NEET UG Physics · Chapter 23

Atoms and Nuclei

Bohr model, hydrogen spectrum, nuclear size, mass defect, binding energy, radioactivity, fission and fusion.

NCERT Class 12NEET: 2 Qs per paper 12 min read

1. Chapter at a Glance

This chapter combines two NCERT chapters: Atoms and Nuclei. The atom part covers Rutherford's α-scattering experiment, the Bohr model of hydrogen, energy levels and the hydrogen spectrum. The nuclei part covers nuclear size and composition, mass defect and binding energy, radioactive decay, and nuclear fission and fusion. Together they give NEET about two questions every year, usually on Bohr's model, spectral lines, binding energy or half-life.

What NEET asks most
  • Radius, speed and energy of the n-th Bohr orbit.
  • Wavelength of spectral lines; number of lines from level n.
  • Nuclear radius R = R₀A1/3 and constant nuclear density.
  • Mass defect, binding energy per nucleon and the BE curve.
  • Half-life, decay constant, mean life and activity.

2. Rutherford's Nuclear Model

  • In the Geiger–Marsden experiment, α-particles were fired at thin gold foil. Most passed straight through, but a very few (about 1 in 8000) bounced back by more than 90°.
  • Conclusion: almost all the mass and all the positive charge of an atom is in a tiny, dense nucleus (about 10⁻¹⁵ m), while the atom (about 10⁻¹⁰ m) is mostly empty space.
  • Distance of closest approach: all the kinetic energy of the α-particle becomes potential energy, so r₀ = k(2e)(Ze)/K.
  • Impact parameter b is the perpendicular distance of the α-particle's initial path from the nucleus. A small b gives a large scattering angle.
  • Rutherford's model could not explain the stability of atoms (an orbiting electron should radiate and spiral in) or line spectra.

3. Bohr Model of Hydrogen

Postulates: (1) electrons move in certain stable orbits without radiating; (2) in these orbits the angular momentum is quantised, mvr = nh/2π; (3) a photon is emitted or absorbed only when an electron jumps between orbits, with hν = Ei − Ef.

Quantity (hydrogen-like, charge Ze)FormulaDependenceNEET Note
Radiusrn = 0.529 n²/Z Å∝ n²/ZBohr radius 0.529 Å
Speedvn = 2.18 × 10⁶ Z/n m s⁻¹∝ Z/nSlower in outer orbits
Total energyEn = −13.6 Z²/n² eV∝ Z²/n²Negative: bound electron
Kinetic / potential energyK = −E, U = 2E—K = +13.6 eV in the ground state
Ionisation energy (ground state)13.6 Z² eV—13.6 eV for H, 54.4 eV for He⁺

4. Hydrogen Spectrum

1/λ = R(1/n₁² − 1/n₂²)    R = 1.097 × 10⁷ m⁻¹    number of lines from level n = n(n − 1)/2
SeriesLower level n₁RegionNEET Note
Lyman1UltravioletLongest line 121.5 nm (2 → 1)
Balmer2VisibleHα 656 nm, Hβ 486 nm; limit 365 nm
Paschen3Infrared—
Brackett4Infrared—
Pfund5Far infrared—
n=1n=2n=3n=4 −13.6−3.4−1.510 Lyman (UV)Balmer (visible)
Energy levels of hydrogen (in eV). Jumps down to n = 1 give the Lyman series (ultraviolet); jumps down to n = 2 give the Balmer series (visible).

5. The Nucleus

  • A nucleus AZX has Z protons and N = A − Z neutrons. Isotopes have the same Z; isobars the same A; isotones the same N.
  • Nuclear radius R = R₀A1/3 with R₀ ≈ 1.2 fm. Since volume ∝ A, nuclear density is the same for all nuclei (about 2.3 × 10¹⁷ kg m⁻³).
  • 1 atomic mass unit (u) = 1.66 × 10⁻²⁷ kg ≈ 931.5 MeV.
  • Nuclear force: very strong, short-range (a few fm), attractive, and charge-independent (acts equally between p–p, n–n and n–p).

6. Mass Defect and Binding Energy

Δm = [Zmp + (A − Z)mn] − M    BE = Δm × 931.5 MeV (Δm in u)
  • The mass of a nucleus is less than the total mass of its separate nucleons. The missing mass (mass defect) has become binding energy (E = mc²).
  • Binding energy per nucleon (BE/A) measures stability. It rises steeply for light nuclei, peaks at about 8.8 MeV near iron (A ≈ 56), and falls slowly for heavy nuclei (about 7.6 MeV for uranium).
  • Heavy nuclei release energy by splitting (fission); light nuclei release energy by joining (fusion). Both move towards the peak of the curve.
  • Q-value of a reaction = (mass of reactants − mass of products) × 931.5 MeV. A positive Q means energy is released.

