·s⁻1

H ground state v ≈ c/137; 1/137 is the fine-structure constant.

  • Speed v ∝ Z/n — Z in the numerator, n in the denominator.
  • Higher n → slower electron (and longer de Broglie wavelength).
  • Independent of where you measure: fixed per orbit.
🔋Orbit energy7/9

Energy of the nth Orbit

Total energy is negative because the electron is bound to the nucleus.

E_n = −13.6 × Z2/n2 eV

H ground state −13.6 eV → ionisation energy = 13.6 eV.

  • E ∝ Z2/n2; negative sign means bound.
  • Energy bookkeeping: K = −E, U = 2E = −2K, E = U/2.
  • First excited state of H (n=2): E = −13.6/4 = −3.4 eV.
🌈Spectra8/9

Rydberg Formula & Spectral Series

Electron jumps to a lower level emit photons of fixed wavelength, grouped into named series.

1/λ = R Z2 (1/n_f2 − 1/n_i2), R = 1.097×107 m⁻1

Lyman (n_f=1) UV; Balmer (n_f=2) visible; Paschen/Brackett/Pfund IR.

  • First line of a series = longest λ, least energy (smallest jump).
  • Series limit (n_i = ∞) = shortest λ of that series.
  • Cascading from level n emits n(n−1)/2 distinct lines.
🌊Why quantised9/9

de Broglie’s Standing Wave

The electron is a wave that must close on itself, which derives Bohr’s quantisation rule.

2πr = nλ , λ = h/mv ⟹ mvr = nh/2π

Full-wavelength condition (not half-wave); nth orbit holds exactly n wavelengths.

  • Only a whole number of wavelengths fits → orbits are discrete.
  • Turns Bohr’s assumed L = nh/2π into a derived result.
  • As n rises, electron slows so λ grows (λ ∝ n).
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📝 Practice Atoms — 10 NEET PYQs
Real previous-year questions · with answers & solutions
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Tap an option to check your answer and see the worked solution. Every question is a real NEET previous-year question.
Q1NEET 2020
For which one of the following is the Bohr model NOT valid?
Correct answer: C. The Bohr model is valid only for one-electron (hydrogen-like) systems: H, He⁺, deuteron, etc. Singly ionised neon (Ne⁺) still has many electrons, so the Bohr model does not apply.
🔎 See the full step-by-step solution in the app →
Q2NEET 2020
The total energy of an electron in the nth stationary orbit of the hydrogen atom can be obtained by:
Correct answer: B. The total energy of the electron in the nth orbit of hydrogen is Eₙ = -(Rhc)/(n²) = -(13.6)/(n²) eV (negative, bound state).
🔎 See the full step-by-step solution in the app →
Q3NEET 2019
The radius of the first permitted Bohr orbit in a hydrogen atom equals 0.51 Å and its ground state energy equals -13.6 eV. If the electron in the hydrogen atom is replaced by a muon (μ⁻) [charge same as electron, mass 207mₑ], the first Bohr radius and ground state energy will be:
Correct answer: C. r ∝ 1/m so r_μ = 0.51 Å/207 = 2.56×10⁻¹³ m. E ∝ m so E_μ = -13.6×207 ≈ -2.8 keV.
🔎 See the full step-by-step solution in the app →
Q4NEET 2017
The ratio of wavelengths of the last line of the Balmer series and the last line of the Lyman series is:
Correct answer: C. Last line of a series corresponds to n₂ = ∞. Balmer: (1)/(λ_B) = R((1)/(4)) = R/4. Lyman: (1)/(λ_L) = R(1) = R. So λ_B/λ_L = 4.
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Q5NEET 2016
When an α-particle of mass m moving with velocity v bombards a heavy nucleus of charge Ze, its distance of closest approach from the nucleus depends on m as:
Correct answer: D. At closest approach all KE converts to PE: (1)/(2)mv² = (1)/(4πε₀)(2Ze²)/(r₀), so r₀ ∝ (1)/(mv²) ∝ (1)/(m) (for fixed v).
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Q6NEET 2016
If an electron in a hydrogen atom jumps from the 3rd orbit to the 2nd orbit, it emits a photon of wavelength λ. When it jumps from the 4th orbit to the 3rd orbit, the corresponding wavelength of the emitted photon will be:
Correct answer: C. (1)/(λ) = R((1)/(4)-(1)/(9)) = (5R)/(36) and (1)/(λ’) = R((1)/(9)-(1)/(16)) = (7R)/(144). Dividing: (λ’)/(λ) = (5/36)/(7/144) = (20)/(7), so λ’ = (20)/(7)λ.
🔎 See the full step-by-step solution in the app →
Q7NEET 2009
In a Rutherford scattering experiment, when a projectile of charge Z₁ and mass M₁ approaches a target nucleus of charge Z₂ and mass M₂, the distance of closest approach is r₀. The energy of the projectile is:
Correct answer: B. At closest approach KE = (1)/(4πε₀)(Z₁Z₂e²)/(r₀), so for fixed r₀ the projectile energy is directly proportional to Z₁ Z₂.
🔎 See the full step-by-step solution in the app →
Q8NEET 2006
In a discharge tube, ionisation of the enclosed gas is produced due to collisions between:
Correct answer: B. In a discharge tube, free electrons accelerated by the applied potential difference gain large kinetic energy and collide with neutral gas atoms/molecules, knocking out further electrons. This electron-neutral collision is the primary ionisation mechanism that sustains the discharge.
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Q9NEET 1994
In a Rutherford scattering experiment, the correct angle for α-scattering for an impact parameter b = 0 is:
Correct answer: D. b = 0 means a head-on collision; the α-particle retraces its path, so it is scattered through 180°.
🔎 See the full step-by-step solution in the app →
Q10NEET 1990
Consider an electron in the nth orbit of a hydrogen atom in the Bohr model. The circumference of the orbit can be expressed in terms of the de Broglie wavelength λ of that electron as:
Correct answer: D. Bohr’s quantisation in de Broglie form: the circumference contains an integral number of wavelengths, 2π rₙ = nλ.
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Frequently Asked Questions

