JEE Main Physics · Chapter 21

Atoms and Nuclei

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

NCERT Class 12JEE Main: 1–2 Qs per shift 8 min read

1. Chapter at a Glance

This chapter explores the structure of the atom and its nucleus. It covers Rutherford's alpha-scattering experiment, Bohr's model of hydrogen and its spectral series, the composition and size of nuclei, mass defect and binding energy, radioactive decay, and nuclear fission and fusion. Together with dual nature, it forms "modern physics" — two or three JEE Main questions every year.

What JEE asks most
  • Bohr model: radius, speed and energy of the nth orbit; hydrogen-like ions.
  • Wavelengths of spectral lines and the number of possible lines.
  • Binding energy per nucleon and Q-values.
  • Radioactive decay law, half-life, mean life, and counting α/β decays.

2. Atomic Models

  • Rutherford's experiment: most alpha particles passed straight through gold foil, a few were deflected by large angles. So the atom is mostly empty with a tiny, dense, positive nucleus. Distance of closest approach: r₀ = 2kZe²/K for an alpha of kinetic energy K.
  • Rutherford's model could not explain the stability of atoms or line spectra.
  • Bohr's postulates: electrons move in certain stationary orbits without radiating; angular momentum is quantised, mvr = nh/2π; radiation is emitted when an electron jumps between orbits, hν = Ei − Ef.

3. Bohr's Model of Hydrogen-like Atoms

QuantityFormulaDependence
Radiusrn = 0.529 n²/Z Å∝ n²/Z
Speedvn = 2.18 × 10⁶ Z/n m s⁻¹∝ Z/n
EnergyEn = −13.6 Z²/n² eV∝ Z²/n²
KE, PEK = −E; U = 2E
1/λ = RZ²(1/n₁² − 1/n₂²)    R = 1.097 × 10⁷ m⁻¹
n=1 (−13.6 eV) n=2 (−3.4) n=3 n=4 ∞ (0) Lyman (UV) Balmer (visible)
Hydrogen energy levels (not to scale near the top). Lyman lines end on n = 1 (ultraviolet); Balmer lines end on n = 2 (visible); Paschen on n = 3 (infrared).
  • Series: Lyman (n₁ = 1, UV), Balmer (n₁ = 2, visible), Paschen (3), Brackett (4), Pfund (5) — the last three in the infrared.
  • Number of lines when electrons fall from level n to the ground state: n(n − 1)/2.
  • Ionisation energy of hydrogen = 13.6 eV; of He⁺ = 54.4 eV; of Li²⁺ = 122.4 eV.

4. The Nucleus

  • Nucleus AZX has Z protons and A − Z neutrons. Isotopes: same Z; isobars: same A; isotones: same N.
  • Radius R = R₀A1/3 (R₀ ≈ 1.2 fm), so nuclear density (~2.3 × 10¹⁷ kg m⁻³) is the same for all nuclei.
  • Nuclear force: strongest force, short range (a few fm), charge-independent, saturating.
  • Mass defect Δm = [Zmp + (A − Z)mn] − Mnucleus; binding energy = Δm c² (1 u ↔ 931.5 MeV).
  • Binding energy per nucleon peaks near A ≈ 56 (Fe, about 8.8 MeV). Heavy nuclei release energy by fission; light nuclei by fusion.
  • Fission of U-235 releases about 200 MeV per nucleus. Fusion powers the Sun (hydrogen → helium).

5. Radioactivity

N = N₀e−λt    T½ = 0.693/λ    τ = 1/λ = 1.44T½    A = λN
  • After n half-lives, fraction left = (1/2)n.
  • α decay: A decreases by 4, Z by 2. β⁻ decay: Z increases by 1 (a neutron becomes a proton, emitting an electron and an antineutrino). β⁺ decay: Z decreases by 1. γ decay: no change in A or Z.
  • Penetrating power: γ > β > α; ionising power: α > β > γ.
  • Activity unit: becquerel (1 decay per second); 1 curie = 3.7 × 10¹⁰ Bq.