7. Radioactivity

N = N₀e−λt = N₀(½)t/T½    T½ = 0.693/λ    τ = 1/λ    A = λN
DecayChange in ZChange in ANEET Note
α (⁴He nucleus)−2−4Least penetrating, most ionising
β⁻ (electron + antineutrino)+10n → p + e⁻ + ν̄
β⁺ (positron + neutrino)−10p → n + e⁺ + ν
γ (photon)00Most penetrating; excited nucleus settles down
  • Activity is measured in becquerel (1 Bq = 1 decay per second); 1 curie = 3.7 × 10¹⁰ Bq.
  • Decay is random for single nuclei but follows a definite law for large numbers. It is not affected by temperature, pressure or chemical state.
  • Mean life τ = 1.44 T½.

8. Nuclear Fission and Fusion

  • Fission: ²³⁵U + n → ¹⁴¹Ba + ⁹²Kr + 3n + about 200 MeV. The extra neutrons can cause a chain reaction.
  • In a nuclear reactor, a moderator (heavy water, graphite) slows neutrons, and control rods (cadmium, boron) absorb extra neutrons to keep the reaction steady.
  • Fusion: light nuclei (hydrogen isotopes) combine at very high temperatures (~10⁷ K), as in the Sun, where hydrogen is converted into helium.
  • Fusion releases more energy per unit mass of fuel than fission.

9. Formula Master Table

QuantityFormula
Closest approachr₀ = k(2e)(Ze)/K
Bohr quantisationmvr = nh/2π
Radiusrn = 0.529n²/Z Å
EnergyEn = −13.6Z²/n² eV
Spectral lines1/λ = RZ²(1/n₁² − 1/n₂²)
Number of linesn(n − 1)/2
Nuclear radiusR = R₀A1/3
Binding energyBE = Δm × 931.5 MeV
Decay lawN = N₀e−λt
Half-life / mean lifeT½ = 0.693/λ ; τ = 1/λ
ActivityA = λN = A₀e−λt

10. Master Summary Table

ConceptKey ideaNEET Note
RutherfordTiny dense nucleusCould not explain stability
Bohr orbitsr ∝ n², v ∝ 1/n, E ∝ −1/n²K = −E, U = 2E
Spectral seriesNamed by lower levelBalmer is visible
Nuclear densitySame for all nucleiBecause R ∝ A1/3
BE per nucleonPeaks near Fe-56Fission and fusion both release energy
Radioactive decayExponentialUnaffected by physical conditions
Mean life1.44 T½τ > T½

11. Mind Map

Atoms and Nuclei
Atomic models
  • Rutherford
  • Bohr postulates
  • r, v, E
Spectra
  • Lyman, Balmer…
  • 1/λ formula
  • n(n − 1)/2 lines
Nucleus
  • R = R₀A1/3
  • Mass defect
  • BE curve
Radioactivity and energy
  • α, β, γ
  • Half-life
  • Fission, fusion

12. Tips and Tricks

Shortcuts
  • Energies of H levels: −13.6, −3.4, −1.51, −0.85 eV for n = 1, 2, 3, 4.
  • After n half-lives, the fraction left is 1/2ⁿ.
  • Number of α decays = ΔA/4; then number of β⁻ decays = 2(number of α) − ΔZ.
  • Ratio of nuclear radii = (ratio of mass numbers)1/3.
Common NEET traps
  • Forgetting that total energy in a Bohr orbit is negative.
  • Thinking a heavier nucleus is denser.
  • Using T½ = λ/0.693 (it is 0.693/λ).
  • Saying heating a sample speeds up radioactive decay.

13. Important One-liners

  • The Bohr model works only for hydrogen-like (one-electron) atoms.
  • The ground-state energy of hydrogen is −13.6 eV.
  • The Balmer series is the only hydrogen series in the visible region.
  • Nuclear density is about 10¹⁴ times the density of water.
  • Nuclear forces are the strongest but shortest-range forces in nature.
  • γ-rays are not deflected by electric or magnetic fields.
  • The Sun's energy comes from nuclear fusion.
  • Cadmium rods control the rate of fission in a reactor.

