Modern Physics — Dual Nature, Atoms & Nuclei - NEET 2026

Modern Physics

Photoelectric effect, Bohr's model, and nuclear physics are the three pillars. Know all formulas and their conditions.

Dual Nature of Matter & Radiation

Photoelectric effect: Light of frequency ν > threshold frequency (ν₀) ejects electrons. Einstein's equation: KE_max = hν - hν₀ = hν - φ (φ = work function). Stopping potential V₀: eV₀ = hν - φ.

Key observations: (1) No electrons below threshold frequency regardless of intensity. (2) KE_max depends on ν, NOT intensity. (3) Number of electrons depends on intensity. (4) Emission is instantaneous. These observations CANNOT be explained by wave theory — established photon nature of light.

de Broglie wavelength: λ = h/mv = h/p. For electron accelerated through V: λ = h/√(2meV) = 1.227/√V nm. For photon: λ = hc/E = c/ν.

Heisenberg's uncertainty principle: Δx · Δp ≥ h/4π. Cannot simultaneously know exact position and momentum.

Bohr's Model of Hydrogen

Postulates: Electrons move in fixed circular orbits (stationary states) without radiating. Energy absorbed/emitted = E = hν when electron jumps between orbits.

Allowed orbits: mvr = nh/2π (angular momentum quantised in multiples of h/2π = ℏ). n = 1, 2, 3... (quantum number).

Energy of nth orbit (H atom): Eₙ = -13.6/n² eV. E₁ = -13.6 eV (ground state), E₂ = -3.4 eV, E₃ = -1.51 eV. Ionisation energy = 13.6 eV.

Radius: rₙ = n²a₀ where a₀ = 0.529 Å (Bohr radius). r₁ = 0.529 Å, r₂ = 2.116 Å.

Spectral series: Lyman (UV, n→1), Balmer (visible, n→2), Paschen (IR, n→3), Brackett (IR, n→4), Pfund (IR, n→5).

Radioactivity

α decay: ₂₃₈U → ₂₃₄Th + ₄He. A decreases by 4, Z decreases by 2. Heaviest nuclei decay by α.

β⁻ decay: Neutron → proton + electron + antineutrino. Z increases by 1, A unchanged. Neutron-rich nuclei.

β⁺ decay: Proton → neutron + positron + neutrino. Z decreases by 1, A unchanged.

γ decay: No change in A or Z — just energy emission from excited nucleus. Always accompanies α or β decay.

Radioactive decay law: N = N₀e^(-λt). Activity A = dN/dt = λN = λN₀e^(-λt). t½ = 0.693/λ = ln2/λ. Mean lifetime τ = 1/λ = 1.44 t½.

Nuclear Physics

Mass defect: Δm = Zm_p + (A-Z)m_n - M_nucleus. Mass is converted to binding energy: BE = Δm × c².

Binding energy per nucleon: BE/A. Peaks at Fe-56 (~8.8 MeV/nucleon) → most stable nucleus. Light nuclei: fusion releases energy (BE/A increases). Heavy nuclei: fission releases energy (BE/A increases).

Nuclear reactions: Fission: ₂₃₅U + n → ₁₄₁Ba + ₉²Kr + 3n + energy. Fusion: ²H + ³H → ⁴He + n + 17.6 MeV. Fusion releases more energy per unit mass but requires extremely high temperature (sun operates by p-p chain fusion).

Modern Physics Facts

Photoelectric Effect

KE_max = hν - φ
eV₀ = hν - φ (stopping potential)
φ = hν₀ = hc/λ₀
h = 6.626 × 10⁻³⁴ J·s
1 eV = 1.6 × 10⁻¹⁹ J

If frequency below threshold: NO emission regardless of intensity. This disproves wave theory.

Bohr's H-atom Energies

E_n = -13.6/n² eV
n=1: -13.6 eV (ground)
n=2: -3.4 eV
n=3: -1.51 eV
n=∞: 0 (ionised)
Ionisation energy: 13.6 eV

Energy is negative = bound state. Higher n = higher energy (less negative) = closer to ionisation

Spectral Series

Lyman: n→1 (UV)
Balmer: n→2 (visible)
Paschen: n→3 (near IR)
Brackett: n→4 (IR)
Pfund: n→5 (far IR)

First line of Balmer series (n=3→2) is H_α line (red, 656 nm) — visible to naked eye

Radioactive Decay

N = N₀(1/2)^(t/t½)
t½ = 0.693/λ
Activity A = λN
Mean life τ = 1/λ = 1.44t½
α: A-4, Z-2
β⁻: A same, Z+1
γ: A and Z unchanged

After n half-lives: N = N₀/2^n. Remaining fraction = (1/2)^n

Worked Examples

EasyWhat is the energy of a photon with wavelength 620 nm? (h = 6.63 × 10⁻³⁴ J·s, c = 3 × 10⁸ m/s)▾

E = hc/λ = (6.63 × 10⁻³⁴ × 3 × 10⁸) / (620 × 10⁻⁹) = 19.89 × 10⁻²⁶ / 620 × 10⁻⁹ = 3.21 × 10⁻¹⁹ J. In eV: E = 3.21 × 10⁻¹⁹ / 1.6 × 10⁻¹⁹ ≈ 2.0 eV.

Answer: ~2 eV (this is red light)

MediumA radioactive isotope has a half-life of 10 years. What fraction remains after 30 years?▾

30 years = 3 half-lives. Remaining = N₀ × (1/2)³ = N₀/8. Fraction remaining = 1/8 = 12.5%.

Answer: 1/8 (12.5%) remains

HardAn electron jumps from n=4 to n=2 in a hydrogen atom. Which spectral series does this transition belong to?▾

Spectral series is determined by the LOWER energy level (final state). The electron ends at n=2. n→2 transitions belong to the BALMER series. ΔE = -13.6(1/4 - 1/16) = -13.6(3/16) = -2.55 eV. This photon has λ ≈ 486 nm (blue-green visible light).

Answer: Balmer series (n=4→2 transition, visible blue-green light)

Mistake DNA

❌ Confusing spectral series by initial vs final level

Series is defined by FINAL level (where electron lands), not initial level. Balmer series = all transitions TO n=2 (from n=3,4,5...). Lyman = TO n=1. Students say "Balmer starts at n=2" — wrong, it ENDS at n=2.

Fix: Lyman → n=1 (final). Balmer → n=2 (final). Paschen → n=3 (final). Final state determines series.

❌ Saying β decay changes mass number

β decay (both β⁻ and β⁺) does NOT change mass number A. Only the atomic number Z changes (±1). Electron/positron have negligible mass compared to nucleons. α decay changes both A (-4) and Z (-2).

Fix: α → A-4, Z-2. β → A unchanged, Z±1. γ → A unchanged, Z unchanged.

Chapter Intelligence

PYQ Frequency

Photoelectric effect: 1 Q/year
Bohr model / energy levels: 1–2 Q/year
Radioactive decay: 1 Q/year
Nuclear fission/fusion: 1 Q/year

2026 Prediction

High: Half-life calculation
Expected: Bohr energy level transition identification
Watch: de Broglie wavelength of particles

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