6. Formula Master Table

QuantityFormula
Bohr radiusrn = 0.529 n²/Z Å
EnergyEn = −13.6 Z²/n² eV
Spectral line1/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; T½ = 0.693/λ
Mean lifeτ = 1/λ = 1.44T½

7. Mind Map

Atoms & Nuclei
Atom models
  • Rutherford scattering
  • Bohr postulates
Hydrogen spectrum
  • r, v, E of orbits
  • Series, number of lines
Nucleus
  • R ∝ A1/3
  • Mass defect, BE
  • Fission, fusion
Radioactivity
  • α, β, γ
  • Half-life, mean life

8. Tips and Tricks

Shortcuts
  • Use λ(nm) = 1240/ΔE(eV) for any transition.
  • Count decays in a chain: number of α = ΔA/4; then β⁻ = 2 × (number of α) − ΔZ.
  • Ratio of radii of two nuclei = (A₁/A₂)1/3.
  • The longest-wavelength line of a series comes from the nearest upper level (e.g. Balmer: 3 → 2).
Mistakes students make
  • Forgetting Z² when using hydrogen formulas for He⁺ or Li²⁺.
  • Confusing half-life with mean life.
  • Thinking a β⁻ electron comes from the atomic shell — it is created in the nucleus.

9. Important One-liners

  • Rutherford's experiment revealed the nucleus.
  • The Balmer series lies in the visible region.
  • Nuclear density is independent of mass number.
  • Iron-56 has nearly the highest binding energy per nucleon.
  • Half-life does not depend on temperature, pressure or chemical state.
  • γ rays are the most penetrating and least ionising radiation.

10. Solved Numericals: Average to Advanced

Five average-level and five advanced-level problems. Click to reveal each answer and its full working. Take hc = 1240 eV nm and 1 u = 931.5 MeV/c².

Average level

N1Average

Find the energy and wavelength of the photon emitted when a hydrogen electron falls from n = 3 to n = 2.

Show answer & solution

Concept: ΔE = 13.6(1/n₁² − 1/n₂²) eV; λ = 1240/ΔE.

  1. ΔE = 13.6(1/4 − 1/9) = 13.6 × 5/36 ≈ 1.89 eV.
  2. λ = 1240/1.89 ≈ 656 nm (red Balmer line, H-alpha).

Answer: about 1.89 eV; about 656 nm

N2Average

Find the radius of the second Bohr orbit of hydrogen.

Show answer & solution

Concept: rn = 0.529 n² Å for Z = 1.

  1. r₂ = 0.529 × 4.
  2. = 2.116 Å.

Answer: 2.116 Å

N3Average

Hydrogen atoms are excited to n = 5. How many different spectral lines can they emit while returning to the ground state?

Show answer & solution

Concept: number of lines = n(n − 1)/2.

  1. = 5 × 4/2.
  2. = 10.

Answer: 10

N4Average

Find the ratio of the nuclear radii of ²⁷Al and ¹²⁵Te.

Show answer & solution

Concept: R ∝ A1/3.

  1. R₁/R₂ = (27/125)1/3.
  2. = 3/5.

Answer: 3 : 5

N5Average

A radioactive sample has a half-life of 10 days. What fraction remains after 30 days?

Show answer & solution

Concept: fraction = (1/2)n, n = t/T½.

  1. n = 30/10 = 3.
  2. Fraction = (1/2)³ = 1/8 = 12.5%.

Answer: 1/8

Advanced level

N6Advanced

Find the binding energy and binding energy per nucleon of ⁴He. Atomic masses: ¹H = 1.007825 u, neutron = 1.008665 u, ⁴He = 4.002603 u.

Show answer & solution

Concept: using atomic masses, electron masses cancel: Δm = 2m(¹H) + 2mn − m(⁴He).

  1. 2 × 1.007825 + 2 × 1.008665 = 4.032980 u.
  2. Δm = 4.032980 − 4.002603 = 0.030377 u.
  3. BE = 0.030377 × 931.5 ≈ 28.3 MeV.
  4. Per nucleon = 28.3/4 ≈ 7.07 MeV.

Answer: about 28.3 MeV; about 7.07 MeV per nucleon

N7Advanced

For the sample of N5 (T½ = 10 days), find the decay constant and mean life. If its initial activity is 800 Bq, what is the activity after 20 days?