14. Solved Numericals: Average to Advanced

Five average and five advanced numericals. Take R = 1.097 × 10⁷ m⁻¹ and 1 u = 931.5 MeV. Click to reveal each answer and working.

Average level

N1Average

Find the energy of the electron in the n = 3 orbit of hydrogen. How much energy is needed to ionise the atom from this level?

Show answer & solution

Concept: En = −13.6/n² eV.

  1. E₃ = −13.6/9 ≈ −1.51 eV.
  2. Ionisation from n = 3 needs +1.51 eV.

Answer: −1.51 eV; 1.51 eV

N2Average

Find the radius of the third Bohr orbit of hydrogen.

Show answer & solution

Concept: rn = 0.529n² Å.

  1. r₃ = 0.529 × 9.
  2. r₃ ≈ 4.76 Å.

Answer: about 4.76 Å

N3Average

Find the wavelength of the Hβ line of hydrogen (transition from n = 4 to n = 2).

Show answer & solution

Concept: 1/λ = R(1/n₁² − 1/n₂²).

  1. 1/λ = R(1/4 − 1/16) = 3R/16.
  2. λ = 16/(3 × 1.097 × 10⁷) ≈ 4.86 × 10⁻⁷ m.
  3. λ ≈ 486 nm (blue-green, visible).

Answer: about 486 nm

N4Average

Find the radii of nuclei with mass numbers 64 and 216 (R₀ = 1.2 fm), and the ratio of their densities.

Show answer & solution

Concept: R = R₀A1/3; density does not depend on A.

  1. A = 64: R = 1.2 × 4 = 4.8 fm.
  2. A = 216: R = 1.2 × 6 = 7.2 fm.
  3. Density ratio = 1 : 1.

Answer: 4.8 fm and 7.2 fm; densities equal

N5Average

A 10 g radioactive sample has a half-life of 5 days. How much remains after 20 days?

Show answer & solution

Concept: N = N₀(½)t/T½.

  1. t/T½ = 20/5 = 4 half-lives.
  2. Remaining = 10/2⁴ = 10/16 = 0.625 g.

Answer: 0.625 g

Advanced level

N6Advanced

Find the binding energy and binding energy per nucleon of ⁷₃Li. Atomic masses: ⁷Li = 7.016003 u, ¹H = 1.007825 u, neutron = 1.008665 u.

Show answer & solution

Concept: Δm = 3mH + 4mn − M(Li) (electron masses cancel when atomic masses are used).

  1. 3 × 1.007825 + 4 × 1.008665 = 3.023475 + 4.034660 = 7.058135 u.
  2. Δm = 7.058135 − 7.016003 = 0.042132 u.
  3. BE = 0.042132 × 931.5 ≈ 39.2 MeV; per nucleon = 39.2/7 ≈ 5.6 MeV.

Answer: about 39.2 MeV; about 5.6 MeV per nucleon

N7Advanced

A radioactive isotope has a half-life of 1386 s. Find its decay constant and mean life. What fraction remains after 4158 s?

Show answer & solution

Concept: λ = 0.693/T½; τ = 1/λ.

  1. λ = 0.693/1386 = 5 × 10⁻⁴ s⁻¹.
  2. τ = 1/λ = 2000 s.
  3. 4158/1386 = 3 half-lives ⇒ fraction left = 1/8.

Answer: 5 × 10⁻⁴ s⁻¹; 2000 s; 1/8

N8Advanced

Find (a) the longest wavelength in the Lyman series, and (b) the shortest wavelength (series limit) in the Balmer series of hydrogen.

Show answer & solution

Concept: longest λ = smallest energy jump; series limit = jump from n = ∞.

  1. (a) Lyman, 2 → 1: 1/λ = R(1 − 1/4) = 3R/4 ⇒ λ = 4/(3 × 1.097 × 10⁷) ≈ 121.5 nm.
  2. (b) Balmer, ∞ → 2: 1/λ = R/4 ⇒ λ = 4/1.097 × 10⁷ ≈ 364.6 nm.

Answer: (a) about 121.5 nm; (b) about 364.6 nm

N9Advanced

Find the energy released in the fusion reaction ²H + ²H → ³He + n. Masses: ²H = 2.014102 u, ³He = 3.016029 u, n = 1.008665 u.

Show answer & solution

Concept: Q = (reactant mass − product mass) × 931.5 MeV.