What did the Rutherford alpha-particle scattering experiment prove?

Firing alpha particles at a thin gold foil showed most passed nearly undeflected while about 1 in 8000 bounced back beyond 90 degrees. This proved that almost all the mass and the entire positive charge of an atom sit in a tiny central nucleus about 10 to the power minus 15 metres across, with the rest of the atom being mostly empty space.

What are the key Bohr-model formulas for a hydrogen-like atom?

The orbit radius is r_n equal to 0.529 times n squared over Z angstrom, the speed is v_n equal to 2.18 times 10 to the 6 times Z over n metres per second, and the energy is E_n equal to minus 13.6 times Z squared over n squared electron volts. Angular momentum is quantised as mvr equal to n times h over 2 pi.

What is the Rydberg formula and which spectral series is in the visible region?

The Rydberg formula is 1 over lambda equal to R times Z squared times the quantity 1 over n_f squared minus 1 over n_i squared, where R is 1.097 times 10 to the 7 per metre. Only the Balmer series, with the electron falling to n_f equal to 2, lies in the visible region; the Lyman series is ultraviolet and the Paschen, Brackett and Pfund series are infrared.

Is the Atoms chapter important for NEET and how many questions come from it?

Yes, Atoms is part of the NEET Physics syllabus and is consistently asked, usually one question per year. Most questions are direct numericals on Bohr-orbit radius, speed, energy, or hydrogen-spectrum wavelengths, so memorising the standard formulas reliably scores marks.

What is the difference between the distance of closest approach and the impact parameter?

The distance of closest approach is how near a head-on alpha particle gets to the nucleus before its kinetic energy fully converts to electrostatic potential energy, and it gives an upper estimate of nuclear size. The impact parameter is the perpendicular off-centre distance of the alpha’s initial path from the nucleus; a small impact parameter gives a large scattering angle while a large one gives almost no deflection.

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