Show answer & solution

Concept: λ = 0.693/T½; τ = 1/λ; activity also halves every half-life.

  1. λ = 0.693/10 = 0.0693 day⁻¹.
  2. τ = 1/0.0693 ≈ 14.4 days.
  3. 20 days = 2 half-lives ⇒ A = 800/4 = 200 Bq.

Answer: 0.0693 day⁻¹; about 14.4 days; 200 Bq

N8Advanced

Each fission of ²³⁵U releases about 200 MeV. Estimate the energy released by the complete fission of 1 g of ²³⁵U.

Show answer & solution

Concept: number of nuclei × energy per fission.

  1. Number of nuclei = (1/235) × 6.022 × 10²³ ≈ 2.56 × 10²¹.
  2. Energy = 2.56 × 10²¹ × 200 MeV ≈ 5.13 × 10²³ MeV.
  3. × 1.6 × 10⁻¹³ J MeV⁻¹ ≈ 8.2 × 10¹⁰ J (about 23 000 kWh).

Answer: about 8.2 × 10¹⁰ J

N9Advanced

(a) Find the ionisation energy of He⁺ from its ground state. (b) Find the wavelength of the first Lyman line of hydrogen.

Show answer & solution

Concept: En = −13.6Z²/n² eV.

  1. (a) He⁺: Z = 2, n = 1 ⇒ E₁ = −54.4 eV, so ionisation energy = 54.4 eV.
  2. (b) Hydrogen 2 → 1: ΔE = 13.6(1 − 1/4) = 10.2 eV.
  3. λ = 1240/10.2 ≈ 121.6 nm (ultraviolet).

Answer: 54.4 eV; about 122 nm

N10Advanced

²³⁸U (Z = 92) decays through a series of α and β⁻ emissions to stable ²⁰⁶Pb (Z = 82). How many α and β⁻ particles are emitted?

Show answer & solution

Concept: only α changes A (by 4); α lowers Z by 2 and β⁻ raises it by 1.

  1. Number of α = (238 − 206)/4 = 8.
  2. After 8 α: Z = 92 − 16 = 76.
  3. To reach 82, β⁻ = 82 − 76 = 6.

Answer: 8 α and 6 β⁻

11. PYQ-pattern Questions with Solutions

These questions are original and modelled on concepts frequently tested in JEE Main.

Q1Easy

Rutherford's alpha-scattering experiment showed that:

  1. Electrons move in fixed orbits
  2. The atom has a tiny, dense, positive nucleus
  3. Energy is quantised
  4. Neutrons exist
Show solution

Rare large-angle deflections could only come from a small, massive, positive centre.

Answer: (B)

Q2Easy

The spectral series of hydrogen lying in the visible region is:

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

Transitions ending at n = 2 give visible lines.

Answer: (B)

Q3Numerical

What percentage of a radioactive sample remains after four half-lives?

Show solution

(1/2)⁴ = 1/16 = 6.25%.

Answer: 6.25

Q4Medium

Nuclear density:

  1. Increases with A
  2. Decreases with A
  3. Is nearly the same for all nuclei
  4. Depends on Z only
Show solution

Volume ∝ R³ ∝ A and mass ∝ A, so density is constant.

Answer: (C)

Q5Hard

Energy is released in nuclear fusion of light nuclei because:

  1. Mass increases
  2. Binding energy per nucleon of the product is larger
  3. Neutrons are absorbed
  4. Coulomb repulsion increases
Show solution

The product nucleus is more tightly bound; the extra binding energy appears as released energy (mass decreases).

Answer: (B)

Q6Numerical

In β⁻ decay, by how much does the atomic number change?

Show solution

A neutron turns into a proton: Z increases by 1 (A unchanged).

Answer: 1

12. Quick Revision Summary

Remember before the exam
  • r ∝ n²/Z, v ∝ Z/n, E = −13.6Z²/n² eV.
  • Lyman (UV), Balmer (visible), Paschen (IR); lines n(n − 1)/2.
  • R = R₀A1/3; BE = Δm × 931.5 MeV; peak near Fe.
  • N = N₀e−λt; T½ = 0.693/λ; τ = 1.44T½.
  • α: A − 4, Z − 2; β⁻: Z + 1; γ: no change.
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