  1. Reactants: 2 × 2.014102 = 4.028204 u.
  2. Products: 3.016029 + 1.008665 = 4.024694 u.
  3. Δm = 0.003510 u ⇒ Q ≈ 3.27 MeV (released, since Q is positive).

Answer: about 3.27 MeV released

N10Advanced

The count rate of a radioactive sample falls from 8000 to 1000 counts per minute in 6 hours. Find its half-life and decay constant. What will the count rate be 2 hours later?

Show answer & solution

Concept: activity halves every half-life.

  1. 8000 → 1000 is a factor of 8 = 2³, so 3 half-lives take 6 h ⇒ T½ = 2 h.
  2. λ = 0.693/2 ≈ 0.347 h⁻¹.
  3. 2 h later is one more half-life: 1000/2 = 500 counts per minute.

Answer: 2 h; about 0.35 h⁻¹; 500 counts per minute

15. NEET-pattern Questions with Detailed Answers

These questions are original and based on concepts that NEET UG asks again and again.

Q1Easy

In Bohr's model, the radius of the n-th orbit is proportional to:

  1. n
  2. n²
  3. 1/n
  4. 1/n²
Show solution

rn = 0.529n²/Z Å, so r ∝ n². The energy goes as −1/n² and the speed as 1/n.

Answer: (B)

Q2Easy

The visible lines of the hydrogen spectrum belong to the:

  1. Lyman series
  2. Balmer series
  3. Paschen series
  4. Pfund series
Show solution

The Balmer series (jumps down to n = 2) is in the visible region. The Lyman series is ultraviolet, and the Paschen and Pfund series are infrared.

Answer: (B)

Q3Medium

The number of spectral lines emitted when hydrogen atoms fall from n = 4 to the ground state is:

  1. 3
  2. 4
  3. 6
  4. 10
Show solution

n(n − 1)/2 = 4 × 3/2 = 6 lines: 4→3, 4→2, 4→1, 3→2, 3→1 and 2→1.

Answer: (C)

Q4Medium

The ratio of the nuclear densities of ⁸O¹⁶ and ⁸²Pb²⁰⁸ is about:

  1. 1 : 13
  2. 1 : 1
  3. 13 : 1
  4. 1 : 2.35
Show solution

Nuclear density does not depend on A, because R³ ∝ A. So the ratio is 1 : 1. Option (D) is the ratio of their radii, (208/16)1/3 ≈ 2.35.

Answer: (B)

Q5Medium

When a nucleus emits one α-particle and then one β⁻ particle, its atomic number:

  1. Decreases by 1
  2. Decreases by 2
  3. Increases by 1
  4. Decreases by 3
Show solution

α decay lowers Z by 2, and β⁻ decay raises Z by 1. The net change is −1. The mass number falls by 4.

Answer: (A)

Q6Medium

Energy is released in both fission and fusion because:

  1. Mass increases
  2. Binding energy per nucleon of the products is greater
  3. Charge is created
  4. Neutrons are destroyed
Show solution

Both processes move towards the peak of the BE/A curve. The products are more tightly bound, and the extra binding energy is released. Mass decreases, not increases.

Answer: (B)

Q7Hard

In the ground state of hydrogen, the kinetic energy of the electron is 13.6 eV. Its potential energy is:

  1. −13.6 eV
  2. −27.2 eV
  3. +27.2 eV
  4. +13.6 eV
Show solution

In a Bohr orbit, K = −E and U = 2E. Here E = −13.6 eV, so U = −27.2 eV. Check: K + U = 13.6 − 27.2 = −13.6 eV ✓.

Answer: (B)

Q8Hard

The mean life of a radioactive sample is 144 days. Its half-life is about:

  1. 100 days
  2. 144 days
  3. 207 days
  4. 72 days
Show solution

T½ = 0.693τ = 0.693 × 144 ≈ 100 days. (C) multiplies by 1.44 instead of dividing, and (D) simply halves τ.

Answer: (A)

16. Quick Revision Summary

Remember before the exam
  • Rutherford: tiny nucleus; r₀ = k(2e)(Ze)/K.
  • Bohr: mvr = nh/2π; r ∝ n²/Z, v ∝ Z/n, E = −13.6Z²/n² eV.
  • 1/λ = R(1/n₁² − 1/n₂²); lines n(n − 1)/2; Balmer visible.
  • R = 1.2A1/3 fm; constant density; BE = Δm × 931.5 MeV; peak at Fe.
  • N = N₀e−λt; T½ = 0.693/λ; τ = 1.44T½; α: −2, −4; β⁻: +1, 0.
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